\(\int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\) [657]

Optimal result

Integrand size = 26, antiderivative size = 90 \[ \int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 i a (e \cos (c+d x))^{3/2}}{3 d}+\frac {2 a (e \cos (c+d x))^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (e \cos (c+d x))^{3/2} \tan (c+d x)}{3 d} \]

[Out]

-2/3*I*a*(e*cos(d*x+c))^(3/2)/d+2/3*a*(e*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)*Ell
ipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(3/2)+2/3*a*(e*cos(d*x+c))^(3/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 90, 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, 3854, 3856, 2720} \[ \int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 i a (e \cos (c+d x))^{3/2}}{3 d}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (e \cos (c+d x))^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \tan (c+d x) (e \cos (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx=\frac {2 a e \sqrt {\cos (c+d x)} \sqrt {e \cos (c+d x)} \left (\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (i \cos (c)+\sin (c))+\sqrt {\cos (c+d x)} (\cos (d x)+i \sin (d x))\right ) (\cos (d x)-i \sin (d x)) (-i+\tan (c+d x))}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 5.50 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.87

method	result	size
default	\(-\frac {2 a \,e^{2} \left (4 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(168\)
parts	\(-\frac {2 a \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, e^{2} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\right )}{3 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}-\frac {2 i a \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}\)	\(209\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \sqrt {2}\, e a \sqrt {e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}}{3 d}+\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) e a \sqrt {e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(224\)

[In]

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

[Out]

-2/3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a*e^2*(4*I*sin(1/2*d*x+1/2*c)^5+4*sin(1/2*d*x+1/2*
c)^4*cos(1/2*d*x+1/2*c)-4*I*sin(1/2*d*x+1/2*c)^3-2*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)+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 = 58, normalized size of antiderivative = 0.64 \[ \int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 \, {\left (i \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} a e e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, \sqrt {2} a e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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