3.3.7 \(\int \frac {(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx\) [207]

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

[Out]

1/2*a^2*cos(f*x+e)*(1-cos(f*x+e)^2)^(-1/2+1/2*q)/(1-b^2*cos(f*x+e)^2/(-a^2+b^2))*((2*a^2-2*b^2+b^2*(1+q)*cos(f
*x+e)^2)*HurwitzLerchPhi(a^2*cos(f*x+e)^2/(a^2-b^2)/(-1+cos(f*x+e)^2),1,1/2-1/2*q)-b^2*(-1+q)*cos(f*x+e)^2*Hur
witzLerchPhi(a^2*cos(f*x+e)^2/(a^2-b^2)/(-1+cos(f*x+e)^2),1,3/2-1/2*q))*sin(f*x+e)*(sin(f*x+e)^2)^(-1/2-1/2*q)
*(g*tan(f*x+e))^q/(a^2-b^2)^2/(-a^2+b^2)/f-a^2*cos(f*x+e)*(1-b^2*cos(f*x+e)^2/(-a^2+b^2))^(-1/2+1/2*q)*Hyperge
ometric2F1(1/2-1/2*q,1/2-1/2*q,3/2-1/2*q,(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)*(sin(f*x+e)^2)^(-1/2-1/2*q)*(g*tan(f*x+e))^q/(a^2-b^2)^2/f/(-1+q)+b^2*cos(f*x+e)*(1-b^2*cos(
f*x+e)^2/(-a^2+b^2))^(-1/2+1/2*q)*Hypergeometric2F1(1/2-1/2*q,1/2-1/2*q,3/2-1/2*q,(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)*(sin(f*x+e)^2)^(-1/2-1/2*q)*(g*tan(f*x+e))^q/(a^2-b
^2)^2/f/(-1+q)-2*a*b*AppellF1(1/2-1/2*q,-1/2*q,2,3/2-1/2*q,cos(f*x+e)^2,b^2*cos(f*x+e)^2/(-a^2+b^2))*cos(f*x+e
)*(g*tan(f*x+e))^q/(a^2-b^2)^2/f/(-1+q)/((sin(f*x+e)^2)^(1/2*q))

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Rubi [F]
time = 0.03, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx \end {gather*}

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*}

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Mathematica [A]
time = 5.29, size = 695, normalized size = 0.94 \begin {gather*} \frac {\cos (e+f x) \sin (e+f x) (g \tan (e+f x))^p \left (a (2+p) \left (\left (a^2+b^2\right ) \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 \, _2F_1\left (2,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (-a^2+b^2\right ) (1+p) F_1\left (\frac {2+p}{2};-\frac {1}{2},2;\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )}{f (1+p) (a+b \sin (e+f x))^2 \left (a (2+p) \left (\left (a^2+b^2\right ) \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 \, _2F_1\left (2,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (-a^2+b^2\right ) F_1\left (\frac {2+p}{2};-\frac {1}{2},2;\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+2 b \left (-a^2+b^2\right ) (1+p) F_1\left (\frac {2+p}{2};-\frac {1}{2},2;\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+\frac {2 b \left (-a^2+b^2\right ) (2+p) \left (\left (-4+\frac {4 b^2}{a^2}\right ) F_1\left (\frac {4+p}{2};-\frac {1}{2},3;\frac {6+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+F_1\left (\frac {4+p}{2};\frac {1}{2},2;\frac {6+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right ) \tan ^3(e+f x)}{4+p}+a (2+p) \left (-2 b^2 \left (-\, _2F_1\left (2,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\frac {1}{\left (1+\left (1-\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )^2}\right )+\left (a^2+b^2\right ) \left (-\, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\frac {1}{1+\left (1-\frac {b^2}{a^2}\right ) \tan ^2(e+f x)}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]*Sin[e + f*x]*(g*Tan[e + f*x])^p*(a*(2 + p)*((a^2 + b^2)*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/
2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 2*b^2*Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, (-1 + b^2/a^2)*Tan[e + f*
x]^2]) + 2*b*(-a^2 + b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 2, (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[
e + f*x]^2]*Tan[e + f*x]))/(f*(1 + p)*(a + b*Sin[e + f*x])^2*(a*(2 + p)*((a^2 + b^2)*Hypergeometric2F1[1, (1 +
p)/2, (3 + p)/2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 2*b^2*Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, (-1 + b^2/
a^2)*Tan[e + f*x]^2]) + 2*b*(-a^2 + b^2)*AppellF1[(2 + p)/2, -1/2, 2, (4 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^
2)*Tan[e + f*x]^2]*Tan[e + f*x] + 2*b*(-a^2 + b^2)*(1 + p)*AppellF1[(2 + p)/2, -1/2, 2, (4 + p)/2, -Tan[e + f*
x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x] + (2*b*(-a^2 + b^2)*(2 + p)*((-4 + (4*b^2)/a^2)*AppellF1[(4
+ p)/2, -1/2, 3, (6 + p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + AppellF1[(4 + p)/2, 1/2, 2, (6 +
p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2])*Tan[e + f*x]^3)/(4 + p) + a*(2 + p)*(-2*b^2*(-Hypergeo
metric2F1[2, (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)^(-2)) +
(a^2 + b^2)*(-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)))))

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Maple [F]
time = 1.08, size = 0, normalized size = 0.00 \[\int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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