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

Optimal. Leaf size=122 \[ -\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

((-4 + 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((4*I)*a^2*S
qrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*a*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2)
)

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Rubi [A]  time = 0.20735, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3545, 3544, 205} \[ -\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4 + 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((4*I)*a^2*S
qrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*a*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2)
)

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))^{5/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}+(2 i a) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\left (4 a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left (8 i a^4\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{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.81702, size = 162, normalized size = 1.33 \[ -\frac{4 i \sqrt{2} a^2 e^{-i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (e^{i (c+d x)} \left (-3+4 e^{2 i (c+d x)}\right )-3 \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{3 d \left (-1+e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/3)*Sqrt[2]*a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))*(-3 + 4*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 = 0.036, size = 370, normalized size = 3. \begin{align*} -{\frac{{a}^{2}}{3\,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+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-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+14\,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 ) }+2\,\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))^(5/2)/tan(d*x+c)^(5/2),x)

[Out]

-1/3/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/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*
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-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+14*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)+2*(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 = 2.37819, size = 1439, normalized size = 11.8 \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))^(5/2)/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)*((-(36*I + 36)*a^2*cos(3*d*x + 3*
c) + (24*I + 24)*a^2*cos(d*x + c) - (36*I - 36)*a^2*sin(3*d*x + 3*c) + (24*I - 24)*a^2*sin(d*x + c))*cos(3/2*a
rctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(36*I - 36)*a^2*cos(3*d*x + 3*c) + (24*I - 24)*a^2*cos(d*
x + c) + (36*I + 36)*a^2*sin(3*d*x + 3*c) - (24*I + 24)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -c
os(2*d*x + 2*c) + 1)))*sqrt(a) + (((36*I + 36)*a^2*cos(2*d*x + 2*c)^2 + (36*I + 36)*a^2*sin(2*d*x + 2*c)^2 - (
72*I + 72)*a^2*cos(2*d*x + 2*c) + (36*I + 36)*a^2)*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*arctan2(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*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)) + ((18*I - 18)*a^2*cos(2*d*x + 2*c)^2 + (18*I - 18)*a^2*sin(2*d*x + 2*c)^2 - (36*I - 36
)*a^2*cos(2*d*x + 2*c) + (18*I - 18)*a^2)*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)*s
in(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^2*cos(d*x + c) + (12*I - 12)*a^2*sin(d*x + c))*cos(2*d*x + 2
*c)^2 + (12*I + 12)*a^2*cos(d*x + c) + ((12*I + 12)*a^2*cos(d*x + c) + (12*I - 12)*a^2*sin(d*x + c))*sin(2*d*x
+ 2*c)^2 + (12*I - 12)*a^2*sin(d*x + c) + (-(24*I + 24)*a^2*cos(d*x + c) - (24*I - 24)*a^2*sin(d*x + c))*cos(
2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((12*I - 12)*a^2*cos(d*x + c) - (12
*I + 12)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (12*I - 12)*a^2*cos(d*x + c) + ((12*I - 12)*a^2*cos(d*x + c) -
(12*I + 12)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (12*I + 12)*a^2*sin(d*x + c) + (-(24*I - 24)*a^2*cos(d*x +
c) + (24*I + 24)*a^2*sin(d*x + c))*cos(2*d*x + 2*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 = 2.31116, size = 1169, normalized size = 9.58 \begin{align*} \frac{8 \, \sqrt{2}{\left (4 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, a^{2}\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 \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}}{\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 )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) + 3 \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}}{\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 )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\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))^(5/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/6*(8*sqrt(2)*(4*a^2*e^(4*I*d*x + 4*I*c) + a^2*e^(2*I*d*x + 2*I*c) - 3*a^2)*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) - 3*sqrt(-32*I*a^5/d^2)*(d*e^(4*
I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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(-32*I
*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 3*sqrt(-32*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*
d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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(-32*I*a^5/d^2)*d*e^(2*I*
d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2))/(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))**(5/2)/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.37809, size = 171, normalized size = 1.4 \begin{align*} -\frac{\left (i - 1\right ) \, \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 )}^{2} a^{3} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} + 5 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - 9 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} + 7 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - 2 i \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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