3.3.11 \(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{13/2}} \, dx\) [211]

Optimal. Leaf size=155 \[ \frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}} \]

[Out]

14/117*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(3/2)+14/39*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^6/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-2/13*I*(a+I*a*tan(d*x+c))^3/d/(
e*sec(d*x+c))^(13/2)-28/117*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(9/2)

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Rubi [A]
time = 0.11, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3578, 3577, 3854, 3856, 2719} \begin {gather*} \frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(13/2),x]

[Out]

(14*a^3*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*a^3*Sin[c + d*x])/
(117*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((2*I)/13)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((28
*I)/117)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(9/2))

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 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[a*((m + n)/(m*d^2)), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && 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 {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{13/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}+\frac {(7 a) \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx}{13 e^2}\\ &=-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {\left (35 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{117 e^4}\\ &=\frac {14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {\left (7 a^3\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{39 e^6}\\ &=\frac {14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {\left (7 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{39 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.50, size = 155, normalized size = 1.00 \begin {gather*} -\frac {a^3 e^{-4 i (c+d x)} \left (-117-34 e^{2 i (c+d x)}+124 e^{4 i (c+d x)}+50 e^{6 i (c+d x)}+9 e^{8 i (c+d x)}+112 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 )\right ) (-i+\tan (c+d x))^3}{936 d e^4 (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(13/2),x]

[Out]

-1/936*(a^3*(-117 - 34*E^((2*I)*(c + d*x)) + 124*E^((4*I)*(c + d*x)) + 50*E^((6*I)*(c + d*x)) + 9*E^((8*I)*(c
+ d*x)) + 112*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c
+ d*x))])*(-I + Tan[c + d*x])^3)/(d*e^4*E^((4*I)*(c + d*x))*(e*Sec[c + d*x])^(5/2))

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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. 379 vs. \(2 (159 ) = 318\).
time = 0.68, size = 380, normalized size = 2.45

method	result	size
risch	\(-\frac {i \left (9 \,{\mathrm e}^{6 i \left (d x +c \right )}+41 \,{\mathrm e}^{4 i \left (d x +c \right )}+83 \,{\mathrm e}^{2 i \left (d x +c \right )}+219\right ) a^{3} \sqrt {2}}{936 d \,e^{6} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {7 i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{3} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{39 d \,e^{6} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(352\)
default	\(-\frac {2 a^{3} \left (36 i \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+36 \left (\cos ^{8}\left (d x +c \right )\right )-13 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-31 \left (\cos ^{6}\left (d x +c \right )\right )+21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-21 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+2 \left (\cos ^{4}\left (d x +c \right )\right )+14 \left (\cos ^{2}\left (d x +c \right )\right )-21 \cos \left (d x +c \right )\right )}{117 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{7} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {13}{2}}}\)	\(380\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x,method=_RETURNVERBOSE)

[Out]

-2/117*a^3/d*(36*I*sin(d*x+c)*cos(d*x+c)^7+36*cos(d*x+c)^8-13*I*sin(d*x+c)*cos(d*x+c)^5+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)*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)-31*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)*sin(d*x+c)*Ellip
ticE(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)+2*cos(d*x+c)^4+14*cos(d*x+c)^2-21*cos(d*x+c))/sin(d*x+c)/cos(d*x+c
)^7/(e/cos(d*x+c))^(13/2)

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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((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="maxima")

[Out]

e^(-13/2)*integrate((I*a*tan(d*x + c) + a)^3/sec(d*x + c)^(13/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 136, normalized size = 0.88 \begin {gather*} \frac {{\left (336 i \, \sqrt {2} a^{3} e^{\left (i \, d x + 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 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 50 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 124 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 34 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 117 i \, a^{3}\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 (-i \, d x - i \, c - \frac {13}{2}\right )}}{936 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="fricas")

[Out]

1/936*(336*I*sqrt(2)*a^3*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) +
sqrt(2)*(-9*I*a^3*e^(8*I*d*x + 8*I*c) - 50*I*a^3*e^(6*I*d*x + 6*I*c) - 124*I*a^3*e^(4*I*d*x + 4*I*c) + 34*I*a
^3*e^(2*I*d*x + 2*I*c) + 117*I*a^3)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-I*d*x - I*c - 1
3/2)/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(13/2),x)

[Out]

Timed out

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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((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*e^(-13/2)/sec(d*x + c)^(13/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(13/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(13/2), x)

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