3.2.59 \(\int \frac {(b \sin (e+f x))^{3/2}}{\sqrt [3]{d \tan (e+f x)}} \, dx\) [159]

Optimal. Leaf size=64 \[ \frac {6 \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {25}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{2/3}}{13 d f} \]

[Out]

6/13*(cos(f*x+e)^2)^(1/3)*hypergeom([1/3, 13/12],[25/12],sin(f*x+e)^2)*(b*sin(f*x+e))^(3/2)*(d*tan(f*x+e))^(2/
3)/d/f

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Rubi [A]
time = 0.07, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2682, 2657} \begin {gather*} \frac {6 \sqrt [3]{\cos ^2(e+f x)} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {25}{12};\sin ^2(e+f x)\right )}{13 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sin[e + f*x])^(3/2)/(d*Tan[e + f*x])^(1/3),x]

[Out]

(6*(Cos[e + f*x]^2)^(1/3)*Hypergeometric2F1[1/3, 13/12, 25/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(2/3))/(13*d*f)

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a*Cos[e + f*
x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b*(a*Sin[e + f*x])^(n + 1))), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {(b \sin (e+f x))^{3/2}}{\sqrt [3]{d \tan (e+f x)}} \, dx &=\frac {\left (b \cos ^{\frac {2}{3}}(e+f x) (d \tan (e+f x))^{2/3}\right ) \int \sqrt [3]{\cos (e+f x)} (b \sin (e+f x))^{7/6} \, dx}{d (b \sin (e+f x))^{2/3}}\\ &=\frac {6 \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {25}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{2/3}}{13 d f}\\ \end {align*}

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Mathematica [A]
time = 50.85, size = 67, normalized size = 1.05 \begin {gather*} \frac {2 d \left (-1+\, _2F_1\left (\frac {1}{12},\frac {3}{4};\frac {13}{12};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) (b \sin (e+f x))^{3/2}}{3 f (d \tan (e+f x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sin[e + f*x])^(3/2)/(d*Tan[e + f*x])^(1/3),x]

[Out]

(2*d*(-1 + Hypergeometric2F1[1/12, 3/4, 13/12, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(3/4))*(b*Sin[e + f*x])^(3/2)
)/(3*f*(d*Tan[e + f*x])^(4/3))

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Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {\left (b \sin \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sin(f*x+e))^(3/2)/(d*tan(f*x+e))^(1/3),x)

[Out]

int((b*sin(f*x+e))^(3/2)/(d*tan(f*x+e))^(1/3),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((b*sin(f*x+e))^(3/2)/(d*tan(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e))^(3/2)/(d*tan(f*x + e))^(1/3), 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((b*sin(f*x+e))^(3/2)/(d*tan(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

integral(sqrt(b*sin(f*x + e))*(d*tan(f*x + e))^(2/3)*b*sin(f*x + e)/(d*tan(f*x + e)), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sin(f*x+e))**(3/2)/(d*tan(f*x+e))**(1/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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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((b*sin(f*x+e))^(3/2)/(d*tan(f*x+e))^(1/3),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e))^(3/2)/(d*tan(f*x + e))^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (b\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sin(e + f*x))^(3/2)/(d*tan(e + f*x))^(1/3),x)

[Out]

int((b*sin(e + f*x))^(3/2)/(d*tan(e + f*x))^(1/3), x)

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