3.3.6 \(\int \frac {(g \tan (e+f x))^p}{a+b \sin (e+f x)} \, dx\) [206]

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

[Out]

a*g*(1-b^2*cos(f*x+e)^2/(-a^2+b^2))^(-1/2+1/2*p)*Hypergeometric2F1(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*cos(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))

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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)} \, dx \end {gather*}

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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(858\) vs. \(2(284)=568\).
time = 13.04, size = 858, normalized size = 3.02 \begin {gather*} \frac {\tan ^{1+p}(e+f x) (g \tan (e+f x))^p \left (\left (a^2-b^2\right ) (1+p) F_1\left (\frac {2+p}{2};-\frac {1}{2},1;\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)+a \left (b (2+p) \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+p) \, _2F_1\left (\frac {1}{2},1+\frac {p}{2};2+\frac {p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b f (1+p) (2+p) (a+b \sin (e+f x)) \left (\frac {\sec ^2(e+f x) \tan ^p(e+f x) \left (\left (a^2-b^2\right ) (1+p) F_1\left (\frac {2+p}{2};-\frac {1}{2},1;\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)+a \left (b (2+p) \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+p) \, _2F_1\left (\frac {1}{2},1+\frac {p}{2};2+\frac {p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b (2+p)}+\frac {\tan ^{1+p}(e+f x) \left (\left (a^2-b^2\right ) (1+p) F_1\left (\frac {2+p}{2};-\frac {1}{2},1;\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+\left (a^2-b^2\right ) (1+p) \tan (e+f x) \left (\frac {2 \left (-1+\frac {b^2}{a^2}\right ) (2+p) F_1\left (1+\frac {2+p}{2};-\frac {1}{2},2;1+\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{4+p}+\frac {(2+p) F_1\left (1+\frac {2+p}{2};\frac {1}{2},1;1+\frac {4+p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{4+p}\right )+a \left (-a (1+p) \, _2F_1\left (\frac {1}{2},1+\frac {p}{2};2+\frac {p}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)-2 a \left (1+\frac {p}{2}\right ) (1+p) \sec ^2(e+f x) \left (-\, _2F_1\left (\frac {1}{2},1+\frac {p}{2};2+\frac {p}{2};-\tan ^2(e+f x)\right )+\frac {1}{\sqrt {1+\tan ^2(e+f x)}}\right )+b (1+p) (2+p) \csc (e+f x) \sec (e+f x) \left (-\, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )+\frac {1}{1-\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}}\right )\right )\right )}{a^2 b (1+p) (2+p)}\right )} \end {gather*}

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, (-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)
*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)*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, (-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]*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^2*b*(1 + p)*(2 + p))))

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Maple [F]
time = 0.32, size = 0, normalized size = 0.00 \[\int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{a +b \sin \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), 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}}{a + b \sin {\left (e + f x \right )}}\, 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)),x)

[Out]

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

[Out]

integrate((g*tan(f*x + e))^p/(b*sin(f*x + e) + a), 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}{a+b\,\sin \left (e+f\,x\right )} \,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)),x)

[Out]

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

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