\(\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx\) [186]

Optimal result

Integrand size = 26, antiderivative size = 94 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 94, 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, 3853, 3856, 2720} \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

(2*a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (((2*I)/5)*a*(e*Sec[c + d*
x])^(5/2))/d + (2*a*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 (e \sec (c+d x))^{5/2}}{5 d}+a \int (e \sec (c+d x))^{5/2} \, dx \\ & = \frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \left (a e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx \\ & = \frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \left (a e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a e^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}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {a (e \sec (c+d x))^{5/2} \left (6 i+10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))\right )}{15 d} \]

[In]

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

[Out]

(a*(e*Sec[c + d*x])^(5/2)*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d)

Maple [A] (verified)

Time = 24.87 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.73

method	result	size
default	\(-\frac {2 a \sqrt {e \sec \left (d x +c \right )}\, e^{2} \left (i \cos \left (d x +c \right ) F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-\tan \left (d x +c \right )\right )}{3 d}+\frac {2 i a \left (e \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}\)	\(163\)
parts	\(-\frac {2 a \sqrt {e \sec \left (d x +c \right )}\, e^{2} \left (i \cos \left (d x +c \right ) F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-\tan \left (d x +c \right )\right )}{3 d}+\frac {2 i a \left (e \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}\)	\(163\)

[In]

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

[Out]

-2/3*a/d*(e*sec(d*x+c))^(1/2)*e^2*(I*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)
*(1/(cos(d*x+c)+1))^(1/2)-tan(d*x+c))+2/5*I*a*(e*sec(d*x+c))^(5/2)/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 = 154, normalized size of antiderivative = 1.64 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, \sqrt {2} {\left (i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-2/15*(sqrt(2)*(5*I*a*e^2*e^(4*I*d*x + 4*I*c) - 12*I*a*e^2*e^(2*I*d*x + 2*I*c) - 5*I*a*e^2)*sqrt(e/(e^(2*I*d*x
+ 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 5*sqrt(2)*(I*a*e^2*e^(4*I*d*x + 4*I*c) + 2*I*a*e^2*e^(2*I*d*x + 2*I*
c) + I*a*e^2)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2
*I*c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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