\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx\) [207]

Optimal result

Integrand size = 28, antiderivative size = 111 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

6/5*I*a^3/d/e^2/(e*sec(d*x+c))^(1/2)-6/5*a^3*(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/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/5*I*a*(a+I*a*tan(d*x+c))^2/d/(e*sec(d*x+c))
^(5/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 111, 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, 3567, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(5/2),x]

[Out]

(((6*I)/5)*a^3)/(d*e^2*Sqrt[e*Sec[c + d*x]]) - (6*a^3*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*S
qrt[e*Sec[c + d*x]]) - (((4*I)/5)*a*(a + I*a*Tan[c + d*x])^2)/(d*(e*Sec[c + d*x])^(5/2))

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 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 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 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))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^2\right ) \int \frac {a+i a \tan (c+d x)}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {4 i a^3 e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}-\sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(5/2),x]

[Out]

(((-4*I)/5)*a^3*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)) - Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1
[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*e^2*(1 + E^((2*I)*(c + d*x)))*Sqrt[e*Sec[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. 323 vs. \(2 (120 ) = 240\).

Time = 14.92 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.92

method	result	size
risch	\(-\frac {2 i \left ({\mathrm e}^{2 i \left (d x +c \right )}-3\right ) a^{3} \sqrt {2}}{5 d \,e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {3 i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{3} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{5 d \,e^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(324\)
default	\(\text {Expression too large to display}\)	\(882\)
parts	\(\text {Expression too large to display}\)	\(1289\)

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*I*(exp(I*(d*x+c))^2-3)/d*a^3*2^(1/2)/e^2/(e*exp(I*(d*x+c))/(exp(I*(d*x+c))^2+1))^(1/2)+3/5*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*(e
xp(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^3*2^(1/2)
/e^2/(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 = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{5 \, d e^{3}} \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/5*(3*I*sqrt(2)*a^3*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(I
*a^3*e^(3*I*d*x + 3*I*c) + I*a^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))/(
d*e^3)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(5/2),x)

[Out]

-I*a**3*(Integral(I/(e*sec(c + d*x))**(5/2), x) + Integral(-3*tan(c + d*x)/(e*sec(c + d*x))**(5/2), x) + Integ
ral(tan(c + d*x)**3/(e*sec(c + d*x))**(5/2), x) + Integral(-3*I*tan(c + d*x)**2/(e*sec(c + d*x))**(5/2), x))

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(5/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(5/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(5/2), x)