3.342 \(\int \frac{(d \tan (e+f x))^{3/2}}{(b \sec (e+f x))^{4/3}} \, dx\)

Optimal. Leaf size=64 \[ \frac{2 \cos ^2(e+f x)^{7/12} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{9}{4};\sin ^2(e+f x)\right )}{5 d f (b \sec (e+f x))^{4/3}} \]

[Out]

(2*(Cos[e + f*x]^2)^(7/12)*Hypergeometric2F1[7/12, 5/4, 9/4, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(5*d*f*(b
*Sec[e + f*x])^(4/3))

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Rubi [A]  time = 0.0597472, 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} \[ \frac{2 \cos ^2(e+f x)^{7/12} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{9}{4};\sin ^2(e+f x)\right )}{5 d f (b \sec (e+f x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(7/12)*Hypergeometric2F1[7/12, 5/4, 9/4, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(5*d*f*(b
*Sec[e + f*x])^(4/3))

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{(d \tan (e+f x))^{3/2}}{(b \sec (e+f x))^{4/3}} \, dx &=\frac{2 \cos ^2(e+f x)^{7/12} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{9}{4};\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{5 d f (b \sec (e+f x))^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.0768728, size = 71, normalized size = 1.11 \[ \frac{3 \left (-\tan ^2(e+f x)\right )^{3/4} \cot ^3(e+f x) (d \tan (e+f x))^{3/2} \, _2F_1\left (-\frac{2}{3},-\frac{1}{4};\frac{1}{3};\sec ^2(e+f x)\right )}{4 f (b \sec (e+f x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.186, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( b\sec \left ( fx+e \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*tan(f*x + e))^(3/2)/(b*sec(f*x + e))^(4/3), 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 )} d \tan \left (f x + e\right )}{b^{2} \sec \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(3/2)/(b*sec(f*x+e))**(4/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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