Integrand size = 33, antiderivative size = 457 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {a^3 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {2 a \sqrt {g \cos (e+f x)}}{b^2 f g}+\frac {2 a^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^3 f \sqrt {g \cos (e+f x)}}+\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b f \sqrt {g \cos (e+f x)}}-\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b f g} \] Output:
a^3*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(5/2)/ (-a^2+b^2)^(3/4)/f/g^(1/2)+a^3*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+ b^2)^(1/4)/g^(1/2))/b^(5/2)/(-a^2+b^2)^(3/4)/f/g^(1/2)+2*a*(g*cos(f*x+e))^ (1/2)/b^2/f/g+2*a^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2) )/b^3/f/(g*cos(f*x+e))^(1/2)+4/3*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+ 1/2*e,2^(1/2))/b/f/(g*cos(f*x+e))^(1/2)-a^4*cos(f*x+e)^(1/2)*EllipticPi(si n(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b^3/(a^2-b*(b-(-a^2+b^2 )^(1/2)))/f/(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^3/(a^2-b*(b+(-a^2+b^2)^(1/2) ))/f/(g*cos(f*x+e))^(1/2)-2/3*(g*cos(f*x+e))^(1/2)*sin(f*x+e)/b/f/g
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 26.23 (sec) , antiderivative size = 1915, normalized size of antiderivative = 4.19 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:
Integrate[Sin[e + f*x]^3/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
Output:
(-2*Cos[e + f*x]*Sin[e + f*x])/(3*b*f*Sqrt[g*Cos[e + f*x]]) + (Sqrt[Cos[e + f*x]]*((-2*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*Ap pellF1[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]*Sqrt[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*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] )) + (3*a*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)* (-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2 )^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTa n[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*( -a^2 + b^2)^(3/4)) + (4*Sqrt[Cos[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(...
Time = 2.68 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.02, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.788, Rules used = {3042, 3388, 3042, 3048, 3042, 3121, 3042, 3120, 3388, 3042, 3045, 15, 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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx\) |
\(\Big \downarrow \) 3388 |
\(\displaystyle \frac {\int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{\sqrt {g \cos (e+f x)}}dx-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\) |
\(\Big \downarrow \) 3388 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}-\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}}d(g \cos (e+f x))}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {b g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2\right )}d(g \cos (e+f x))}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \int \frac {1}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}+\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\) |
Input:
Int[Sin[e + f*x]^3/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
Output:
-((a*((-2*Sqrt[g*Cos[e + f*x]])/(b*f*g) - (a*((2*Sqrt[Cos[e + f*x]]*Ellipt icF[(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]]*Ellipt icPi[(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]]*Ellipti cPi[(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))/b))/b) + ((4*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Co s[e + f*x]]*Sin[e + f*x])/(3*f*g))/b
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)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
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[d/b Int[(g *Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Simp[a*(d/b) Int[(g*C os[e + f*x])^p*((d*Sin[e + f*x])^(n - 1)/(a + b*Sin[e + f*x])), x], x] /; F reeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && GtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1072\) vs. \(2(403)=806\).
Time = 3.73 (sec) , antiderivative size = 1073, normalized size of antiderivative = 2.35
Input:
int(sin(f*x+e)^3/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x,method=_RETURNVER BOSE)
Output:
(-16*a*(-1/8/b^2/g*(g*(-1+2*cos(1/2*f*x+1/2*e)^2))^(1/2)-1/4*a^2/b^2*(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*arc tan(-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)+1/2*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2* e)^2)^(1/2)/b^3*(sin(1/2*f*x+1/2*e)^2-1)*(8*EllipticF(cos(1/2*f*x+1/2*e),2 ^(1/2))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*b^2 -8*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2) *(sin(1/2*f*x+1/2*e)^2)^(1/2)*b^2-sum(1/(2*_alpha^2-1)*(8*(sin(1/2*f*x+1/2 *e)^2)^(1/2)*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2* e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2) ^(1/2)*_alpha^4*b^2-8*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-1+2*sin(1/2*f*x+1/2*e )^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/ 2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*_alpha^2*b^2-a^2*2^(1/2)*_alp ha*arctanh(1/2/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)/(g *(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(4*a^2-3*b^2)*g*2^(1/2)*(16*sin(1/2 *f*x+1/2*e)^2*_alpha^2*a^2-12*sin(1/2*f*x+1/2*e)^2*_alpha^2*b^2-4*_alph...
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= "fricas")
Output:
Timed out
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**3/(g*cos(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)
Output:
Timed out
\[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= "maxima")
Output:
integrate(sin(f*x + e)^3/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)), x)
\[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= "giac")
Output:
integrate(sin(f*x + e)^3/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)), x)
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:
int(sin(e + f*x)^3/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))),x)
Output:
int(sin(e + f*x)^3/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))), x)
\[ \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )^{3}}{\cos \left (f x +e \right ) \sin \left (f x +e \right ) b +\cos \left (f x +e \right ) a}d x \right )}{g} \] Input:
int(sin(f*x+e)^3/(g*cos(f*x+e))^(1/2)/(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)*sin(e + f* x)*b + cos(e + f*x)*a),x))/g