3.2.55 \(\int \frac {\sqrt {b \sin (e+f x)}}{\sqrt [3]{d \tan (e+f x)}} \, dx\) [155]

Optimal. Leaf size=64 \[ \frac {6 \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{3},\frac {7}{12};\frac {19}{12};\sin ^2(e+f x)\right ) \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3}}{7 d f} \]

[Out]

6/7*(cos(f*x+e)^2)^(1/3)*hypergeom([1/3, 7/12],[19/12],sin(f*x+e)^2)*(b*sin(f*x+e))^(1/2)*(d*tan(f*x+e))^(2/3)
/d/f

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Rubi [A]
time = 0.06, 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)} \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {7}{12};\frac {19}{12};\sin ^2(e+f x)\right )}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(Cos[e + f*x]^2)^(1/3)*Hypergeometric2F1[1/3, 7/12, 19/12, Sin[e + f*x]^2]*Sqrt[b*Sin[e + f*x]]*(d*Tan[e +
f*x])^(2/3))/(7*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 {\sqrt {b \sin (e+f x)}}{\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)} \sqrt [6]{b \sin (e+f x)} \, 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 {7}{12};\frac {19}{12};\sin ^2(e+f x)\right ) \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3}}{7 d f}\\ \end {align*}

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Mathematica [A]
time = 10.38, size = 66, normalized size = 1.03 \begin {gather*} \frac {6 \, _2F_1\left (\frac {7}{12},\frac {5}{4};\frac {19}{12};-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3}}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Hypergeometric2F1[7/12, 5/4, 19/12, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4)*Sqrt[b*Sin[e + f*x]]*(d*Tan[e +
f*x])^(2/3))/(7*d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b \sin {\left (e + f x \right )}}}{\sqrt [3]{d \tan {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sqrt(b*sin(f*x + e))/(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 {\sqrt {b\,\sin \left (e+f\,x\right )}}{{\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))^(1/2)/(d*tan(e + f*x))^(1/3),x)

[Out]

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

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