\(\int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx\) [661]

Optimal result

Integrand size = 26, antiderivative size = 96 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \]

[Out]

2/5*I*a/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)^(5/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)*sin(d*x+c)/d/(e*cos(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3567, 3853, 3856, 2720} \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \sin (c+d x) \cos (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \]

[In]

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

[Out]

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

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 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 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 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 {\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}} \\ & = \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {a \int (e \sec (c+d x))^{5/2} \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}} \\ & = \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}}+\frac {\left (a e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx}{3 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}} \\ & = \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}}+\frac {\left (a \cos ^{\frac {5}{2}}(c+d x)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 (e \cos (c+d x))^{5/2}} \\ & = \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {a \left (6 i+10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))\right )}{15 d (e \cos (c+d x))^{5/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (107 ) = 214\).

Time = 6.70 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.71

method	result	size
parts	\(-\frac {2 a \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 e^{2} \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 i a}{5 d \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(260\)
default	\(-\frac {2 \left (20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}-3 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{15 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(283\)

[In]

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

[Out]

-2/3*a*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^2-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2))/e^2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)
^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/
(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d+2/5*I*a/d/(e*cos(d*x+c))^(5/2)

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.98 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} {\left (5 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} - 12 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, {\left (i \, \sqrt {2} a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, \sqrt {2} a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, \sqrt {2} a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2} a\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )}} \]

[In]

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

[Out]

-2/15*(2*sqrt(1/2)*(5*I*a*e^(5*I*d*x + 5*I*c) - 12*I*a*e^(3*I*d*x + 3*I*c) - 5*I*a*e^(I*d*x + I*c))*sqrt(e*e^(
2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1/2*I*c) + 5*(I*sqrt(2)*a*e^(6*I*d*x + 6*I*c) + 3*I*sqrt(2)*a*e^(4*I*d*x
+ 4*I*c) + 3*I*sqrt(2)*a*e^(2*I*d*x + 2*I*c) + I*sqrt(2)*a)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c
)))/(d*e^3*e^(6*I*d*x + 6*I*c) + 3*d*e^3*e^(4*I*d*x + 4*I*c) + 3*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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