3.2.51 \(\int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx\) [151]

Optimal. Leaf size=64 \[ \frac {6 \cos ^2(e+f x)^{5/4} \, _2F_1\left (\frac {13}{12},\frac {5}{4};\frac {25}{12};\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{13 d f \sqrt [3]{b \sin (e+f x)}} \]

[Out]

6/13*(cos(f*x+e)^2)^(5/4)*hypergeom([13/12, 5/4],[25/12],sin(f*x+e)^2)*(d*tan(f*x+e))^(5/2)/d/f/(b*sin(f*x+e))
^(1/3)

________________________________________________________________________________________

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 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac {13}{12},\frac {5}{4};\frac {25}{12};\sin ^2(e+f x)\right )}{13 d f \sqrt [3]{b \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[13/12, 5/4, 25/12, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(13*d*f
*(b*Sin[e + f*x])^(1/3))

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 {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx &=\frac {\left (b \cos ^{\frac {5}{2}}(e+f x) (d \tan (e+f x))^{5/2}\right ) \int \frac {(b \sin (e+f x))^{7/6}}{\cos ^{\frac {3}{2}}(e+f x)} \, dx}{d (b \sin (e+f x))^{5/2}}\\ &=\frac {6 \cos ^2(e+f x)^{5/4} \, _2F_1\left (\frac {13}{12},\frac {5}{4};\frac {25}{12};\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{13 d f \sqrt [3]{b \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 50.54, size = 63, normalized size = 0.98 \begin {gather*} -\frac {2 d \left (-1+\sqrt [4]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{12},\frac {1}{4};\frac {13}{12};\sin ^2(e+f x)\right )\right ) \sqrt {d \tan (e+f x)}}{f \sqrt [3]{b \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________