3.3.43 \(\int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx\) [243]

Optimal. Leaf size=150 \[ \frac {42 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

2/13*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(7/2)+14/65*sin(d*x+c)/a^2/d/e/(e*sec(d*x+c))^(3/2)+42/65*(cos(1/2*d*x+
1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x
+c))^(1/2)+4/13*I*e^2/d/(e*sec(d*x+c))^(9/2)/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.09, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3854, 3856, 2719} \begin {gather*} \frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}+\frac {42 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(42*EllipticE[(c + d*x)/2, 2])/(65*a^2*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sin[c + d*x])/(13
*a^2*d*(e*Sec[c + d*x])^(7/2)) + (14*Sin[c + d*x])/(65*a^2*d*e*(e*Sec[c + d*x])^(3/2)) + (((4*I)/13)*e^2)/(d*(
e*Sec[c + d*x])^(9/2)*(a^2 + I*a^2*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 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 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))^2} \, dx &=\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (9 e^2\right ) \int \frac {1}{(e \sec (c+d x))^{9/2}} \, dx}{13 a^2}\\ &=\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{13 a^2}\\ &=\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {21 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{65 a^2 e^2}\\ &=\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {21 \int \sqrt {\cos (c+d x)} \, dx}{65 a^2 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {42 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.46, size = 149, normalized size = 0.99 \begin {gather*} \frac {(\cos (2 (c+d x))-i \sin (2 (c+d x))) \left (88 i+416 i \cos (2 (c+d x))-8 i \cos (4 (c+d x))-\frac {224 i e^{4 i (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-356 \sin (2 (c+d x))+18 \sin (4 (c+d x))\right )}{520 a^2 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])^2),x]

[Out]

((Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(88*I + (416*I)*Cos[2*(c + d*x)] - (8*I)*Cos[4*(c + d*x)] - ((224*I)*
E^((4*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - 35
6*Sin[2*(c + d*x)] + 18*Sin[4*(c + d*x)]))/(520*a^2*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. 385 vs. \(2 (156 ) = 312\).
time = 1.12, size = 386, normalized size = 2.57

method	result	size
default	\(-\frac {2 \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \left (-10 i \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+10 \left (\cos ^{8}\left (d x +c \right )\right )-5 \left (\cos ^{6}\left (d x +c \right )\right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+2 \left (\cos ^{4}\left (d x +c \right )\right )+14 \left (\cos ^{2}\left (d x +c \right )\right )-21 \cos \left (d x +c \right )\right )}{65 a^{2} d \,e^{5} \sin \left (d x +c \right )^{5}}\)	\(386\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

-2/65/a^2/d*(e/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*cos(d*x+c)^2*(-10*I*cos(d*x+c)^7*sin(d*x+c
)+10*cos(d*x+c)^8-5*cos(d*x+c)^6+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)*si
n(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-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)*(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)-21*I*sin(d*x+c)*EllipticF(I*(-1+co
s(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)+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))^2,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 = 131, normalized size = 0.87 \begin {gather*} \frac {{\left (672 i \, \sqrt {2} e^{\left (7 i \, d x + 7 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 (-13 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 373 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 474 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 118 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 35 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 (-7 i \, d x - 7 i \, c - \frac {5}{2}\right )}}{1040 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/1040*(672*I*sqrt(2)*e^(7*I*d*x + 7*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))
+ sqrt(2)*(-13*I*e^(10*I*d*x + 10*I*c) + 373*I*e^(8*I*d*x + 8*I*c) + 474*I*e^(6*I*d*x + 6*I*c) + 118*I*e^(4*I*
d*x + 4*I*c) + 35*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^(-7*I*
d*x - 7*I*c - 5/2)/(a^2*d)

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

[Out]

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

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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))^2,x, algorithm="giac")

[Out]

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