3.326 \(\int \frac{1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2}{3 b f \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2}}-\frac{8 \sqrt{d \sec (e+f x)}}{3 b d^2 f \sqrt{b \tan (e+f x)}} \]

[Out]

2/(3*b*f*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]) - (8*Sqrt[d*Sec[e + f*x]])/(3*b*d^2*f*Sqrt[b*Tan[e + f*x
]])

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Rubi [A]  time = 0.108494, antiderivative size = 67, normalized size of antiderivative = 0.93, 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 = {2609, 2605} \[ -\frac{8 (b \tan (e+f x))^{3/2}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{2}{b f \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

-2/(b*f*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]) - (8*(b*Tan[e + f*x])^(3/2))/(3*b^3*f*(d*Sec[e + f*x])^(3
/2))

Rule 2609

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))/(b*f*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rule 2605

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))/(b*f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx &=-\frac{2}{b f (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}-\frac{4 \int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{3/2}} \, dx}{b^2}\\ &=-\frac{2}{b f (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}-\frac{8 (b \tan (e+f x))^{3/2}}{3 b^3 f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.176048, size = 52, normalized size = 0.72 \[ \frac{(\cos (2 (e+f x))-7) \sec ^2(e+f x)}{3 b f \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

((-7 + Cos[2*(e + f*x)])*Sec[e + f*x]^2)/(3*b*f*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])

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Maple [A]  time = 0.147, size = 60, normalized size = 0.8 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) \left ( -4+ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/f*sin(f*x+e)*(-4+cos(f*x+e)^2)/(d/cos(f*x+e))^(3/2)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/cos(f*x+e)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67619, size = 161, normalized size = 2.24 \begin{align*} \frac{2 \,{\left (\cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{d}{\cos \left (f x + e\right )}}}{3 \, b^{2} d^{2} f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(cos(f*x + e)^3 - 4*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b^2*d^2*f*sin(f*
x + e))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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