3.4.33 \(\int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx\) [333]

Optimal. Leaf size=132 \[ -\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {8 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 d^2 f \sqrt {b \tan (e+f x)}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}} \]

[Out]

8/3*(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)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))
*(d*sec(f*x+e))^(1/2)*sin(f*x+e)^(1/2)/b^2/d^2/f/(b*tan(f*x+e))^(1/2)-4/3*(b*tan(f*x+e))^(1/2)/b^3/f/(d*sec(f*
x+e))^(3/2)-2/3/b/f/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.12, antiderivative size = 132, 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 = {2689, 2692, 2696, 2721, 2720} \begin {gather*} -\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac {8 \sqrt {\sin (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {d \sec (e+f x)}}{3 b^2 d^2 f \sqrt {b \tan (e+f x)}}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(5/2)),x]

[Out]

-2/(3*b*f*(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)) - (8*EllipticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[d*Sec[e +
f*x]]*Sqrt[Sin[e + f*x]])/(3*b^2*d^2*f*Sqrt[b*Tan[e + f*x]]) - (4*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*(d*Sec[e + f*
x])^(3/2))

Rule 2689

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*(n + 1))), x] - Dist[(m + n + 1)/(b^2*(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] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rule 2692

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*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx &=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx}{b^2}\\ &=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac {4 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 b^2 d^2}\\ &=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac {\left (4 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{3 b^2 d^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac {\left (4 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{3 b^2 d^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {8 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 d^2 f \sqrt {b \tan (e+f x)}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.55, size = 112, normalized size = 0.85 \begin {gather*} \frac {\csc ^2(e+f x) \sqrt {b \tan (e+f x)} \left (-\tan ^2(e+f x)\right )^{3/4} \left (-8 \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\sec ^2(e+f x)\right )+\left (-1+\cos (2 (e+f x))+2 \csc ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{3 b^3 f (d \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(5/2)),x]

[Out]

(Csc[e + f*x]^2*Sqrt[b*Tan[e + f*x]]*(-Tan[e + f*x]^2)^(3/4)*(-8*Hypergeometric2F1[1/4, 3/4, 5/4, Sec[e + f*x]
^2] + (-1 + Cos[2*(e + f*x)] + 2*Csc[e + f*x]^2)*(-Tan[e + f*x]^2)^(1/4)))/(3*b^3*f*(d*Sec[e + f*x])^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 336, normalized size = 2.55

method	result	size
default	\(-\frac {\left (4 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+4 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+2 \cos \left (f x +e \right ) \sqrt {2}\right ) \sin \left (f x +e \right ) \sqrt {2}}{3 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4}}\)	\(336\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*(4*I*(-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))*sin(f*
x+e)*cos(f*x+e)+4*I*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*sin(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)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),
1/2*2^(1/2))-2^(1/2)*cos(f*x+e)^3+2*cos(f*x+e)*2^(1/2))*sin(f*x+e)/(d/cos(f*x+e))^(3/2)/(b*sin(f*x+e)/cos(f*x+
e))^(5/2)/cos(f*x+e)^4*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(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((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 = 0.12, size = 165, normalized size = 1.25 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {-2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 \, \sqrt {2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2}\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 )}}\right )}}{3 \, {\left (b^{3} d^{2} f \cos \left (f x + e\right )^{2} - b^{3} d^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(3/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4372 deep

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

[Out]

integrate(1/((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 \frac {1}{{\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(1/((b*tan(e + f*x))^(5/2)*(d/cos(e + f*x))^(3/2)),x)

[Out]

int(1/((b*tan(e + f*x))^(5/2)*(d/cos(e + f*x))^(3/2)), x)

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