\(\int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx\) [199]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}} \]

[Out]

2/9*a^2*sin(d*x+c)/d/e^3/(e*sec(d*x+c))^(3/2)+2/3*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
E(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^4/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/9*I*(a^2+I*a^2*tan(d*x+c))/d/(e*se
c(d*x+c))^(9/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3854, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(9/2),x]

[Out]

(2*a^2*EllipticE[(c + d*x)/2, 2])/(3*d*e^4*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*a^2*Sin[c + d*x])/(9*
d*e^3*(e*Sec[c + d*x])^(3/2)) - (((4*I)/9)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(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 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*} \text {integral}& = -\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {\left (5 a^2\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{9 e^2} \\ & = \frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {a^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{3 e^4} \\ & = \frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}+\frac {a^2 \int \sqrt {\cos (c+d x)} \, dx}{3 e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{9 d e^3 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.56 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\frac {i a^2 \left (9-4 e^{2 i (c+d x)}-e^{4 i (c+d x)}-\frac {8 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{18 \sqrt {2} d e^4 \sqrt {\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(9/2),x]

[Out]

((I/18)*a^2*(9 - 4*E^((2*I)*(c + d*x)) - E^((4*I)*(c + d*x)) - (8*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/2, 3
/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))]))/(Sqrt[2]*d*e^4*Sqrt[(e*E^(I*(c + d*x)))/(1 + E
^((2*I)*(c + d*x)))])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (126 ) = 252\).

Time = 19.86 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.91

method	result	size
risch	\(-\frac {i \left ({\mathrm e}^{4 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+15\right ) a^{2} \sqrt {2}}{36 d \,e^{4} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{2} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \,e^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(337\)
default	\(-\frac {2 i a^{2} \left (2 \left (\cos ^{5}\left (d x +c \right )\right )+3 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-3 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 \left (\cos ^{4}\left (d x +c \right )\right )+2 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+6 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-6 F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+i \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 i \sin \left (d x +c \right )\right )}{9 e^{4} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(477\)
parts	\(\text {Expression too large to display}\)	\(933\)

[In]

int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/36*I*(exp(I*(d*x+c))^4+4*exp(I*(d*x+c))^2+15)/d*a^2*2^(1/2)/e^4/(e*exp(I*(d*x+c))/(exp(I*(d*x+c))^2+1))^(1/
2)-1/3*I/d*(-2*(e*exp(I*(d*x+c))^2+e)/e/(exp(I*(d*x+c))*(e*exp(I*(d*x+c))^2+e))^(1/2)+I*(-I*(exp(I*(d*x+c))+I)
)^(1/2)*2^(1/2)*(I*(exp(I*(d*x+c))-I))^(1/2)*(I*exp(I*(d*x+c)))^(1/2)/(e*exp(I*(d*x+c))^3+e*exp(I*(d*x+c)))^(1
/2)*(-2*I*EllipticE((-I*(exp(I*(d*x+c))+I))^(1/2),1/2*2^(1/2))+I*EllipticF((-I*(exp(I*(d*x+c))+I))^(1/2),1/2*2
^(1/2))))*a^2*2^(1/2)/e^4/(exp(I*(d*x+c))^2+1)/(e*exp(I*(d*x+c))/(exp(I*(d*x+c))^2+1))^(1/2)*(e*exp(I*(d*x+c))
*(exp(I*(d*x+c))^2+1))^(1/2)

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.13 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\frac {{\left (24 i \, \sqrt {2} a^{2} \sqrt {e} e^{\left (i \, d x + i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, a^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{36 \, d e^{5}} \]

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/36*(24*I*sqrt(2)*a^2*sqrt(e)*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*
c))) + sqrt(2)*(-I*a^2*e^(6*I*d*x + 6*I*c) - 5*I*a^2*e^(4*I*d*x + 4*I*c) + 5*I*a^2*e^(2*I*d*x + 2*I*c) + 9*I*a
^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-I*d*x - I*c)/(d*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**2/(e*sec(d*x+c))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(9/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(9/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(9/2), x)