3.3.33 \(\int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx\) [233]

Optimal. Leaf size=183 \[ -\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

22/45*e^5*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/a^2/d+22/63*e^3*(e*sec(d*x+c))^(9/2)*sin(d*x+c)/a^2/d-22/15*e^8*(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/cos(d*x+c)^(1/2)/(e*se
c(d*x+c))^(1/2)+22/15*e^7*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/a^2/d-4/7*I*e^2*(e*sec(d*x+c))^(11/2)/d/(a^2+I*a^2*t
an(d*x+c))

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Rubi [A]
time = 0.11, antiderivative size = 183, 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 {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sin (c+d x) \sqrt {e \sec (c+d x)}}{15 a^2 d}+\frac {22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{45 a^2 d}+\frac {22 e^3 \sin (c+d x) (e \sec (c+d x))^{9/2}}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*e^8*EllipticE[(c + d*x)/2, 2])/(15*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (22*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(15*a^2*d) + (22*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(45*a^2*d) + (22*e^3*(e*Sec[c
+ d*x])^(9/2)*Sin[c + d*x])/(63*a^2*d) - (((4*I)/7)*e^2*(e*Sec[c + d*x])^(11/2))/(d*(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 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))^2} \, dx &=-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^2\right ) \int (e \sec (c+d x))^{11/2} \, dx}{7 a^2}\\ &=\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{9 a^2}\\ &=\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{15 a^2}\\ &=\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (11 e^8\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (11 e^8\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \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 = 2.73, size = 302, normalized size = 1.65 \begin {gather*} \frac {(e \sec (c+d x))^{15/2} (\cos (d x)+i \sin (d x))^2 \left (\frac {22 i \sqrt {2} e^{3 i c-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {1}{56} \csc (c) \sec ^{\frac {9}{2}}(c+d x) (\cos (2 c)+i \sin (2 c)) (1260 \cos (d x)+1050 \cos (2 c+d x)+1078 \cos (2 c+3 d x)+77 \cos (4 c+3 d x)+231 \cos (4 c+5 d x)+720 i \sin (d x)-720 i \sin (2 c+d x))\right )}{45 d \sec ^{\frac {11}{2}}(c+d x) (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sec[c + d*x])^(15/2)*(Cos[d*x] + I*Sin[d*x])^2*(((22*I)*Sqrt[2]*E^((3*I)*c - I*d*x)*Sqrt[E^(I*(c + d*x))/(
1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1
+ E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(-1 + E^((2*I)*c)) + (Csc[c]*Sec[c + d
*x]^(9/2)*(Cos[2*c] + I*Sin[2*c])*(1260*Cos[d*x] + 1050*Cos[2*c + d*x] + 1078*Cos[2*c + 3*d*x] + 77*Cos[4*c +
3*d*x] + 231*Cos[4*c + 5*d*x] + (720*I)*Sin[d*x] - (720*I)*Sin[2*c + d*x]))/56))/(45*d*Sec[c + d*x]^(11/2)*(a
+ I*a*Tan[c + d*x])^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. 383 vs. \(2 (185 ) = 370\).
time = 0.73, size = 384, normalized size = 2.10

method	result	size
default	\(\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{5}\left (d x +c \right )\right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 \left (\cos ^{5}\left (d x +c \right )\right )+154 \left (\cos ^{4}\left (d x +c \right )\right )-90 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+112 \left (\cos ^{2}\left (d x +c \right )\right )-35\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {15}{2}} \left (\cos ^{3}\left (d x +c \right )\right )}{315 a^{2} d \sin \left (d x +c \right )^{5}}\)	\(384\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

2/315/a^2/d*(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)^5*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-231*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)^5*sin(d*x+c)+231*I*cos(d*x+c)^4*(
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)*sin(d*x+c)
-231*I*cos(d*x+c)^4*(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)-231*cos(d*x+c)^5+154*cos(d*x+c)^4-90*I*cos(d*x+c)*sin(d*x+c)+112*cos(d*x+c)^2-35)*(e/cos
(d*x+c))^(15/2)*cos(d*x+c)^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))^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 = 0.13, size = 242, normalized size = 1.32 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (231 i \, e^{\left (9 i \, d x + 9 i \, c + \frac {15}{2}\right )} + 1078 i \, e^{\left (7 i \, d x + 7 i \, c + \frac {15}{2}\right )} + 1980 i \, e^{\left (5 i \, d x + 5 i \, c + \frac {15}{2}\right )} + 1770 i \, e^{\left (3 i \, d x + 3 i \, c + \frac {15}{2}\right )} + 77 i \, e^{\left (i \, d x + 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^{\frac {15}{2}} + i \, \sqrt {2} e^{\left (8 i \, d x + 8 i \, c + \frac {15}{2}\right )} + 4 i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c + \frac {15}{2}\right )} + 6 i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c + \frac {15}{2}\right )} + 4 i \, \sqrt {2} e^{\left (2 i \, d x + 2 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 )}}{315 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\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))^2,x, algorithm="fricas")

[Out]

-2/315*(sqrt(2)*(231*I*e^(9*I*d*x + 9*I*c + 15/2) + 1078*I*e^(7*I*d*x + 7*I*c + 15/2) + 1980*I*e^(5*I*d*x + 5*
I*c + 15/2) + 1770*I*e^(3*I*d*x + 3*I*c + 15/2) + 77*I*e^(I*d*x + 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^(15/2) + I*sqrt(2)*e^(8*I*d*x + 8*I*c + 15/2) + 4*I*sqrt(2)*e^(6*I*d
*x + 6*I*c + 15/2) + 6*I*sqrt(2)*e^(4*I*d*x + 4*I*c + 15/2) + 4*I*sqrt(2)*e^(2*I*d*x + 2*I*c + 15/2))*weierstr
assZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(a^2*d*e^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x +
6*I*c) + 6*a^2*d*e^(4*I*d*x + 4*I*c) + 4*a^2*d*e^(2*I*d*x + 2*I*c) + a^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((e*sec(d*x+c))**(15/2)/(a+I*a*tan(d*x+c))**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((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

[Out]

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

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