\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx\) [210]

Optimal result

Integrand size = 28, antiderivative size = 155 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}} \]

[Out]

10/77*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(1/2)+10/77*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticF(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^6-2/11*I*(a+I*a*tan(d*x+c))^3/d/(e
*sec(d*x+c))^(11/2)-20/77*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(7/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3578, 3577, 3854, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(11/2),x]

[Out]

(10*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*d*e^6) + (10*a^3*Sin[c + d*x])/
(77*d*e^5*Sqrt[e*Sec[c + d*x]]) - (((2*I)/11)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(11/2)) - (((20*I)
/77)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(7/2))

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}{11 d (e \sec (c+d x))^{11/2}}+\frac {(5 a) \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx}{11 e^2} \\ & = -\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (15 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{77 e^4} \\ & = \frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (5 a^3\right ) \int \sqrt {e \sec (c+d x)} \, dx}{77 e^6} \\ & = \frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (5 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 e^6} \\ & = \frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {a^3 \sqrt {e \sec (c+d x)} \left (-46 i \cos (c+d x)-22 i \cos (3 (c+d x))-15 \sin (c+d x)+20 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-15 \sin (3 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{154 d e^6 (\cos (d x)+i \sin (d x))^3} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(11/2),x]

[Out]

(a^3*Sqrt[e*Sec[c + d*x]]*((-46*I)*Cos[c + d*x] - (22*I)*Cos[3*(c + d*x)] - 15*Sin[c + d*x] + 20*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 15*Sin[3*(c + d*x)])*(Cos[3*(c + 2*d
*x)] + I*Sin[3*(c + 2*d*x)]))/(154*d*e^6*(Cos[d*x] + I*Sin[d*x])^3)

Maple [A] (verified)

Time = 20.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {2 a^{3} \left (28 i \left (\cos ^{5}\left (d x +c \right )\right )-28 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+5 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-11 i \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 \sin \left (d x +c \right )\right )}{77 e^{5} d \sqrt {e \sec \left (d x +c \right )}}\)	\(201\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (7 \,{\mathrm e}^{4 i \left (d x +c \right )}+24 \,{\mathrm e}^{2 i \left (d x +c \right )}+37\right ) a^{3} \sqrt {2}}{308 d \,e^{5} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {10 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{3} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{77 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(260\)
parts	\(-\frac {2 a^{3} \left (-7 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-9 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-15 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {2 i a^{3} \left (7 \left (\cos ^{5}\left (d x +c \right )\right )-11 \left (\cos ^{3}\left (d x +c \right )\right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {6 i a^{3}}{11 d \left (e \sec \left (d x +c \right )\right )^{\frac {11}{2}}}+\frac {2 a^{3} \left (21 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+10 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+10 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}\)	\(420\)

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/77*a^3/e^5/d/(e*sec(d*x+c))^(1/2)*(28*I*cos(d*x+c)^5-28*sin(d*x+c)*cos(d*x+c)^4+5*I*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)-11*I*cos(d*x+c)^3+5*I*sec(d*x+c)*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)-3*cos(d*x+c)^2
*sin(d*x+c)-5*sin(d*x+c))

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {-40 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-7 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 31 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 61 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 37 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{308 \, d e^{6}} \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

1/308*(-40*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-7*I*a^3*e^(6*I*d*x +
6*I*c) - 31*I*a^3*e^(4*I*d*x + 4*I*c) - 61*I*a^3*e^(2*I*d*x + 2*I*c) - 37*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) +
1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(11/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(11/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(11/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(11/2), x)