Integrand size = 23, antiderivative size = 138 \[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\frac {(g \tan (e+f x))^{1+p}}{a^2 f g (1+p)}-\frac {2 \cos ^2(e+f x)^{\frac {5+p}{2}} \operatorname {Hypergeometric2F1}\left (\frac {2+p}{2},\frac {5+p}{2},\frac {4+p}{2},\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac {2 (g \tan (e+f x))^{3+p}}{a^2 f g^3 (3+p)} \]
(g*tan(f*x+e))^(p+1)/a^2/f/g/(p+1)-2*(cos(f*x+e)^2)^(5/2+1/2*p)*hypergeom( [1+1/2*p, 5/2+1/2*p],[2+1/2*p],sin(f*x+e)^2)*sec(f*x+e)^3*(g*tan(f*x+e))^( 2+p)/a^2/f/g^2/(2+p)+2*(g*tan(f*x+e))^(3+p)/a^2/f/g^3/(3+p)
Leaf count is larger than twice the leaf count of optimal. \(626\) vs. \(2(138)=276\).
Time = 6.13 (sec) , antiderivative size = 626, normalized size of antiderivative = 4.54 \[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\frac {2 \left (\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )^p \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+p}{2},2+p,\frac {3+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+p}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1+p}{2},3+p,\frac {3+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+p}+\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1+p}{2},4+p,\frac {3+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+p}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {2+p}{2},2+p,\frac {4+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{2+p}+\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {2+p}{2},3+p,\frac {4+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{2+p}-\frac {8 \operatorname {Hypergeometric2F1}\left (\frac {2+p}{2},4+p,\frac {4+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{2+p}+\frac {\operatorname {Hypergeometric2F1}\left (2+p,\frac {3+p}{2},\frac {5+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{3+p}-\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {3+p}{2},3+p,\frac {5+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{3+p}+\frac {12 \operatorname {Hypergeometric2F1}\left (\frac {3+p}{2},4+p,\frac {5+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{3+p}+\frac {2 \operatorname {Hypergeometric2F1}\left (3+p,\frac {4+p}{2},\frac {6+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^3\left (\frac {1}{2} (e+f x)\right )}{4+p}-\frac {8 \operatorname {Hypergeometric2F1}\left (\frac {4+p}{2},4+p,\frac {6+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^3\left (\frac {1}{2} (e+f x)\right )}{4+p}+\frac {2 \operatorname {Hypergeometric2F1}\left (4+p,\frac {5+p}{2},\frac {7+p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^4\left (\frac {1}{2} (e+f x)\right )}{5+p}\right ) (g \tan (e+f x))^p}{f (a+a \sin (e+f x))^2} \]
(2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])^4*Tan[(e + f*x)/2]*(Hypergeometric2F1[(1 + p)/2, 2 + p, (3 + p)/2, Tan[ (e + f*x)/2]^2]/(1 + p) - (2*Hypergeometric2F1[(1 + p)/2, 3 + p, (3 + p)/2 , Tan[(e + f*x)/2]^2])/(1 + p) + (2*Hypergeometric2F1[(1 + p)/2, 4 + p, (3 + p)/2, Tan[(e + f*x)/2]^2])/(1 + p) - (2*Hypergeometric2F1[(2 + p)/2, 2 + p, (4 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2])/(2 + p) + (6*Hyperge ometric2F1[(2 + p)/2, 3 + p, (4 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/ 2])/(2 + p) - (8*Hypergeometric2F1[(2 + p)/2, 4 + p, (4 + p)/2, Tan[(e + f *x)/2]^2]*Tan[(e + f*x)/2])/(2 + p) + (Hypergeometric2F1[2 + p, (3 + p)/2, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3 + p) - (6*Hypergeom etric2F1[(3 + p)/2, 3 + p, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2] ^2)/(3 + p) + (12*Hypergeometric2F1[(3 + p)/2, 4 + p, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3 + p) + (2*Hypergeometric2F1[3 + p, (4 + p)/2, (6 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^3)/(4 + p) - (8*Hype rgeometric2F1[(4 + p)/2, 4 + p, (6 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f* x)/2]^3)/(4 + p) + (2*Hypergeometric2F1[4 + p, (5 + p)/2, (7 + p)/2, Tan[( e + f*x)/2]^2]*Tan[(e + f*x)/2]^4)/(5 + p))*(g*Tan[e + f*x])^p)/(f*(a + a* Sin[e + f*x])^2)
Time = 0.45 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3190, 2009}
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.
| \(\displaystyle \int \frac {(g \tan (e+f x))^p}{(a \sin (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(g \tan (e+f x))^p}{(a \sin (e+f x)+a)^2}dx\) |
| \(\Big \downarrow \) 3190 |
| \(\displaystyle \frac {\int \left (a^2 \sec ^4(e+f x) (g \tan (e+f x))^p+a^2 \sec ^2(e+f x) \tan ^2(e+f x) (g \tan (e+f x))^p-2 a^2 \sec ^3(e+f x) \tan (e+f x) (g \tan (e+f x))^p\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^2 (g \tan (e+f x))^{p+3}}{f g^3 (p+3)}-\frac {2 a^2 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {p+5}{2}} (g \tan (e+f x))^{p+2} \operatorname {Hypergeometric2F1}\left (\frac {p+2}{2},\frac {p+5}{2},\frac {p+4}{2},\sin ^2(e+f x)\right )}{f g^2 (p+2)}+\frac {a^2 (g \tan (e+f x))^{p+1}}{f g (p+1)}}{a^4}\) |
((a^2*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p)) - (2*a^2*(Cos[e + f*x]^2)^(( 5 + p)/2)*Hypergeometric2F1[(2 + p)/2, (5 + p)/2, (4 + p)/2, Sin[e + f*x]^ 2]*Sec[e + f*x]^3*(g*Tan[e + f*x])^(2 + p))/(f*g^2*(2 + p)) + (2*a^2*(g*Ta n[e + f*x])^(3 + p))/(f*g^3*(3 + p)))/a^4
3.2.27.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x _)])^(p_.), x_Symbol] :> Simp[a^(2*m) Int[ExpandIntegrand[(g*Tan[e + f*x] )^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x] /; F reeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
\[\int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{\left (a +a \sin \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (g \tan {\left (e + f x \right )}\right )^{p}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
\[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx=\int \frac {{\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]