3.4.8 \(\int (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2} \, dx\) [308]

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

[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 = 0.12, 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 = {2691, 2693, 2696, 2721, 2719} \begin {gather*} \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} \end {gather*}

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 2691

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 2693

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 2696

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 2719

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 12.57, size = 93, normalized size = 0.71 \begin {gather*} \frac {b^3 d^2 \left (3-5 \sec ^2(e+f x)+2 \sec ^4(e+f x)-3 \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{6 f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.42, size = 581, normalized size = 4.44

method	result	size
default	\(\frac {\left (3 \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 )+\sin \left (f x +e \right )+i}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\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 ) \left (\cos ^{4}\left (f x +e \right )\right )-6 \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 )+\sin \left (f x +e \right )+i}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\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 ) \left (\cos ^{4}\left (f x +e \right )\right )+3 \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 )+\sin \left (f x +e \right )+i}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\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 ) \left (\cos ^{3}\left (f x +e \right )\right )-6 \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 )+\sin \left (f x +e \right )+i}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\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 ) \left (\cos ^{3}\left (f x +e \right )\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}}\)	\(581\)

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,method=_RETURNVERBOSE)

[Out]

1/12/f*(3*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*((-I*cos(f*x+e)+sin(f*x+e)+I)/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))*cos(f*x+
e)^4-6*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*((-I*cos(f*x+e)+sin(f*x+e)+I)/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))*cos(f*x+e)^
4+3*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*((-I*cos(f*x+e)+sin(f*x+e)+I)/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))*cos(f*x+e)^3-6
*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*((-I*cos(f*x+e)+sin(f*x+e)+I)/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))*cos(f*x+e)^3+3*2^
(1/2)*cos(f*x+e)^3-5*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 167, normalized size = 1.27 \begin {gather*} \frac {3 i \, \sqrt {-2 i \, b d} b^{2} d \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {2 i \, b d} b^{2} d \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (3 \, b^{2} d \cos \left (f x + e\right )^{2} - 2 \, b^{2} d\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 )}} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{2}} \end {gather*}

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]

1/12*(3*I*sqrt(-2*I*b*d)*b^2*d*cos(f*x + e)^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I
*sin(f*x + e))) - 3*I*sqrt(2*I*b*d)*b^2*d*cos(f*x + e)^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f
*x + e) - I*sin(f*x + e))) - 2*(3*b^2*d*cos(f*x + e)^2 - 2*b^2*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos
(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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