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

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

[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)

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Rubi [A]  time = 0.0830445, 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} \[ \frac{6 \sqrt [3]{\cos ^2(e+f x)} \sqrt{b \sin (e+f x)} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{7}{12};\frac{19}{12};\sin ^2(e+f x)\right )}{7 d f} \]

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 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{\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*}

Mathematica [A]  time = 0.318824, size = 66, normalized size = 1.03 \[ \frac{6 \sqrt [4]{\sec ^2(e+f x)} \sqrt{b \sin (e+f x)} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{19}{12};-\tan ^2(e+f x)\right )}{7 d f} \]

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 = 0.275, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{b\sin \left ( fx+e \right ) }{\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))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sin \left (f x + e\right )}}{\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))^(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., 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}}}{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))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sin{\left (e + f x \right )}}}{\sqrt [3]{d \tan{\left (e + f x \right )}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sin \left (f x + e\right )}}{\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))^(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)