\(\int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx\) [660]

Optimal result

Integrand size = 26, antiderivative size = 89 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03, 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, 2719} \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x) \cos (c+d x)}{d (e \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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))^{3/2} (a+i a \tan (c+d x)) \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \\ & = \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {a \int (e \sec (c+d x))^{3/2} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \\ & = \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}}-\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \\ & = \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}}-\frac {\left (a \cos ^{\frac {3}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{(e \cos (c+d x))^{3/2}} \\ & = \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.02 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.13 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=\frac {(\cos (d x)-i \sin (d x)) \left (\frac {6 \cos (c+d x) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) (1-i \tan (c))}{\sqrt {\sin ^2(d x+\arctan (\tan (c)))}}+(\csc (c)-i \sec (c)) \left (-3 \cos (c+d x) (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c))))+\frac {4 (i+3 \cos (d x) \cos (c+d x) \csc (c)) \tan (c)}{\sqrt {\sec ^2(c)}}\right )\right ) (a+i a \tan (c+d x))}{6 d e \sqrt {e \cos (c+d x)} \sqrt {\sec ^2(c)}} \]

[In]

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

[Out]

((Cos[d*x] - I*Sin[d*x])*((6*Cos[c + d*x]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*
Sin[d*x + ArcTan[Tan[c]]]*(1 - I*Tan[c]))/Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2] + (Csc[c] - I*Sec[c])*(-3*Cos[c +
d*x]*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]]) + (4*(I + 3*Cos[d*x]*Cos[c + d*x]*Csc[c
])*Tan[c])/Sqrt[Sec[c]^2]))*(a + I*a*Tan[c + d*x]))/(6*d*e*Sqrt[e*Cos[c + d*x]]*Sqrt[Sec[c]^2])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (104 ) = 208\).

Time = 5.44 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.40

method	result	size
default	\(\frac {2 \left (12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-6 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}-i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e d}\)	\(214\)
parts	\(-\frac {2 a \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 i a}{3 d \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(217\)

[In]

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

[Out]

2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e*(12*sin(1/2*d*x+1/2*c)
^4*cos(1/2*d*x+1/2*c)-6*EllipticE(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)^(1/2)*sin(1/2*d*x+1/2*c)^2-6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*EllipticE(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)^(1/2)-I*sin(1/2*d*x+1/2*c))*a/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 = 151, normalized size of antiderivative = 1.70 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} {\left (3 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 3 \, {\left (i \, \sqrt {2} a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, \sqrt {2} a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2} a\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{3 \, {\left (d e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)