Integrand size = 31, antiderivative size = 432 \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {a b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^2 b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^2 b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 (a-b \sin (e+f x))}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \] Output:
a*b^(3/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/(- a^2+b^2)^(7/4)/f/g^(5/2)+a*b^(3/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(- a^2+b^2)^(1/4)/g^(1/2))/(-a^2+b^2)^(7/4)/f/g^(5/2)-2/3*b*cos(f*x+e)^(1/2)* InverseJacobiAM(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^2*b*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))/(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^2*b*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))/(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*(a-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 = 22.42 (sec) , antiderivative size = 1183, normalized size of antiderivative = 2.74 \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:
Integrate[Sin[e + f*x]/((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)) + (b*Cos[e + f*x]^(5/2)*((-4*a*(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]* Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sq rt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqr t[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* Cos[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*b*(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, 9/4, Cos[e + ...
Time = 1.86 (sec) , antiderivative size = 413, normalized size of antiderivative = 0.96, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.581, Rules used = {3042, 3345, 27, 3042, 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.
| \(\displaystyle \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle \frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {2 \int -\frac {2 a b-b^2 \sin (e+f x)}{2 \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{3 g^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 a b-b^2 \sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a b-b^2 \sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle \frac {3 a b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx-b \int \frac {1}{\sqrt {g \cos (e+f x)}}dx}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 a b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx-b \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {3 a b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx-\frac {b \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{\sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 a b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx-\frac {b \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{\sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {3 a b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle \frac {3 a b \left (\frac {b g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2\right )}d(g \cos (e+f x))}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {3 a b \left (\frac {2 b g \int \frac {1}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {3 a b \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {3 a b \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {3 a b \left (-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 a b \left (-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {3 a b \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 a b \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {3 a b \left (\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}+\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )-\frac {2 b \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \sqrt {g \cos (e+f x)}}}{3 g^2 \left (a^2-b^2\right )}+\frac {2 (a-b \sin (e+f x))}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}\) |
Input:
Int[Sin[e + f*x]/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
Output:
((-2*b*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(f*Sqrt[g*Cos[e + f*x ]]) + 3*a*b*((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]]) ))/(3*(a^2 - b^2)*g^2) + (2*(a - b*Sin[e + f*x]))/(3*(a^2 - b^2)*f*g*(g*Co s[e + f*x])^(3/2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[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_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1174\) vs. \(2(383)=766\).
Time = 5.21 (sec) , antiderivative size = 1175, normalized size of antiderivative = 2.72
Input:
int(sin(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBO SE)
Output:
(-4/g^2*a*(-1/24*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)-1/24*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)+b^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*c os(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)-8*(g*(-1+2*cos(1/2*f* x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2-1)*b/g^2*(1/ 2/(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-sin(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)))+1/2/(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*cos(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)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*Ell ipticE(cos(1/2*f*x+1/2*e),2^(1/2)))-1/16/(a-b)/(a+b)*sum(_alpha/(2*_alp...
Timed out. \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="f ricas")
Output:
Timed out
Timed out. \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)/(g*cos(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)
Output:
Timed out
\[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="m axima")
Output:
integrate(sin(f*x + e)/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
\[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="g iac")
Output:
integrate(sin(f*x + e)/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)
Timed out. \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int \frac {\sin \left (e+f\,x\right )}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:
int(sin(e + f*x)/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)
Output:
int(sin(e + f*x)/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)
\[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{3} a}d x \right )}{g^{3}} \] Input:
int(sin(f*x+e)/(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))/(cos(e + f*x)**3*sin(e + f* x)*b + cos(e + f*x)**3*a),x))/g**3