\(\int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx\) [190]

Optimal result

Integrand size = 26, antiderivative size = 96 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 i a}{3 d (e \sec (c+d x))^{3/2}}+\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 a \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}} \]

[Out]

-2/3*I*a/d/(e*sec(d*x+c))^(3/2)+2/3*a*sin(d*x+c)/d/e/(e*sec(d*x+c))^(1/2)+2/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3567, 3854, 3856, 2720} \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx=\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {2 i a}{3 d (e \sec (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx=\frac {2 a \left (-i \cos (c+d x)+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{\sqrt {\cos (c+d x)}}+\sin (c+d x)\right )}{3 d e \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.75 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {2 a \left (i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}{3 e d \sqrt {e \sec \left (d x +c \right )}}\)	\(152\)
parts	\(-\frac {2 a \left (i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-\sin \left (d x +c \right )\right )}{3 d \sqrt {e \sec \left (d x +c \right )}\, e}-\frac {2 i a}{3 d \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(163\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} a \sqrt {2}}{3 d e \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(228\)

[In]

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

[Out]

2/3*a/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)-I*cos(d*x+c)+sin(d*x+c))

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.79 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 2 i \, \sqrt {2} a \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )}{3 \, d e^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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