3.312 \(\int \frac{(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{6 b^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 d^2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}} \]

[Out]

(6*b^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]
) - (2*b*(b*Tan[e + f*x])^(3/2))/(5*f*(d*Sec[e + f*x])^(5/2))

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Rubi [A]  time = 0.120262, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2610, 2616, 2640, 2639} \[ \frac{6 b^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 d^2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(5/2),x]

[Out]

(6*b^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]
) - (2*b*(b*Tan[e + f*x])^(3/2))/(5*f*(d*Sec[e + f*x])^(5/2))

Rule 2610

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(n - 1))/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

Rule 2616

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx &=-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac{\left (3 b^2\right ) \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac{\left (3 b^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{b \sin (e+f x)} \, dx}{5 d^2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac{\left (3 b^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\sin (e+f x)} \, dx}{5 d^2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}\\ &=\frac{6 b^2 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 d^2 f \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}-\frac{2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.738884, size = 81, normalized size = 0.84 \[ \frac{b^3 \left (-6 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-1\right )}{5 d^2 f \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(5/2),x]

[Out]

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

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Maple [C]  time = 0.213, size = 564, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(5/2),x)

[Out]

1/5/f*2^(1/2)*(-6*cos(f*x+e)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f
*x+e))^(1/2),1/2*2^(1/2))*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x
+e))^(1/2)+3*cos(f*x+e)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-
(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(
1/2))+cos(f*x+e)^3*2^(1/2)-6*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(
f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-I*(cos(f*x+e)-1)/sin(f*x
+e))^(1/2)+3*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*El
lipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)-4*cos
(f*x+e)*2^(1/2)+3*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(5/2)/(d/cos(f*x+e))^(5/2)/sin(f*x+e)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))*b^2*tan(f*x + e)^2/(d^3*sec(f*x + e)^3), 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((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(5/2), x)