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

Optimal. Leaf size=64 \[ \frac{6 \cos ^2(e+f x)^{2/3} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, _2F_1\left (\frac{2}{3},\frac{17}{12};\frac{29}{12};\sin ^2(e+f x)\right )}{17 d f} \]

[Out]

(6*(Cos[e + f*x]^2)^(2/3)*Hypergeometric2F1[2/3, 17/12, 29/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(4/3))/(17*d*f)

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Rubi [A]  time = 0.0919098, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2602, 2577} \[ \frac{6 \cos ^2(e+f x)^{2/3} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, _2F_1\left (\frac{2}{3},\frac{17}{12};\frac{29}{12};\sin ^2(e+f x)\right )}{17 d f} \]

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)^(2/3)*Hypergeometric2F1[2/3, 17/12, 29/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(4/3))/(17*d*f)

Rule 2602

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]

Rule 2577

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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int (b \sin (e+f x))^{3/2} \sqrt [3]{d \tan (e+f x)} \, dx &=\frac{\left (b \cos ^{\frac{4}{3}}(e+f x) (d \tan (e+f x))^{4/3}\right ) \int \frac{(b \sin (e+f x))^{11/6}}{\sqrt [3]{\cos (e+f x)}} \, dx}{d (b \sin (e+f x))^{4/3}}\\ &=\frac{6 \cos ^2(e+f x)^{2/3} \, _2F_1\left (\frac{2}{3},\frac{17}{12};\frac{29}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3}}{17 d f}\\ \end{align*}

Mathematica [A]  time = 0.476477, size = 72, normalized size = 1.12 \[ \frac{6 \cos (e+f x) \sec ^2(e+f x)^{7/4} (b \sin (e+f x))^{5/2} \sqrt [3]{d \tan (e+f x)} \, _2F_1\left (\frac{17}{12},\frac{7}{4};\frac{29}{12};-\tan ^2(e+f x)\right )}{17 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Cos[e + f*x]*Hypergeometric2F1[17/12, 7/4, 29/12, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(7/4)*(b*Sin[e + f*x])^
(5/2)*(d*Tan[e + f*x])^(1/3))/(17*b*f)

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Maple [F]  time = 0.153, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\sqrt [3]{d\tan \left ( fx+e \right ) }\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b \sin \left (f x + e\right )\right )^{\frac{3}{2}} \left (d \tan \left (f x + e\right )\right )^{\frac{1}{3}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sin \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac{1}{3}} b \sin \left (f x + e\right ), x\right ) \end{align*}

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))^(1/3)*b*sin(f*x + e), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError