\(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx\) [221]

Optimal result

Integrand size = 28, antiderivative size = 156 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}} \]

[Out]

2/117*a^4*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(3/2)+2/39*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^6/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/13*I*a*(a+I*a*tan(d*x+c))^3/d/(
e*sec(d*x+c))^(13/2)-4/117*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(9/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3854, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]

[Out]

(2*a^4*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*a^4*Sin[c + d*x])/(1
17*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((4*I)/13)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((4*
I)/117)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(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 a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}+\frac {a^2 \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx}{13 e^2} \\ & = -\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {\left (5 a^4\right ) \int \frac {1}{(e \sec (c+d x))^{5/2}} \, dx}{117 e^4} \\ & = \frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {a^4 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{39 e^6} \\ & = \frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac {a^4 \int \sqrt {\cos (c+d x)} \, dx}{39 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (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 = 8.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=-\frac {i a^4 e^{i (c+d x)} \left (31+59 e^{2 i (c+d x)}+37 e^{4 i (c+d x)}+9 e^{6 i (c+d x)}+8 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{468 d e^7} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]

[Out]

((-1/468*I)*a^4*E^(I*(c + d*x))*(31 + 59*E^((2*I)*(c + d*x)) + 37*E^((4*I)*(c + d*x)) + 9*E^((6*I)*(c + d*x))
+ 8*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]]
)/(d*e^7)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (160 ) = 320\).

Time = 26.29 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.26

method	result	size
risch	\(-\frac {i \left (9 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+31 \,{\mathrm e}^{2 i \left (d x +c \right )}+24\right ) a^{4} \sqrt {2}}{468 d \,e^{6} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{4} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{39 d \,e^{6} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(352\)
default	\(-\frac {2 i a^{4} \left (72 \left (\cos ^{7}\left (d x +c \right )\right )+72 \left (\cos ^{6}\left (d x +c \right )\right )+i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-52 \left (\cos ^{5}\left (d x +c \right )\right )-16 i \sin \left (d x +c \right ) \left (\cos ^{4}\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 )-52 \left (\cos ^{4}\left (d x +c \right )\right )+72 i \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \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}}-16 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}}+3 i \sin \left (d x +c \right )+i \cos \left (d x +c \right ) \sin \left (d x +c \right )+72 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{117 e^{6} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(531\)
parts	\(\text {Expression too large to display}\)	\(1528\)

[In]

int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x,method=_RETURNVERBOSE)

[Out]

-1/468*I*(9*exp(I*(d*x+c))^6+28*exp(I*(d*x+c))^4+31*exp(I*(d*x+c))^2+24)/d*a^4*2^(1/2)/e^6/(e*exp(I*(d*x+c))/(
exp(I*(d*x+c))^2+1))^(1/2)-1/39*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^4*2^(1/2)/e^6/(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 = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {24 i \, \sqrt {2} a^{4} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-9 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} - 37 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 31 i \, a^{4} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{468 \, d e^{7}} \]

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="fricas")

[Out]

1/468*(24*I*sqrt(2)*a^4*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*
(-9*I*a^4*e^(7*I*d*x + 7*I*c) - 37*I*a^4*e^(5*I*d*x + 5*I*c) - 59*I*a^4*e^(3*I*d*x + 3*I*c) - 31*I*a^4*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))/(d*e^7)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(13/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(13/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(13/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2), x)