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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0943887, 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} \[ \frac{6 \sqrt [3]{\cos ^2(e+f x)} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{13}{12};\frac{25}{12};\sin ^2(e+f x)\right )}{13 d f} \]

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 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 \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*}

Mathematica [A]  time = 0.730304, size = 67, normalized size = 1.05 \[ \frac{2 d (b \sin (e+f x))^{3/2} \left (\sec ^2(e+f x)^{3/4} \, _2F_1\left (\frac{1}{12},\frac{3}{4};\frac{13}{12};-\tan ^2(e+f x)\right )-1\right )}{3 f (d \tan (e+f x))^{4/3}} \]

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))

________________________________________________________________________________________

Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}}}\,{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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sin \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac{2}{3}} b \sin \left (f x + e\right )}{d \tan \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))^(2/3)*b*sin(f*x + e)/(d*tan(f*x + e)), x)

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}}}\,{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="giac")

[Out]

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