\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx\) [212]

Optimal result

Integrand size = 28, antiderivative size = 186 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}} \]

[Out]

6/55*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(5/2)+2/11*a^3*sin(d*x+c)/d/e^7/(e*sec(d*x+c))^(1/2)+2/11*a^3*(cos(1/
2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))
^(1/2)/d/e^8-2/15*I*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(15/2)-12/55*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d
*x+c))^(11/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3578, 3577, 3854, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*d*e^8) + (6*a^3*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^3*Sin[c + d*x])/(11*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((2*I)/15)*(a + I*a*
Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((12*I)/55)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x]
)^(11/2))

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac {(3 a) \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2} \\ & = -\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (21 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4} \\ & = \frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (3 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6} \\ & = \frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {a^3 \int \sqrt {e \sec (c+d x)} \, dx}{11 e^8} \\ & = \frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac {\left (a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{11 e^8} \\ & = \frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.91 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {a^3 \sqrt {e \sec (c+d x)} \left (-332 i \cos (c+d x)-154 i \cos (3 (c+d x))+22 i \cos (5 (c+d x))-114 \sin (c+d x)+240 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-81 \sin (3 (c+d x))+33 \sin (5 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{1320 d e^8 (\cos (d x)+i \sin (d x))^3} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(a^3*Sqrt[e*Sec[c + d*x]]*((-332*I)*Cos[c + d*x] - (154*I)*Cos[3*(c + d*x)] + (22*I)*Cos[5*(c + d*x)] - 114*Si
n[c + d*x] + 240*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 81*Sin
[3*(c + d*x)] + 33*Sin[5*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d*x)]))/(1320*d*e^8*(Cos[d*x] + I*Si
n[d*x])^3)

Maple [A] (verified)

Time = 24.65 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {2 a^{3} \left (-44 i \left (\cos ^{7}\left (d x +c \right )\right )+44 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+15 i \left (\cos ^{5}\left (d x +c \right )\right )+7 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+9 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+15 \sin \left (d x +c \right )\right )}{165 e^{7} d \sqrt {e \sec \left (d x +c \right )}}\)	\(217\)
parts	\(-\frac {2 a^{3} \left (-77 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )-91 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+195 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+195 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-117 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-195 \sin \left (d x +c \right )\right )}{1155 d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}-\frac {2 i a^{3} \left (11 \left (\cos ^{7}\left (d x +c \right )\right )-15 \left (\cos ^{5}\left (d x +c \right )\right )\right )}{165 d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}-\frac {2 i a^{3}}{5 d \left (e \sec \left (d x +c \right )\right )^{\frac {15}{2}}}+\frac {2 a^{3} \left (77 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )-14 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+30 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+30 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-18 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-30 \sin \left (d x +c \right )\right )}{385 d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}\)	\(452\)

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x,method=_RETURNVERBOSE)

[Out]

2/165*a^3/e^7/d/(e*sec(d*x+c))^(1/2)*(-44*I*cos(d*x+c)^7+44*sin(d*x+c)*cos(d*x+c)^6+15*I*cos(d*x+c)^5+7*sin(d*
x+c)*cos(d*x+c)^4+15*I*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c
)+1))^(1/2)+15*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c
)+cot(d*x+c)),I)+9*cos(d*x+c)^2*sin(d*x+c)+15*sin(d*x+c))

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {{\left (-480 i \, \sqrt {2} a^{3} \sqrt {e} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-11 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 73 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 218 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 446 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 235 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 55 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2640 \, d e^{8}} \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="fricas")

[Out]

1/2640*(-480*I*sqrt(2)*a^3*sqrt(e)*e^(2*I*d*x + 2*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(
-11*I*a^3*e^(10*I*d*x + 10*I*c) - 73*I*a^3*e^(8*I*d*x + 8*I*c) - 218*I*a^3*e^(6*I*d*x + 6*I*c) - 446*I*a^3*e^(
4*I*d*x + 4*I*c) - 235*I*a^3*e^(2*I*d*x + 2*I*c) + 55*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x +
1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(d*e^8)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(15/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2), x)