3.207 \(\int \frac{(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=737 \[ -\frac{2 a b \cos (e+f x) \sin ^2(e+f x)^{-q/2} (g \tan (e+f x))^q F_1\left (\frac{1-q}{2};-\frac{q}{2},2;\frac{3-q}{2};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{f (q-1) \left (a^2-b^2\right )^2}+\frac{a^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-q-1)} \left (1-\cos ^2(e+f x)\right )^{\frac{q-1}{2}} (g \tan (e+f x))^q \left (1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )^{\frac{3-q}{2}+\frac{q-1}{2}-2} \left (\left (2 \left (a^2-b^2\right )+b^2 (q+1) \cos ^2(e+f x)\right ) \Phi \left (-\frac{a^2 \cot ^2(e+f x)}{a^2-b^2},1,\frac{1-q}{2}\right )-b^2 (q-1) \cos ^2(e+f x) \Phi \left (-\frac{a^2 \cot ^2(e+f x)}{a^2-b^2},1,\frac{3-q}{2}\right )\right )}{2 f \left (a^2-b^2\right )^2 \left (b^2-a^2\right )}-\frac{a^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-q-1)} (g \tan (e+f x))^q \left (1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )^{\frac{q-1}{2}} \, _2F_1\left (\frac{1-q}{2},\frac{1-q}{2};\frac{3-q}{2};\frac{\cos ^2(e+f x)-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}{1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}\right )}{f (q-1) \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-q-1)} (g \tan (e+f x))^q \left (1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )^{\frac{q-1}{2}} \, _2F_1\left (\frac{1-q}{2},\frac{1-q}{2};\frac{3-q}{2};\frac{\cos ^2(e+f x)-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}{1-\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}}\right )}{f (q-1) \left (a^2-b^2\right )^2} \]

[Out]

(a^2*Cos[e + f*x]*(1 - Cos[e + f*x]^2)^((-1 + q)/2)*(1 - (b^2*Cos[e + f*x]^2)/(-a^2 + b^2))^(-2 + (3 - q)/2 +
(-1 + q)/2)*((2*(a^2 - b^2) + b^2*(1 + q)*Cos[e + f*x]^2)*HurwitzLerchPhi[-((a^2*Cot[e + f*x]^2)/(a^2 - b^2)),
1, (1 - q)/2] - b^2*(-1 + q)*Cos[e + f*x]^2*HurwitzLerchPhi[-((a^2*Cot[e + f*x]^2)/(a^2 - b^2)), 1, (3 - q)/2
])*Sin[e + f*x]*(Sin[e + f*x]^2)^((-1 - q)/2)*(g*Tan[e + f*x])^q)/(2*(a^2 - b^2)^2*(-a^2 + b^2)*f) - (a^2*Cos[
e + f*x]*(1 - (b^2*Cos[e + f*x]^2)/(-a^2 + b^2))^((-1 + q)/2)*Hypergeometric2F1[(1 - q)/2, (1 - q)/2, (3 - q)/
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]*
(Sin[e + f*x]^2)^((-1 - q)/2)*(g*Tan[e + f*x])^q)/((a^2 - b^2)^2*f*(-1 + q)) + (b^2*Cos[e + f*x]*(1 - (b^2*Cos
[e + f*x]^2)/(-a^2 + b^2))^((-1 + q)/2)*Hypergeometric2F1[(1 - q)/2, (1 - q)/2, (3 - q)/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]*(Sin[e + f*x]^2)^((-1
- q)/2)*(g*Tan[e + f*x])^q)/((a^2 - b^2)^2*f*(-1 + q)) - (2*a*b*AppellF1[(1 - q)/2, -q/2, 2, (3 - q)/2, Cos[e
+ f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]*(g*Tan[e + f*x])^q)/((a^2 - b^2)^2*f*(-1 + q)*(Sin[e
+ f*x]^2)^(q/2))

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Rubi [F]  time = 0.045093, 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 = {} \[ \int \frac{(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x])^2,x]

[Out]

Defer[Int][(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x])^2, x]

Rubi steps

\begin{align*} \int \frac{(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx &=\int \frac{(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx\\ \end{align*}

