3.308 \(\int (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=131 \[ \frac{b^2 d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.173201, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2611, 2613, 2616, 2640, 2639} \[ \frac{b^2 d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(5/2),x]

[Out]

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

Rule 2611

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 + n - 1)), x] - Dist[(b^2*(n - 1))/(m + 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] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && 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 (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2} \, dx &=\frac{b (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}{3 f}-\frac{1}{2} b^2 \int (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)} \, dx\\ &=-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}{3 f}+\frac{1}{4} \left (b^2 d^2\right ) \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{d \sec (e+f x)}} \, dx\\ &=-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}{3 f}+\frac{\left (b^2 d^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{b \sin (e+f x)} \, dx}{4 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}{3 f}+\frac{\left (b^2 d^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\sin (e+f x)} \, dx}{4 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}\\ &=\frac{b^2 d^2 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{2 f \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}-\frac{b d^2 (b \tan (e+f x))^{3/2}}{2 f \sqrt{d \sec (e+f x)}}+\frac{b (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [C]  time = 2.33937, size = 93, normalized size = 0.71 \[ \frac{b^3 d^2 \left (-3 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )+2 \sec ^4(e+f x)-5 \sec ^2(e+f x)+3\right )}{6 f \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(5/2),x]

[Out]

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

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Maple [C]  time = 0.179, size = 593, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/f*2^(1/2)*(3*cos(f*x+e)^4*((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))/si
n(f*x+e))^(1/2)*(-I*(cos(f*x+e)-1)/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))-6*cos(f*x+e)^4*((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)*(-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))+3*cos(f*x+e)^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)*(-I*(cos(f*x+e)-1)/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))-6*cos(f*x+e)^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)*(-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))+3*cos(f*x+e)^3*2^(1/2)-5*cos(f*x+e)^2*2^(1/2)+2*2^(1/2))*cos(f*x+e)*(d/cos(f*x+e))^(3/2)*(b*sin(f*x+e)/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 \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \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((d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))*b^2*d*sec(f*x + e)*tan(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*sec(f*x+e))**(3/2)*(b*tan(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 \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \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((d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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