3.206 \(\int \frac{(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=284 \[ \frac{b \cos (e+f x) \sin ^2(e+f x)^{-p/2} (g \tan (e+f x))^p F_1\left (\frac{1-p}{2};-\frac{p}{2},1;\frac{3-p}{2};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{f (p-1) \left (b^2-a^2\right )}+\frac{a g \sin ^2(e+f x)^{\frac{1-p}{2}} (g \tan (e+f x))^{p-1} \left (1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )^{\frac{p-1}{2}} \, _2F_1\left (\frac{1-p}{2},\frac{1-p}{2};\frac{3-p}{2};\frac{\cos ^2(e+f x)-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}{1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}\right )}{f (p-1) \left (a^2-b^2\right )} \]

[Out]

(a*g*(1 - (b^2*Cos[e + f*x]^2)/(-a^2 + b^2))^((-1 + p)/2)*Hypergeometric2F1[(1 - p)/2, (1 - p)/2, (3 - p)/2, (
Cos[e + f*x]^2 - (b^2*Cos[e + f*x]^2)/(-a^2 + b^2))/(1 - (b^2*Cos[e + f*x]^2)/(-a^2 + b^2))]*(Sin[e + f*x]^2)^
((1 - p)/2)*(g*Tan[e + f*x])^(-1 + p))/((a^2 - b^2)*f*(-1 + p)) + (b*AppellF1[(1 - p)/2, -p/2, 1, (3 - p)/2, C
os[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]*(g*Tan[e + f*x])^p)/((-a^2 + b^2)*f*(-1 + p)*(S
in[e + f*x]^2)^(p/2))

________________________________________________________________________________________

Rubi [F]  time = 0.0480098, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x]),x]

[Out]

Defer[Int][(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x]), x]

Rubi steps

\begin{align*} \int \frac{(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx &=\int \frac{(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx\\ \end{align*}

Mathematica [B]  time = 13.4983, size = 864, normalized size = 3.04 \[ \frac{\tan ^{p+1}(e+f x) (g \tan (e+f x))^p \left (\left (a^2-b^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},1;\frac{p+4}{2};-\tan ^2(e+f x),\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan (e+f x)+a \left (b (p+2) \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\left (\frac{b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )-a (p+1) \, _2F_1\left (\frac{1}{2},\frac{p}{2}+1;\frac{p}{2}+2;-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b f (p+1) (p+2) (a+b \sin (e+f x)) \left (\frac{\sec ^2(e+f x) \left (\left (a^2-b^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},1;\frac{p+4}{2};-\tan ^2(e+f x),\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan (e+f x)+a \left (b (p+2) \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\left (\frac{b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )-a (p+1) \, _2F_1\left (\frac{1}{2},\frac{p}{2}+1;\frac{p}{2}+2;-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right ) \tan ^p(e+f x)}{a^2 b (p+2)}+\frac{\left (\left (a^2-b^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},1;\frac{p+4}{2};-\tan ^2(e+f x),\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \sec ^2(e+f x)+\left (a^2-b^2\right ) (p+1) \tan (e+f x) \left (\frac{2 \left (b^2-a^2\right ) (p+2) F_1\left (\frac{p+2}{2}+1;-\frac{1}{2},2;\frac{p+4}{2}+1;-\tan ^2(e+f x),\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan (e+f x) \sec ^2(e+f x)}{a^2 (p+4)}+\frac{(p+2) F_1\left (\frac{p+2}{2}+1;\frac{1}{2},1;\frac{p+4}{2}+1;-\tan ^2(e+f x),\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan (e+f x) \sec ^2(e+f x)}{p+4}\right )+a \left (-a (p+1) \, _2F_1\left (\frac{1}{2},\frac{p}{2}+1;\frac{p}{2}+2;-\tan ^2(e+f x)\right ) \sec ^2(e+f x)-2 a \left (\frac{p}{2}+1\right ) (p+1) \left (\frac{1}{\sqrt{\tan ^2(e+f x)+1}}-\, _2F_1\left (\frac{1}{2},\frac{p}{2}+1;\frac{p}{2}+2;-\tan ^2(e+f x)\right )\right ) \sec ^2(e+f x)+b (p+1) (p+2) \csc (e+f x) \left (\frac{1}{1-\left (\frac{b^2}{a^2}-1\right ) \tan ^2(e+f x)}-\, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\left (\frac{b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )\right ) \sec (e+f x)\right )\right ) \tan ^{p+1}(e+f x)}{a^2 b (p+1) (p+2)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x]),x]

[Out]

(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, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e + f*x] + a*(b*(2 + p)*Hypergeometric2F1[1, (1 + p)/2, (3 + p
)/2, (-1 + b^2/a^2)*Tan[e + f*x]^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)
*AppellF1[(2 + p)/2, -1/2, 1, (4 + p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e + f*x] + a*
(b*(2 + p)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - a*(1 + p)*Hypergeometri
c2F1[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, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e
+ f*x]^2 + (a^2 - b^2)*(1 + p)*Tan[e + f*x]*((2*(-a^2 + b^2)*(2 + p)*AppellF1[1 + (2 + p)/2, -1/2, 2, 1 + (4 +
p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(a^2*(4 + p)) + ((2 +
p)*AppellF1[1 + (2 + p)/2, 1/2, 1, 1 + (4 + p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^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]*Sec[e + f*x]*(-Hypergeometric2F1[1, (1 + p)/2,
(3 + p)/2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (1 - (-1 + b^2/a^2)*Tan[e + f*x]^2)^(-1)))))/(a^2*b*(1 + p)*(2 + p
))))

________________________________________________________________________________________

Maple [F]  time = 0.362, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\tan \left ( fx+e \right ) \right ) ^{p}}{a+b\sin \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x)

[Out]

int((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan{\left (e + f x \right )}\right )^{p}}{a + b \sin{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))**p/(a+b*sin(f*x+e)),x)

[Out]

Integral((g*tan(e + f*x))**p/(a + b*sin(e + f*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan \left (f x + e\right )\right )^{p}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), x)