3.3.61 \(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx\) [261]

Optimal. Leaf size=163 \[ \frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

2/117*e^3*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(3/2)+2/39*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+4/13*I*e^2/a/d/(e*sec(d*x+c))^(1/
2)/(a+I*a*tan(d*x+c))^3+4/117*I*e^4/d/(e*sec(d*x+c))^(5/2)/(a^4+I*a^4*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 163, 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^4}{117 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(2*e^2*EllipticE[(c + d*x)/2, 2])/(39*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e^3*Sin[c + d*x])/(1
17*a^4*d*(e*Sec[c + d*x])^(3/2)) + (((4*I)/13)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (((4
*I)/117)*e^4)/(d*(e*Sec[c + d*x])^(5/2)*(a^4 + I*a^4*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 {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{13 a^2}\\ &=\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (5 e^4\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{117 a^4}\\ &=\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{39 a^4}\\ &=\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {e^2 \int \sqrt {\cos (c+d x)} \, dx}{39 a^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.69, size = 142, normalized size = 0.87 \begin {gather*} \frac {i e^{-i d x} \sec ^2(c+d x) (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x)) \left (28+40 \cos (2 (c+d x))+\frac {24 e^{4 i (c+d x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+22 i \sin (2 (c+d x))\right )}{234 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/234)*Sec[c + d*x]^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])*(28 + 40*Cos[2*(c + d*x)] + (24*E^((4*I
)*(c + d*x))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (22*I)*S
in[2*(c + d*x)]))/(a^4*d*E^(I*d*x)*(-I + Tan[c + d*x])^4)

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (167 ) = 334\).
time = 1.03, size = 398, normalized size = 2.44

method	result	size
default	\(\frac {2 \left (72 i \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-72 \left (\cos ^{8}\left (d x +c \right )\right )-52 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+88 \left (\cos ^{6}\left (d x +c \right )\right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-17 \left (\cos ^{4}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )}{117 a^{4} d \sin \left (d x +c \right )^{5}}\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

2/117/a^4/d*(72*I*sin(d*x+c)*cos(d*x+c)^7-72*cos(d*x+c)^8-52*I*sin(d*x+c)*cos(d*x+c)^5+3*I*(1/(1+cos(d*x+c)))^
(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-3*I*(1
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*
sin(d*x+c)+88*cos(d*x+c)^6+3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(
d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*Ell
ipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-17*cos(d*x+c)^4-2*cos(d*x+c)^2+3*cos(d*x+c))*(1+cos(d*x+c))^2*(-1+cos(d
*x+c))^2*(e/cos(d*x+c))^(3/2)*cos(d*x+c)/sin(d*x+c)^5

________________________________________________________________________________________

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((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 127, normalized size = 0.78 \begin {gather*} \frac {{\left (24 i \, \sqrt {2} e^{\left (7 i \, d x + 7 i \, c + \frac {3}{2}\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^{\frac {3}{2}} + 24 i \, e^{\left (8 i \, d x + 8 i \, c + \frac {3}{2}\right )} + 55 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {3}{2}\right )} + 59 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {3}{2}\right )} + 37 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\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\right )}}{468 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/468*(24*I*sqrt(2)*e^(7*I*d*x + 7*I*c + 3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c
))) + sqrt(2)*(9*I*e^(3/2) + 24*I*e^(8*I*d*x + 8*I*c + 3/2) + 55*I*e^(6*I*d*x + 6*I*c + 3/2) + 59*I*e^(4*I*d*x
+ 4*I*c + 3/2) + 37*I*e^(2*I*d*x + 2*I*c + 3/2))*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)/(a^4*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral((e*sec(c + d*x))**(3/2)/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x)
+ 1), x)/a**4

________________________________________________________________________________________

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((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(e^(3/2)*sec(d*x + c)^(3/2)/(I*a*tan(d*x + c) + a)^4, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i)^4, x)

________________________________________________________________________________________