3.314 \(\int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {4 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{15 d^4 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}} \]

[Out]

-4/15*b^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)/d^4/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)-2/9*b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*
x+e))^(9/2)+2/15*b*(b*tan(f*x+e))^(3/2)/d^2/f/(d*sec(f*x+e))^(5/2)

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Rubi [A]  time = 0.18, 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 = {2610, 2612, 2616, 2640, 2639} \[ \frac {4 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{15 d^4 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

(4*b^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(15*d^4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]
]) - (2*b*(b*Tan[e + f*x])^(3/2))/(9*f*(d*Sec[e + f*x])^(9/2)) + (2*b*(b*Tan[e + f*x])^(3/2))/(15*d^2*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 2612

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] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[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 \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx &=-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {b^2 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx}{3 d^2}\\ &=-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}+\frac {\left (2 b^2\right ) \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{15 d^4}\\ &=-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}+\frac {\left (2 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{15 d^4 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}+\frac {\left (2 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{15 d^4 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ &=\frac {4 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)}}{15 d^4 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.99, size = 99, normalized size = 0.76 \[ -\frac {b^2 \sin (2 (e+f x)) \sqrt {b \tan (e+f x)} \left (12 \sqrt [4]{-\tan ^2(e+f x)} \csc ^2(e+f x) \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right )+5 \cos (2 (e+f x))-1\right )}{90 d^4 f \sqrt {d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

-1/90*(b^2*Sin[2*(e + f*x)]*Sqrt[b*Tan[e + f*x]]*(-1 + 5*Cos[2*(e + f*x)] + 12*Csc[e + f*x]^2*Hypergeometric2F
1[-1/4, 1/4, 3/4, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(1/4)))/(d^4*f*Sqrt[d*Sec[e + f*x]])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} b^{2} \tan \left (f x + e\right )^{2}}{d^{5} \sec \left (f x + e\right )^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(9/2),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^5*sec(f*x + e)^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(9/2), x)

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maple [C]  time = 0.65, size = 586, normalized size = 4.47 \[ \frac {\left (5 \sqrt {2}\, \left (\cos ^{5}\left (f x +e \right )\right )+6 \cos \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 \cos \left (f x +e \right )-i-\sin \left (f x +e \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 ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}-12 \cos \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 \cos \left (f x +e \right )-i-\sin \left (f x +e \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 ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}-8 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+6 \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\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 \cos \left (f x +e \right )-i-\sin \left (f x +e \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 )-12 \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\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 \cos \left (f x +e \right )-i-\sin \left (f x +e \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 \cos \left (f x +e \right ) \sqrt {2}+6 \sqrt {2}\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {2}}{45 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{3} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(9/2),x)

[Out]

1/45/f*(5*2^(1/2)*cos(f*x+e)^5+6*cos(f*x+e)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-s
in(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))*(-I*(-1+cos
(f*x+e))/sin(f*x+e))^(1/2)-12*cos(f*x+e)*((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*(-1+cos(f*
x+e))/sin(f*x+e))^(1/2)-8*cos(f*x+e)^3*2^(1/2)+6*(-I*(-1+cos(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)-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))-12*(-I*(-1+cos(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)-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))-3*cos(f*x+e)*2^(1/2)+6*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(5/2)/cos(f*x+e)^2/sin(f*x+e)^3/(d
/cos(f*x+e))^(9/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(9/2),x)

[Out]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(9/2), 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((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(9/2),x)

[Out]

Timed out

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