3.198 \(\int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=119 \[ -\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]

[Out]

((-2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((2*I)*a*Sqr
t[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2))

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Rubi [A]  time = 0.224434, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3548, 3545, 3544, 205} \[ -\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]

[Out]

((-2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((2*I)*a*Sqr
t[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2))

Rule 3548

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*m*(c^2 + d^2)), x] + Dist[a/(a*c - b*d), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*
d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -1]

Rule 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}+i \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-(2 a) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left (4 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.67922, size = 160, normalized size = 1.34 \[ -\frac{2 i \sqrt{2} a e^{-i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (e^{i (c+d x)} \left (-3+5 e^{2 i (c+d x)}\right )-3 \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{3 d \left (-1+e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]

[Out]

(((-2*I)/3)*Sqrt[2]*a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))*(-3 + 5*E^((2*I
)*(c + d*x))) - 3*(-1 + E^((2*I)*(c + d*x)))^(3/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(
d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.034, size = 368, normalized size = 3.1 \begin{align*} -{\frac{a}{6\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 3\,i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-3\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a+12\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}a+16\,i\sqrt{ia}\sqrt{-ia}\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+4\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x)

[Out]

-1/6/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(3/2)*(3*I*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-3*(I*a)^(1/2)*2^(1/2)*ln((
2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*
a+12*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/
2)*tan(d*x+c)^2*a+16*I*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+4*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^
(1/2)

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Maxima [B]  time = 3.06068, size = 1353, normalized size = 11.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-1/9*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(18*I + 18)*a*cos(3*d*x + 3*c)
+ (6*I + 6)*a*cos(d*x + c) - (18*I - 18)*a*sin(3*d*x + 3*c) + (6*I - 6)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2
*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(18*I - 18)*a*cos(3*d*x + 3*c) + (6*I - 6)*a*cos(d*x + c) + (18*I + 1
8)*a*sin(3*d*x + 3*c) - (6*I + 6)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*s
qrt(a) + (((18*I + 18)*a*cos(2*d*x + 2*c)^2 + (18*I + 18)*a*sin(2*d*x + 2*c)^2 - (36*I + 36)*a*cos(2*d*x + 2*c
) + (18*I + 18)*a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*ar
ctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*c
os(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + ((9*I -
9)*a*cos(2*d*x + 2*c)^2 + (9*I - 9)*a*sin(2*d*x + 2*c)^2 - (18*I - 18)*a*cos(2*d*x + 2*c) + (9*I - 9)*a)*log(
cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(
1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c)
+ 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d
*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x +
2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((((12*I + 12
)*a*cos(d*x + c) + (12*I - 12)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I + 12)*a*cos(d*x + c) + (12*I - 12)*
a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(24*I + 24)*a*cos(d*x + c) - (24*I - 24)*a*sin(d*x + c))*cos(2*d*x + 2*
c) + (12*I + 12)*a*cos(d*x + c) + (12*I - 12)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2
*c) + 1)) + (((12*I - 12)*a*cos(d*x + c) - (12*I + 12)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I - 12)*a*cos
(d*x + c) - (12*I + 12)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(24*I - 24)*a*cos(d*x + c) + (24*I + 24)*a*sin(
d*x + c))*cos(2*d*x + 2*c) + (12*I - 12)*a*cos(d*x + c) - (12*I + 12)*a*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*
x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1
)^(5/4)*d)

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Fricas [B]  time = 2.33991, size = 1142, normalized size = 9.6 \begin{align*} \frac{4 \, \sqrt{2}{\left (5 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 3 \,{\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 )} \sqrt{-\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + \sqrt{-\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right ) + 3 \,{\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 )} \sqrt{-\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - \sqrt{-\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right )}{6 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/6*(4*sqrt(2)*(5*a*e^(4*I*d*x + 4*I*c) + 2*a*e^(2*I*d*x + 2*I*c) - 3*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr
t((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(
2*I*d*x + 2*I*c) + d)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt(-8*I*a^3/d^2)*
d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) + 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(
-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(-8*I*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-
2*I*d*x - 2*I*c)/a))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.36152, size = 169, normalized size = 1.42 \begin{align*} \frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i - 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} + \left (5 i - 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - \left (9 i - 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} + \left (7 i - 7\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - \left (2 i - 2\right ) \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

2*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^3*log(sqrt(I*a*tan(d*x + c) + a))/(-(I -
1)*(I*a*tan(d*x + c) + a)^4 + (5*I - 5)*(I*a*tan(d*x + c) + a)^3*a - (9*I - 9)*(I*a*tan(d*x + c) + a)^2*a^2 +
(7*I - 7)*(I*a*tan(d*x + c) + a)*a^3 - (2*I - 2)*a^4)