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

Optimal. Leaf size=64 \[ \frac{2 \cos ^2(e+f x)^{7/12} (d \tan (e+f x))^{3/2} \, _2F_1\left (\frac{7}{12},\frac{3}{4};\frac{7}{4};\sin ^2(e+f x)\right )}{3 d f \sqrt [3]{b \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0484381, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2617} \[ \frac{2 \cos ^2(e+f x)^{7/12} (d \tan (e+f x))^{3/2} \, _2F_1\left (\frac{7}{12},\frac{3}{4};\frac{7}{4};\sin ^2(e+f x)\right )}{3 d f \sqrt [3]{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

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

Mathematica [A]  time = 0.0929996, size = 62, normalized size = 0.97 \[ -\frac{3 d \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{6},\frac{1}{4};\frac{5}{6};\sec ^2(e+f x)\right )}{f \sqrt [3]{b \sec (e+f x)} \sqrt{d \tan (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*d*Hypergeometric2F1[-1/6, 1/4, 5/6, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(1/4))/(f*(b*Sec[e + f*x])^(1/3)*Sqr
t[d*Tan[e + f*x]])

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Maple [F]  time = 0.241, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [3]{b\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \tan \left (f x + e\right )}}{\left (b \sec \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((d*tan(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))/(b*sec(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{\left (b \sec \left (f x + e\right )\right )^{\frac{2}{3}} \sqrt{d \tan \left (f x + e\right )}}{b \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \tan \left (f x + e\right )}}{\left (b \sec \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((d*tan(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/3),x, algorithm="giac")

[Out]

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