Integrand size = 33, antiderivative size = 528 \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {a^3 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {2 a}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {2 a^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {4 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {2 b \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \] Output:
a^3*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(1/2)/ (-a^2+b^2)^(7/4)/f/g^(5/2)+a^3*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+ b^2)^(1/4)/g^(1/2))/b^(1/2)/(-a^2+b^2)^(7/4)/f/g^(5/2)+2/3*a/(a^2-b^2)/f/g /(g*cos(f*x+e))^(3/2)-2*a^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e ,2^(1/2))/b/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)+4/3*b*cos(f*x+e)^(1/2)*In verseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)+ a^4*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2) ),2^(1/2))/b/(a^2-b^2)/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+e))^( 1/2)+a^4*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^ (1/2)),2^(1/2))/b/(a^2-b^2)/(a^2-b*(b+(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+ e))^(1/2)-2/3*b*sin(f*x+e)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 23.33 (sec) , antiderivative size = 1193, normalized size of antiderivative = 2.26 \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:
Integrate[Sin[e + f*x]^3/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
Output:
(2*Cos[e + f*x]*(a - b*Sin[e + f*x]))/(3*(a^2 - b^2)*f*(g*Cos[e + f*x])^(5 /2)) - (Cos[e + f*x]^(5/2)*((4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a* (a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2 )/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^ 2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f* x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*S qrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqr t[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt [b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] - Log[Sqrt[- a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*C os[e + f*x]]))/(-a^2 + b^2)^(3/4))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2] *(a + b*Sin[e + f*x])) - (2*(3*a^2 - 2*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2 ])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[ e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]]*Sqrt[1 - Cos[e + f*x]^2])/((- 5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x ]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, ...
Time = 2.70 (sec) , antiderivative size = 494, normalized size of antiderivative = 0.94, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {3042, 3381, 3042, 3045, 15, 3046, 3042, 3121, 3042, 3120, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 218, 221, 3042, 3286, 3042, 3284}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}dx\) |
\(\Big \downarrow \) 3381 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}+\frac {a \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}+\frac {a \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {a \int \frac {1}{(g \cos (e+f x))^{5/2}}d(g \cos (e+f x))}{f g \left (a^2-b^2\right )}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{5/2}}dx}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {g \cos (e+f x)}}dx}{3 g^2}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 g^2}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle -\frac {a^2 \left (\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \left (\frac {\int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {a^2 \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {b g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2\right )}d(g \cos (e+f x))}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \int \frac {1}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}+\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{3 f g (g \cos (e+f x))^{3/2}}-\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f g^2 \sqrt {g \cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
Input:
Int[Sin[e + f*x]^3/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
Output:
(2*a)/(3*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2)) - (a^2*((2*Sqrt[Cos[e + f *x]]*EllipticF[(e + f*x)/2, 2])/(b*f*Sqrt[g*Cos[e + f*x]]) - (a*((2*b*g*(- 1/2*ArcTan[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(Sqrt[b]*(-a ^2 + b^2)^(3/4)*g^(3/2)) - ArcTanh[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*g^(3/2))))/f + (a*Sqrt[Cos[e + f *x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a*Sqrt[Cos[e + f* x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])))/b))/((a^2 - b^2)*g^ 2) - (b*((-4*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*f*g^2*Sqrt[g *Cos[e + f*x]]) + (2*Sin[e + f*x])/(3*f*g*(g*Cos[e + f*x])^(3/2))))/(a^2 - b^2)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Sin[e + f *x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q) Int[1/( Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f) Subst[ Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S imp[a/(2*q) Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / ; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* (x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int [(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b Int[(g*Cos[e + f*x])^p/( a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a*(d^2/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ b*(d/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Simp[a^2*(d^2/(g^2*(a^2 - b^2))) Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[ e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g }, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1 ]
Leaf count of result is larger than twice the leaf count of optimal. \(1207\) vs. \(2(472)=944\).
Time = 5.37 (sec) , antiderivative size = 1208, normalized size of antiderivative = 2.29
Input:
int(sin(f*x+e)^3/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x,method=_RETURNVER BOSE)
Output:
(16/g^2*a*(1/96*2^(1/2)/(a^2-b^2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(2^( 1/2)+cos(1/2*f*x+1/2*e))/g/(2*2^(1/2)*cos(1/2*f*x+1/2*e)-2*sin(1/2*f*x+1/2 *e)^2+3)+1/96*2^(1/2)/(a^2-b^2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-2^(1 /2)+cos(1/2*f*x+1/2*e))/g/(2*2^(1/2)*cos(1/2*f*x+1/2*e)+2*sin(1/2*f*x+1/2* e)^2-3)-1/4*a^2/(a-b)/(a+b)*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(-ln((-(g^2* (a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)-2*g*cos(1/ 2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+g)/((g^2*(a^2-b^2)/b^2)^(1/4)*(2* g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2 -b^2)/b^2)^(1/2)+g))-2*arctan(2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1 /2*f*x+1/2*e)^2-g)^(1/2)+1)+2*arctan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2 *g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1))/(16*a^2-16*b^2)/g)+32*(g*(-1+2*cos(1/ 2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)*b/g^2*(-1/8*(sin(1/2*f*x+1/2*e )^2-1)/(a^2-b^2)*(-1/6*cos(1/2*f*x+1/2*e)/g*(-g*(2*sin(1/2*f*x+1/2*e)^4-si n(1/2*f*x+1/2*e)^2))^(1/2)/(cos(1/2*f*x+1/2*e)^2-1/2)^2+1/3*(sin(1/2*f*x+1 /2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^ 4-sin(1/2*f*x+1/2*e)^2))^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2)))-1/8* (sin(1/2*f*x+1/2*e)^2-1)/(a^2-b^2)/sin(1/2*f*x+1/2*e)^2/g/(-1+2*sin(1/2*f* x+1/2*e)^2)*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*co s(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2-(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(s in(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2)))+1/64*...
Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm= "fricas")
Output:
Timed out
Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**3/(g*cos(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)
Output:
Timed out
\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm= "maxima")
Output:
integrate(sin(f*x + e)^3/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm= "giac")
Output:
integrate(sin(f*x + e)^3/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:
int(sin(e + f*x)^3/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)
Output:
int(sin(e + f*x)^3/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)
\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )^{3}}{\cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{3} a}d x \right )}{g^{3}} \] Input:
int(sin(f*x+e)^3/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x)
Output:
(sqrt(g)*int((sqrt(cos(e + f*x))*sin(e + f*x)**3)/(cos(e + f*x)**3*sin(e + f*x)*b + cos(e + f*x)**3*a),x))/g**3