3.3.55 \(\int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx\) [255]

Optimal. Leaf size=192 \[ \frac {154 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

-154/15*e^5*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/a^4/d+154/5*e^8*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(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)-154/5*e^7*sin(d*x+c)*(e*sec(d*
x+c))^(1/2)/a^4/d+4*I*e^2*(e*sec(d*x+c))^(11/2)/a/d/(a+I*a*tan(d*x+c))^3+44/3*I*e^4*(e*sec(d*x+c))^(7/2)/d/(a^
4+I*a^4*tan(d*x+c))

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Rubi [A]
time = 0.13, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3853, 3856, 2719} \begin {gather*} \frac {154 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 e^7 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 a^4 d}-\frac {154 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^4 d}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(154*e^8*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (154*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(5*a^4*d) - (154*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(15*a^4*d) + ((4*I)*e^2*(e*Sec
[c + d*x])^(11/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + (((44*I)/3)*e^4*(e*Sec[c + d*x])^(7/2))/(d*(a^4 + I*a^4*Ta
n[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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {\left (11 e^2\right ) \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (77 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^4}\\ &=-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (77 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^4}\\ &=-\frac {154 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (77 e^8\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^4}\\ &=-\frac {154 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (77 e^8\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {154 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \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 = 1.59, size = 124, normalized size = 0.65 \begin {gather*} -\frac {i e^5 (e \sec (c+d x))^{5/2} \left (-1133 \cos (c+d x)+77 e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 (117 \cos (3 (c+d x))+33 i \sin (c+d x)+37 i \sin (3 (c+d x)))\right )}{30 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-1/30*I)*e^5*(e*Sec[c + d*x])^(5/2)*(-1133*Cos[c + d*x] + (77*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric
2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) - 3*(117*Cos[3*(c + d*x)] + (33*I)*Sin[c + d*x] + (3
7*I)*Sin[3*(c + d*x)])))/(a^4*d)

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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. 400 vs. \(2 (194 ) = 388\).
time = 0.88, size = 401, normalized size = 2.09

method	result	size
default	\(-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {15}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-231 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+231 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+231 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-120 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+120 \left (\cos ^{4}\left (d x +c \right )\right )-231 \left (\cos ^{3}\left (d x +c \right )\right )-20 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+114 \left (\cos ^{2}\left (d x +c \right )\right )-3\right )}{15 a^{4} d \sin \left (d x +c \right )^{5}}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-2/15/a^4/d*cos(d*x+c)^5*(e/cos(d*x+c))^(15/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(-231*I*(1/(1+cos(d*x+c)))^(
1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+231*I
*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c)
)/sin(d*x+c),I)-231*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*Ellip
ticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+231*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+
c)^2*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-120*I*cos(d*x+c)^3*sin(d*x+c)+120*cos(d*x+c)^4-231*c
os(d*x+c)^3-20*I*cos(d*x+c)*sin(d*x+c)+114*cos(d*x+c)^2-3)/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((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 179, normalized size = 0.93 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-120 i \, e^{\frac {15}{2}} - 231 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {15}{2}\right )} - 616 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {15}{2}\right )} - 517 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {15}{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}} + 231 \, {\left (-i \, \sqrt {2} e^{\left (5 i \, d x + 5 i \, c + \frac {15}{2}\right )} - 2 i \, \sqrt {2} e^{\left (3 i \, d x + 3 i \, c + \frac {15}{2}\right )} - i \, \sqrt {2} e^{\left (i \, d x + i \, c + \frac {15}{2}\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-2/15*(sqrt(2)*(-120*I*e^(15/2) - 231*I*e^(6*I*d*x + 6*I*c + 15/2) - 616*I*e^(4*I*d*x + 4*I*c + 15/2) - 517*I*
e^(2*I*d*x + 2*I*c + 15/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 231*(-I*sqrt(2)*e^(5*I*d*x
+ 5*I*c + 15/2) - 2*I*sqrt(2)*e^(3*I*d*x + 3*I*c + 15/2) - I*sqrt(2)*e^(I*d*x + I*c + 15/2))*weierstrassZeta(
-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^4*d*e^(3*I*d*x + 3*I*c)
+ a^4*d*e^(I*d*x + I*c))

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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((e*sec(d*x+c))**(15/2)/(a+I*a*tan(d*x+c))**4,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((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(e^(15/2)*sec(d*x + c)^(15/2)/(I*a*tan(d*x + c) + a)^4, x)

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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 )}^{15/2}}{{\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))^(15/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(15/2)/(a + a*tan(c + d*x)*1i)^4, x)

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