3.7.69 \(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx\) [669]

Optimal. Leaf size=92 \[ -\frac {2 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d (e \cos (c+d x))^{5/2}}+\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

-2/3*cos(d*x+c)^(5/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^
2/d/(e*cos(d*x+c))^(5/2)+4/3*I*cos(d*x+c)^2/d/(e*cos(d*x+c))^(5/2)/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3596, 3581, 3856, 2720} \begin {gather*} -\frac {2 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d (e \cos (c+d x))^{5/2}}+\frac {4 i \cos ^2(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(-2*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*(e*Cos[c + d*x])^(5/2)) + (((4*I)/3)*Cos[c + d*x]^2
)/(d*(e*Cos[c + d*x])^(5/2)*(a^2 + I*a^2*Tan[c + d*x]))

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 3581

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*d^2*
(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {1}{(e \cos (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx &=\frac {\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {e^2 \int \sqrt {e \sec (c+d x)} \, dx}{3 a^2 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\cos ^{\frac {5}{2}}(c+d x) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2 (e \cos (c+d x))^{5/2}}\\ &=-\frac {2 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d (e \cos (c+d x))^{5/2}}+\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 116, normalized size = 1.26 \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} (\cos (d x)+i \sin (d x))^2 \left (F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 c)+i \sin (2 c))+2 \sqrt {\cos (c+d x)} (-i \cos (c-d x)+\sin (c-d x))\right )}{3 a^2 d (e \cos (c+d x))^{5/2} (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2*(EllipticF[(c + d*x)/2, 2]*(Cos[2*c] + I*Sin[2*c]) + 2*Sqrt[Co
s[c + d*x]]*((-I)*Cos[c - d*x] + Sin[c - d*x])))/(3*a^2*d*(e*Cos[c + d*x])^(5/2)*(-I + Tan[c + d*x])^2)

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Maple [A]
time = 1.26, size = 171, normalized size = 1.86

method	result	size
default	\(-\frac {2 \left (-8 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 e^{2} a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-2/3/e^2/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-8*I*sin(1/2*d*x+1/2*c)^5+8*sin(1/2*d*x+1
/2*c)^4*cos(1/2*d*x+1/2*c)+8*I*sin(1/2*d*x+1/2*c)^3-4*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*I*sin(1/2*d*x+1/2*c))/d

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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/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(-5/2)*integrate(1/((I*a*tan(d*x + c) + a)^2*cos(d*x + c)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 72, normalized size = 0.78 \begin {gather*} -\frac {2 \, {\left (-i \, \sqrt {2} e^{\left (i \, d x + i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) - 2 i \, \sqrt {\frac {1}{2}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c - \frac {5}{2}\right )}}{3 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(-I*sqrt(2)*e^(I*d*x + I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) - 2*I*sqrt(1/2)*sqrt(e^(2*I*d*x +
2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c))*e^(-I*d*x - I*c - 5/2)/(a^2*d)

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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/(e*cos(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 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/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(e^(-5/2)/((I*a*tan(d*x + c) + a)^2*cos(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

int(1/((e*cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^2), x)

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