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]

-1/2*b^2*d^2*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x)
,2^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)+1/3*b*(d*sec(f*x+e))^(3/2)*(b*tan(f*x+e
))^(3/2)/f-1/2*b*d^2*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [C]  time = 2.35, size = 93, normalized size = 0.71 \[ \frac {b^3 d^2 \left (-3 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right )+2 \sec ^4(e+f x)-5 \sec ^2(e+f x)+3\right )}{6 f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} b^{2} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2}, x\right ) \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [C]  time = 0.59, size = 593, normalized size = 4.53 \[ -\frac {\left (6 \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-3 \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+6 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-3 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-3 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+5 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-2 \sqrt {2}\right ) \cos \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {2}}{12 f \sin \left (f x +e \right )^{3}} \]

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*(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*(-1+cos(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))-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))/sin(f*x+e))
^(1/2)*(-I*(-1+cos(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))+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*(-1+cos(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
))-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*(-1+cos(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))
-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*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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