3.4.50 \(\int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx\) [350]

Optimal. Leaf size=62 \[ -\frac {3 \cos ^2(e+f x)^{7/12} \, _2F_1\left (-\frac {1}{6},\frac {7}{12};\frac {5}{6};\sin ^2(e+f x)\right ) (b \sec (e+f x))^{3/2}}{d f \sqrt [3]{d \tan (e+f x)}} \]

[Out]

-3*(cos(f*x+e)^2)^(7/12)*hypergeom([-1/6, 7/12],[5/6],sin(f*x+e)^2)*(b*sec(f*x+e))^(3/2)/d/f/(d*tan(f*x+e))^(1
/3)

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Rubi [A]
time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2697} \begin {gather*} -\frac {3 \cos ^2(e+f x)^{7/12} (b \sec (e+f x))^{3/2} \, _2F_1\left (-\frac {1}{6},\frac {7}{12};\frac {5}{6};\sin ^2(e+f x)\right )}{d f \sqrt [3]{d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2697

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)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx &=-\frac {3 \cos ^2(e+f x)^{7/12} \, _2F_1\left (-\frac {1}{6},\frac {7}{12};\frac {5}{6};\sin ^2(e+f x)\right ) (b \sec (e+f x))^{3/2}}{d f \sqrt [3]{d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 64, normalized size = 1.03 \begin {gather*} \frac {2 d \, _2F_1\left (\frac {3}{4},\frac {7}{6};\frac {7}{4};\sec ^2(e+f x)\right ) (b \sec (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{7/6}}{3 f (d \tan (e+f x))^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Hypergeometric2F1[3/4, 7/6, 7/4, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(3/2)*(-Tan[e + f*x]^2)^(7/6))/(3*f*(d*
Tan[e + f*x])^(7/3))

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {\left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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