3.2.57 \(\int (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, dx\) [157]

Optimal. Leaf size=64 \[ \frac {6 \cos ^2(e+f x)^{7/6} \, _2F_1\left (\frac {7}{6},\frac {23}{12};\frac {35}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3}}{23 d f} \]

[Out]

6/23*(cos(f*x+e)^2)^(7/6)*hypergeom([7/6, 23/12],[35/12],sin(f*x+e)^2)*(b*sin(f*x+e))^(3/2)*(d*tan(f*x+e))^(7/
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 \cos ^2(e+f x)^{7/6} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3} \, _2F_1\left (\frac {7}{6},\frac {23}{12};\frac {35}{12};\sin ^2(e+f x)\right )}{23 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sin[e + f*x])^(3/2)*(d*Tan[e + f*x])^(4/3),x]

[Out]

(6*(Cos[e + f*x]^2)^(7/6)*Hypergeometric2F1[7/6, 23/12, 35/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(7/3))/(23*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 (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, dx &=\frac {\left (b \cos ^{\frac {7}{3}}(e+f x) (d \tan (e+f x))^{7/3}\right ) \int \frac {(b \sin (e+f x))^{17/6}}{\cos ^{\frac {4}{3}}(e+f x)} \, dx}{d (b \sin (e+f x))^{7/3}}\\ &=\frac {6 \cos ^2(e+f x)^{7/6} \, _2F_1\left (\frac {7}{6},\frac {23}{12};\frac {35}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3}}{23 d f}\\ \end {align*}

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Mathematica [A]
time = 50.71, size = 85, normalized size = 1.33 \begin {gather*} \frac {3 d \left (-\, _2F_1\left (\frac {11}{12},\frac {7}{4};\frac {23}{12};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)+\sqrt [4]{\sec ^2(e+f x)}\right ) (b \sin (e+f x))^{3/2} \sqrt [3]{d \tan (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sin[e + f*x])^(3/2)*(d*Tan[e + f*x])^(4/3),x]

[Out]

(3*d*(-(Hypergeometric2F1[11/12, 7/4, 23/12, -Tan[e + f*x]^2]*Sec[e + f*x]^2) + (Sec[e + f*x]^2)^(1/4))*(b*Sin
[e + f*x])^(3/2)*(d*Tan[e + f*x])^(1/3))/(f*(Sec[e + f*x]^2)^(1/4))

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Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \left (b \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (d \tan \left (f x +e \right )\right )^{\frac {4}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sin(f*x+e))^(3/2)*(d*tan(f*x+e))^(4/3),x)

[Out]

int((b*sin(f*x+e))^(3/2)*(d*tan(f*x+e))^(4/3),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))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e))^(3/2)*(d*tan(f*x + e))^(4/3), 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))^(4/3),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(4/3),x)

[Out]

Timed out

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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((b*sin(f*x+e))^(3/2)*(d*tan(f*x+e))^(4/3),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming sageVARb near 0Simplification assuming sageVARf near 0Simplification assuming sageV
ARx near 0S

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (b\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{4/3} \,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))^(4/3),x)

[Out]

int((b*sin(e + f*x))^(3/2)*(d*tan(e + f*x))^(4/3), x)

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