3.3.31 \(\int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx\) [231]

Optimal. Leaf size=114 \[ \frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \]

[Out]

14/45*sin(d*x+c)/a/d/e/(e*sec(d*x+c))^(3/2)+14/15*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si
n(1/2*d*x+1/2*c),2^(1/2))/a/d/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+2/9*I/d/(e*sec(d*x+c))^(5/2)/(a+I*a*ta
n(d*x+c))

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Rubi [A]
time = 0.08, antiderivative size = 114, 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 = {3583, 3854, 3856, 2719} \begin {gather*} \frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(15*a*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*Sin[c + d*x])/(45*a*
d*e*(e*Sec[c + d*x])^(3/2)) + ((2*I)/9)/(d*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]))

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 3583

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

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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 \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx &=\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{9 a}\\ &=\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac {7 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{15 a e^2}\\ &=\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac {7 \int \sqrt {\cos (c+d x)} \, dx}{15 a e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.16, size = 134, normalized size = 1.18 \begin {gather*} \frac {\left (106+104 \cos (2 (c+d x))-2 \cos (4 (c+d x))-56 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+70 i \sin (2 (c+d x))-7 i \sin (4 (c+d x))\right ) (i+\tan (c+d x))}{180 a d e^2 \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((106 + 104*Cos[2*(c + d*x)] - 2*Cos[4*(c + d*x)] - 56*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + (70*I)*Sin[2*(c + d*x)] - (7*I)*Sin[4*(c + d*x)])*(I + Tan
[c + d*x]))/(180*a*d*e^2*Sqrt[e*Sec[c + d*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (124 ) = 248\).
time = 1.13, size = 376, normalized size = 3.30

method	result	size
default	\(-\frac {2 \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-5 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+5 \left (\cos ^{6}\left (d x +c \right )\right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+2 \left (\cos ^{4}\left (d x +c \right )\right )+14 \left (\cos ^{2}\left (d x +c \right )\right )-21 \cos \left (d x +c \right )\right )}{45 a d \,e^{5} \sin \left (d x +c \right )^{5}}\)	\(376\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45/a/d*cos(d*x+c)^2*(e/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(-5*I*cos(d*x+c)^5*sin(d*x+c)+5
*cos(d*x+c)^6-21*I*sin(d*x+c)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin
(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+2*cos(d*x+c)^4+14*cos(d*x+c)^2-21*cos(d*x+c))/
e^5/sin(d*x+c)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 120, normalized size = 1.05 \begin {gather*} \frac {{\left (336 i \, \sqrt {2} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (-9 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 174 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 212 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 34 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c - \frac {5}{2}\right )}}{360 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(336*I*sqrt(2)*e^(5*I*d*x + 5*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) +
sqrt(2)*(-9*I*e^(8*I*d*x + 8*I*c) + 174*I*e^(6*I*d*x + 6*I*c) + 212*I*e^(4*I*d*x + 4*I*c) + 34*I*e^(2*I*d*x +
2*I*c) + 5*I)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-5*I*d*x - 5*I*c - 5/2)/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )} - i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(1/((e*sec(c + d*x))**(5/2)*tan(c + d*x) - I*(e*sec(c + d*x))**(5/2)), x)/a

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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*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(e^(-5/2)/((I*a*tan(d*x + c) + a)*sec(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 (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,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)),x)

[Out]

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

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