\(\int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx\) [201]

Optimal result

Integrand size = 28, antiderivative size = 202 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d} \]

[Out]

10/21*I*a^3*(e*sec(d*x+c))^(7/2)/d+2/3*a^3*e*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/d-2*a^3*e^4*(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))/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+2*a^3
*e^3*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d+2/11*I*a*(e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^2/d+10/33*I*(e*sec(d*x
+c))^(7/2)*(a^3+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3579, 3567, 3853, 3856, 2719} \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^3 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{3 d}+\frac {10 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}{33 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{7/2}}{11 d} \]

[In]

Int[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-2*a^3*e^4*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((10*I)/21)*a^3*(e*Sec[c
+ d*x])^(7/2))/d + (2*a^3*e^3*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d + (2*a^3*e*(e*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(3*d) + (((2*I)/11)*a*(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2)/d + (((10*I)/33)*(e*Sec[c + d*x])
^(7/2)*(a^3 + I*a^3*Tan[c + d*x]))/d

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 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3579

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

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*} \text {integral}& = \frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {1}{11} (15 a) \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\frac {1}{3} \left (5 a^2\right ) \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx \\ & = \frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\frac {1}{3} \left (5 a^3\right ) \int (e \sec (c+d x))^{7/2} \, dx \\ & = \frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\left (a^3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx \\ & = \frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}-\left (a^3 e^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx \\ & = \frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}-\frac {\left (a^3 e^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = -\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.84 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.80 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3 e^3 \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (-908 \cos (c+d x)-858 \cos (3 (c+d x))-154 \cos (5 (c+d x))+\frac {77}{2} e^{-5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{11/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-38 i \sin (c+d x)-451 i \sin (3 (c+d x))-77 i \sin (5 (c+d x))\right ) (-i+\tan (c+d x))}{1848 d} \]

[In]

Integrate[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

-1/1848*(a^3*e^3*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(-908*Cos[c + d*x] - 858*Cos[3*(c + d*x)] - 154*Cos[5*(c
+ d*x)] + (77*(1 + E^((2*I)*(c + d*x)))^(11/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/(2*E^((
5*I)*(c + d*x))) - (38*I)*Sin[c + d*x] - (451*I)*Sin[3*(c + d*x)] - (77*I)*Sin[5*(c + d*x)])*(-I + Tan[c + d*x
]))/d

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (201 ) = 402\).

Time = 11.29 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.42

\[-\frac {2 i e^{3} a^{3} \sqrt {e \sec \left (d x +c \right )}\, \left (231 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-231 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+462 \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-462 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+231 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-231 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-231 i \sin \left (d x +c \right ) \left (\tan ^{2}\left (d x +c \right )\right )-77 i \left (\tan ^{3}\left (d x +c \right )\right )-308 i \left (\tan ^{3}\left (d x +c \right )\right ) \sec \left (d x +c \right )+231 i \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )-132 \left (\sec ^{2}\left (d x +c \right )\right )-132 \left (\sec ^{3}\left (d x +c \right )\right )+21 \left (\sec ^{4}\left (d x +c \right )\right )+21 \left (\sec ^{5}\left (d x +c \right )\right )\right )}{231 d \left (\cos \left (d x +c \right )+1\right )}\]

[In]

int((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x)

[Out]

-2/231*I*e^3*a^3/d*(e*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)*(231*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-231*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+462*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-462*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I
)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+231*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-231*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/
(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-231*I*sin(d*x+c)*tan(d*x+c)^2-77*I*tan(d*x+c)^3-308*I*
tan(d*x+c)^3*sec(d*x+c)+231*I*tan(d*x+c)*sec(d*x+c)^3-132*sec(d*x+c)^2-132*sec(d*x+c)^3+21*sec(d*x+c)^4+21*sec
(d*x+c)^5)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.56 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (231 i \, a^{3} e^{3} e^{\left (11 i \, d x + 11 i \, c\right )} + 1309 i \, a^{3} e^{3} e^{\left (9 i \, d x + 9 i \, c\right )} + 946 i \, a^{3} e^{3} e^{\left (7 i \, d x + 7 i \, c\right )} + 870 i \, a^{3} e^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 407 i \, a^{3} e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 77 i \, a^{3} e^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (i \, a^{3} e^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 i \, a^{3} e^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 i \, a^{3} e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 i \, a^{3} e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, a^{3} e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} e^{3}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{231 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/231*(sqrt(2)*(231*I*a^3*e^3*e^(11*I*d*x + 11*I*c) + 1309*I*a^3*e^3*e^(9*I*d*x + 9*I*c) + 946*I*a^3*e^3*e^(7
*I*d*x + 7*I*c) + 870*I*a^3*e^3*e^(5*I*d*x + 5*I*c) + 407*I*a^3*e^3*e^(3*I*d*x + 3*I*c) + 77*I*a^3*e^3*e^(I*d*
x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 231*sqrt(2)*(I*a^3*e^3*e^(10*I*d*x + 10*
I*c) + 5*I*a^3*e^3*e^(8*I*d*x + 8*I*c) + 10*I*a^3*e^3*e^(6*I*d*x + 6*I*c) + 10*I*a^3*e^3*e^(4*I*d*x + 4*I*c) +
5*I*a^3*e^3*e^(2*I*d*x + 2*I*c) + I*a^3*e^3)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d
*x + I*c))))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x +
4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(7/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^3, x)

Giac [F]

\[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^3, x)