Integrand size = 23, antiderivative size = 284 \[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\frac {a g \left (1-\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )^{\frac {1}{2} (-1+p)} \operatorname {Hypergeometric2F1}\left (\frac {1-p}{2},\frac {1-p}{2},\frac {3-p}{2},\frac {\cos ^2(e+f x)-\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}}{1-\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}}\right ) \sin ^2(e+f x)^{\frac {1-p}{2}} (g \tan (e+f x))^{-1+p}}{\left (a^2-b^2\right ) f (-1+p)}+\frac {b \operatorname {AppellF1}\left (\frac {1-p}{2},-\frac {p}{2},1,\frac {3-p}{2},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-p/2} (g \tan (e+f x))^p}{\left (-a^2+b^2\right ) f (-1+p)} \] Output:
a*g*(1-b^2*cos(f*x+e)^2/(-a^2+b^2))^(-1/2+1/2*p)*hypergeom([1/2-1/2*p, 1/2 -1/2*p],[3/2-1/2*p],(cos(f*x+e)^2-b^2*cos(f*x+e)^2/(-a^2+b^2))/(1-b^2*cos( f*x+e)^2/(-a^2+b^2)))*(sin(f*x+e)^2)^(1/2-1/2*p)*(g*tan(f*x+e))^(-1+p)/(a^ 2-b^2)/f/(-1+p)+b*AppellF1(1/2-1/2*p,-1/2*p,1,3/2-1/2*p,cos(f*x+e)^2,b^2*c os(f*x+e)^2/(-a^2+b^2))*cos(f*x+e)*(g*tan(f*x+e))^p/(-a^2+b^2)/f/(-1+p)/(( sin(f*x+e)^2)^(1/2*p))
Leaf count is larger than twice the leaf count of optimal. \(858\) vs. \(2(284)=568\).
Time = 14.41 (sec) , antiderivative size = 858, normalized size of antiderivative = 3.02 \[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
Integrate[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x]),x]
Output:
(Tan[e + f*x]^(1 + p)*(g*Tan[e + f*x])^p*((a^2 - b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 1, (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2 ]*Tan[e + f*x] + a*(b*(2 + p)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, ( (-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - a*(1 + p)*Hypergeometric2F1[1/2, 1 + p /2, 2 + p/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a^2*b*f*(1 + p)*(2 + p)*(a + b*Sin[e + f*x])*((Sec[e + f*x]^2*Tan[e + f*x]^p*((a^2 - b^2)*(1 + p)*App ellF1[(2 + p)/2, -1/2, 1, (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x] + a*(b*(2 + p)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - a*(1 + p)*Hypergeometric2F1[1 /2, 1 + p/2, 2 + p/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a^2*b*(2 + p)) + ( Tan[e + f*x]^(1 + p)*((a^2 - b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 1, (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2 + ( a^2 - b^2)*(1 + p)*Tan[e + f*x]*((2*(-1 + b^2/a^2)*(2 + p)*AppellF1[1 + (2 + p)/2, -1/2, 2, 1 + (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f *x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(4 + p) + ((2 + p)*AppellF1[1 + (2 + p )/2, 1/2, 1, 1 + (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2 ]*Sec[e + f*x]^2*Tan[e + f*x])/(4 + p)) + a*(-(a*(1 + p)*Hypergeometric2F1 [1/2, 1 + p/2, 2 + p/2, -Tan[e + f*x]^2]*Sec[e + f*x]^2) - 2*a*(1 + p/2)*( 1 + p)*Sec[e + f*x]^2*(-Hypergeometric2F1[1/2, 1 + p/2, 2 + p/2, -Tan[e + f*x]^2] + 1/Sqrt[1 + Tan[e + f*x]^2]) + b*(1 + p)*(2 + p)*Csc[e + f*x]*...
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.
\(\displaystyle \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)}dx\) |
\(\Big \downarrow \) 3211 |
\(\displaystyle \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)}dx\) |
Input:
Int[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x]),x]
Output:
$Aborted
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Unintegrable[(a + b*Sin[e + f*x])^m*(g*Tan[e + f* x])^p, x] /; FreeQ[{a, b, e, f, g, m, p}, x]
\[\int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{a +b \sin \left (f x +e \right )}d x\]
Input:
int((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x)
Output:
int((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x)
\[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:
integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="fricas")
Output:
integral((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)
\[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\int \frac {\left (g \tan {\left (e + f x \right )}\right )^{p}}{a + b \sin {\left (e + f x \right )}}\, dx \] Input:
integrate((g*tan(f*x+e))**p/(a+b*sin(f*x+e)),x)
Output:
Integral((g*tan(e + f*x))**p/(a + b*sin(e + f*x)), x)
\[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:
integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="maxima")
Output:
integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)
\[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:
integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="giac")
Output:
integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)
Timed out. \[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p}{a+b\,\sin \left (e+f\,x\right )} \,d x \] Input:
int((g*tan(e + f*x))^p/(a + b*sin(e + f*x)),x)
Output:
int((g*tan(e + f*x))^p/(a + b*sin(e + f*x)), x)
\[ \int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx=g^{p} \left (\int \frac {\tan \left (f x +e \right )^{p}}{\sin \left (f x +e \right ) b +a}d x \right ) \] Input:
int((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x)
Output:
g**p*int(tan(e + f*x)**p/(sin(e + f*x)*b + a),x)