\(\int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx\) [196]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3577, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

(-2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d*e^2) - (((4*I)/3)*(a^2 + I*a^2
*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(3/2))

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

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sec ^2(c+d x) \left (2 i \cos (c+d x)+\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)-i \sin (c+d x))\right ) (\cos (c+3 d x)+i \sin (c+3 d x))}{3 d (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^2} \]

[In]

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

[Out]

(-2*a^2*Sec[c + d*x]^2*((2*I)*Cos[c + d*x] + Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] - I*Si
n[c + d*x]))*(Cos[c + 3*d*x] + I*Sin[c + 3*d*x]))/(3*d*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])^2)

Maple [A] (verified)

Time = 9.08 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.84

method	result	size
default	\(-\frac {2 a^{2} \left (i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 i \cos \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{3 e d \sqrt {e \sec \left (d x +c \right )}}\)	\(156\)
risch	\(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{2} \sqrt {2}}{3 d e \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{2} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(232\)
parts	\(-\frac {2 a^{2} \left (i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-\sin \left (d x +c \right )\right )}{3 d \sqrt {e \sec \left (d x +c \right )}\, e}-\frac {4 i a^{2}}{3 d \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {2 a^{2} \left (2 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right )\right )}{3 d \sqrt {e \sec \left (d x +c \right )}\, e}\)	\(311\)

[In]

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

[Out]

-2/3*a^2/e/d/(e*sec(d*x+c))^(1/2)*(I*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)+I*sec(d*x+c)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*cos(d*x+c)-2*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 = 82, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} a^{2} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{3 \, d e^{2}} \]

[In]

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

[Out]

-2/3*(-I*sqrt(2)*a^2*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(I*a^2*e^(2*I*d*x + 2*I*c)
+ I*a^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^2)

Sympy [F]

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

[In]

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

[Out]

-a**2*(Integral(-1/(e*sec(c + d*x))**(3/2), x) + Integral(tan(c + d*x)**2/(e*sec(c + d*x))**(3/2), x) + Integr
al(-2*I*tan(c + d*x)/(e*sec(c + d*x))**(3/2), x))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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