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]

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 = 0.05, 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 = {} \[ \int \frac {(g \tan (e+f x))^p}{(a+b \sin (e+f x))^2} \, dx \]

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*}

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Mathematica [A]  time = 14.40, size = 866, normalized size = 1.18 \[ \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};\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )-2 b^2 \, _2F_1\left (2,\frac {p+1}{2};\frac {p+3}{2};\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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),\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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};\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )-2 b^2 \, _2F_1\left (2,\frac {p+1}{2};\frac {p+3}{2};\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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),\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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),\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+2 b \left (b^2-a^2\right ) (p+1) \tan (e+f x) \left (\frac {4 \left (\frac {b^2}{a^2}-1\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),\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\right ) \tan (e+f x) \sec ^2(e+f x)}{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),\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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-\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 )-2 b^2 (p+1) \csc (e+f x) \sec (e+f x) \left (\frac {1}{\left (1-\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\right )^2}-\, _2F_1\left (2,\frac {p+1}{2};\frac {p+3}{2};\left (\frac {b^2}{a^2}-1\right ) \tan ^2(e+f x)\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, (-
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]))/(a^3*(a^2 - b^2)*f*(1 + p)*(2 + p)*(a + b*Sin[e + f*x])^2*((Sec[e + f*x]^2*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*Hy
pergeometric2F1[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]))/(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, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2 + 2*b*(-a^2 + b^2)*(1 + p)*Tan[e + f*x]*((4*(-1 + b^2/a^2)
*(2 + p)*AppellF1[1 + (2 + p)/2, -1/2, 3, 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, 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)) + a*(2 + p)*(-2*b^2*(1 + p)*Csc[e + f*x]*Sec
[e + f*x]*(-Hypergeometric2F1[2, (1 + p)/2, (3 + p)/2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (1 - (-1 + b^2/a^2)*Ta
n[e + f*x]^2)^(-2)) + (a^2 + b^2)*(1 + 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^3*(a^2 - b^2)*(1 + p)*(2 +
p))))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (g \tan \left (f x + e\right )\right )^{p}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}, x\right ) \]

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

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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maple [F]  time = 1.57, 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 \[ \int \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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

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

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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