\(\int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\) [668]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac {4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, 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 = {3596, 3581, 3856, 2719} \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac {4 i \cos ^2(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{3/2}} \]

[In]

Int[1/((e*Cos[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(3/2)) + (((4*I)/5)*Cos[c + d*x]^2)
/(d*(e*Cos[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x]))

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 3581

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*d^2*
(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[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 {\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \\ & = \frac {4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^2 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \\ & = \frac {4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\cos ^{\frac {3}{2}}(c+d x) \int \sqrt {\cos (c+d x)} \, dx}{5 a^2 (e \cos (c+d x))^{3/2}} \\ & = \frac {2 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac {4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 3.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \cos ^2(c+d x) \left (1+2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt [4]{1-i \tan (c+d x)} (1+i \tan (c+d x))-i \tan (c+d x)\right )}{5 a^2 d (e \cos (c+d x))^{3/2} (-i+\tan (c+d x))} \]

[In]

Integrate[1/((e*Cos[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Cos[c + d*x]^2*(1 + 2^(3/4)*Hypergeometric2F1[-1/4, 1/4, 3/4, (1 + I*Tan[c + d*x])/2]*(1 - I*Tan[c + d*x])^
(1/4)*(1 + I*Tan[c + d*x]) - I*Tan[c + d*x]))/(5*a^2*d*(e*Cos[c + d*x])^(3/2)*(-I + Tan[c + d*x]))

Maple [A] (verified)

Time = 4.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.24

method	result	size
default	\(\frac {-\frac {32 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {48 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {32 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {24 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{5}+\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{e \,a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(206\)

[In]

int(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/5/e/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-16*I*sin(1/2*d*x+1/2*c)^7+16*sin(1/2*d*x+1/
2*c)^6*cos(1/2*d*x+1/2*c)+24*I*sin(1/2*d*x+1/2*c)^5-16*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-12*I*sin(1/2*d*
x+1/2*c)^3+4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)+2*I*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (-2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - i \, \sqrt {2} \sqrt {e} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{5 \, a^{2} d e^{2}} \]

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/5*(sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(-2*I*e^(2*I*d*x + 2*I*c) - I)*e^(-1/2*I*d*x - 1/2*I*c) - I*sq
rt(2)*sqrt(e)*e^(2*I*d*x + 2*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))*e^(-2*I
*d*x - 2*I*c)/(a^2*d*e^2)

Sympy [F(-1)]

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

[In]

integrate(1/(e*cos(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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