Mathematica [A]  time = 14.2268, size = 908, normalized size = 1.23 \[ \frac{\tan ^{p+1}(e+f x) (g \tan (e+f x))^p \left (a (p+2) \left (\left (a^2+b^2\right ) \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )-2 b^2 \, _2F_1\left (2,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )\right )+2 b \left (b^2-a^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},2;\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)\right )}{a^3 \left (a^2-b^2\right ) f (p+1) (p+2) (a+b \sin (e+f x))^2 \left (\frac{\sec ^2(e+f x) \left (a (p+2) \left (\left (a^2+b^2\right ) \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )-2 b^2 \, _2F_1\left (2,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )\right )+2 b \left (b^2-a^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},2;\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)\right ) \tan ^p(e+f x)}{a^3 \left (a^2-b^2\right ) (p+2)}+\frac{\left (2 b \left (b^2-a^2\right ) (p+1) F_1\left (\frac{p+2}{2};-\frac{1}{2},2;\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)+2 b \left (b^2-a^2\right ) (p+1) \tan (e+f x) \left (\frac{4 \left (b^2-a^2\right ) (p+2) F_1\left (\frac{p+2}{2}+1;-\frac{1}{2},3;\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},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)}{p+4}\right )+a (p+2) \left (\left (a^2+b^2\right ) (p+1) \csc (e+f x) \sec (e+f x) \left (\frac{1}{1-\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}}-\, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )\right )-2 b^2 (p+1) \csc (e+f x) \sec (e+f x) \left (\frac{1}{\left (1-\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )^2}-\, _2F_1\left (2,\frac{p+1}{2};\frac{p+3}{2};\frac{\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right )\right )\right )\right ) \tan ^{p+1}(e+f x)}{a^3 \left (a^2-b^2\right ) (p+1) (p+2)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Tan[e + f*x])^p/(a + b*Sin[e + f*x])^2,x]

[Out]

(Tan[e + f*x]^(1 + p)*(g*Tan[e + f*x])^p*(a*(2 + p)*((a^2 + b^2)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, ((
-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 2*b^2*Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*Tan[e + f*x]
^2)/a^2]) + 2*b*(-a^2 + b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 2, (4 + p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Ta
n[e + f*x]^2)/a^2]*Tan[e + f*x]))/(a^3*(a^2 - b^2)*f*(1 + p)*(2 + p)*(a + b*Sin[e + f*x])^2*((Sec[e + f*x]^2*T
an[e + f*x]^p*(a*(2 + p)*((a^2 + b^2)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*Tan[e + f*x]^2)
/a^2] - 2*b^2*Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]) + 2*b*(-a^2 + b^2
)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 2, (4 + p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e +
f*x]))/(a^3*(a^2 - b^2)*(2 + p)) + (Tan[e + f*x]^(1 + p)*(2*b*(-a^2 + b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 2
, (4 + p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2 + 2*b*(-a^2 + b^2)*(1 + p)*Tan
[e + f*x]*((4*(-a^2 + b^2)*(2 + p)*AppellF1[1 + (2 + p)/2, -1/2, 3, 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, 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])/(4 + p)) + a*
(2 + p)*(-2*b^2*(1 + p)*Csc[e + f*x]*Sec[e + f*x]*(-Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*T
an[e + f*x]^2)/a^2] + (1 - ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2)^(-2)) + (a^2 + b^2)*(1 + p)*Csc[e + f*x]*Sec[e +
f*x]*(-Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (1 - ((-a^2 + b^2)*Tan
[e + f*x]^2)/a^2)^(-1)))))/(a^3*(a^2 - b^2)*(1 + p)*(2 + p))))

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Maple [F]  time = 0.677, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\tan \left ( fx+e \right ) \right ) ^{p}}{ \left ( a+b\sin \left ( fx+e \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int((g*tan(f*x+e))^p/(a+b*sin(f*x+e))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a)^2, x)

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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^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(-(g*tan(f*x + e))^p/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan{\left (e + f x \right )}\right )^{p}}{\left (a + b \sin{\left (e + f x \right )}\right )^{2}}\, 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))**2,x)

[Out]

Integral((g*tan(e + f*x))**p/(a + b*sin(e + f*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a)^2, x)