3.199 \(\int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=198 \[ -\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}} \]

[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*a^2)/(5*d
*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*a^2)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d
*x]]) - (((4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) + (12*a*Sqrt[a + I*a*Tan[c + d*x]])/(5
*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.596871, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3553, 3596, 3598, 12, 3544, 205} \[ -\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),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*a^2)/(5*d
*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*a^2)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d
*x]]) - (((4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) + (12*a*Sqrt[a + I*a*Tan[c + d*x]])/(5
*d*Sqrt[Tan[c + d*x]])

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

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

Rule 3598

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{7}{2}}(c+d x)} \, dx &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2}{5} \int \frac{-\frac{9 i a^2}{2}+\frac{11}{2} a^2 \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-3 i a^3+2 a^3 \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a^2}\\ &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{9 a^4}{2}+3 i a^4 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^3}\\ &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}}-\frac{8 \int \frac{15 i a^5 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}}-(2 i a) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}}-\frac{\left (4 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 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2}{5 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{12 a \sqrt{a+i a \tan (c+d x)}}{5 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.54333, size = 166, normalized size = 0.84 \[ \frac{2 i a e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)} \left (e^{i (c+d x)} \left (-10 e^{2 i (c+d x)}+9 e^{4 i (c+d x)}+5\right )-5 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{5 d \left (-1+e^{2 i (c+d x)}\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]

[Out]

(((2*I)/5)*a*(1 + E^((2*I)*(c + d*x)))*(E^(I*(c + d*x))*(5 - 10*E^((2*I)*(c + d*x)) + 9*E^((4*I)*(c + d*x))) -
5*(-1 + E^((2*I)*(c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]
]*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^3)

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Maple [B]  time = 0.038, size = 410, normalized size = 2.1 \begin{align*}{\frac{a}{10\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 5\,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 ) ^{3}a+5\,\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 ) ^{3}a+24\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}+20\,\ln \left ( 1/2\,{\frac{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}{\sqrt{ia}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-8\,i\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}-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{5}{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)^(7/2),x)

[Out]

1/10/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(5/2)*(5*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)^3*a+5*(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)^3*
a+24*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)^2+20*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)^3*a-8*I*tan(d*x+c)*
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(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 = 3.80521, size = 1607, normalized size = 8.12 \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)^(7/2),x, algorithm="maxima")

[Out]

-1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(450*I - 450)*a*cos(3*d*x +
3*c) + (480*I - 480)*a*cos(d*x + c) + (450*I + 450)*a*sin(3*d*x + 3*c) - (480*I + 480)*a*sin(d*x + c))*cos(3/2
*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((450*I + 450)*a*cos(3*d*x + 3*c) - (480*I + 480)*a*cos(d
*x + c) + (450*I - 450)*a*sin(3*d*x + 3*c) - (480*I - 480)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -
cos(2*d*x + 2*c) + 1)))*sqrt(a) + (((450*I - 450)*a*cos(2*d*x + 2*c)^2 + (450*I - 450)*a*sin(2*d*x + 2*c)^2 -
(900*I - 900)*a*cos(2*d*x + 2*c) + (450*I - 450)*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*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)) + (-(225*I + 225)*a*cos(2*d*x + 2*c)^2 - (225*I + 225)*a*sin(2*d*x + 2*c)^2 + (450*I +
450)*a*cos(2*d*x + 2*c) - (225*I + 225)*a)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + si
n(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) + (((450*I - 450)*a*cos(5*d*x + 5*c) - (150*I - 150)*a*cos(3*d*x + 3*c) + (60
*I - 60)*a*cos(d*x + c) - (450*I + 450)*a*sin(5*d*x + 5*c) + (150*I + 150)*a*sin(3*d*x + 3*c) - (60*I + 60)*a*
sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((-(90*I - 90)*a*cos(d*x + c) + (90*
I + 90)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(90*I - 90)*a*cos(d*x + c) + (90*I + 90)*a*sin(d*x + c))*sin(2*
d*x + 2*c)^2 + ((180*I - 180)*a*cos(d*x + c) - (180*I + 180)*a*sin(d*x + c))*cos(2*d*x + 2*c) - (90*I - 90)*a*
cos(d*x + c) + (90*I + 90)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(450*
I + 450)*a*cos(5*d*x + 5*c) + (150*I + 150)*a*cos(3*d*x + 3*c) - (60*I + 60)*a*cos(d*x + c) - (450*I - 450)*a*
sin(5*d*x + 5*c) + (150*I - 150)*a*sin(3*d*x + 3*c) - (60*I - 60)*a*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x +
2*c), -cos(2*d*x + 2*c) + 1)) + (((90*I + 90)*a*cos(d*x + c) + (90*I - 90)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2
+ ((90*I + 90)*a*cos(d*x + c) + (90*I - 90)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(180*I + 180)*a*cos(d*x + c
) - (180*I - 180)*a*sin(d*x + c))*cos(2*d*x + 2*c) + (90*I + 90)*a*cos(d*x + c) + (90*I - 90)*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 = 2.4233, size = 1296, normalized size = 6.55 \begin{align*} \frac{\sqrt{2}{\left (36 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 4 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 20 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 20 i \, 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 )} + 5 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 )} + i \, \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 ) - 5 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 )} - i \, \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 )}{10 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^(7/2),x, algorithm="fricas")

[Out]

1/10*(sqrt(2)*(36*I*a*e^(6*I*d*x + 6*I*c) - 4*I*a*e^(4*I*d*x + 4*I*c) - 20*I*a*e^(2*I*d*x + 2*I*c) + 20*I*a)*s
qrt(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)
+ 5*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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) + I*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) -
5*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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) - I*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^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.3738, size = 193, normalized size = 0.97 \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^{4} \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 )}^{5} + \left (6 i + 6\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - \left (14 i + 14\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + \left (16 i + 16\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - \left (9 i + 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + \left (2 i + 2\right ) \, a^{5}} \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)^(7/2),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^4*log(sqrt(I*a*tan(d*x + c) + a))/(-(I +
1)*(I*a*tan(d*x + c) + a)^5 + (6*I + 6)*(I*a*tan(d*x + c) + a)^4*a - (14*I + 14)*(I*a*tan(d*x + c) + a)^3*a^2
+ (16*I + 16)*(I*a*tan(d*x + c) + a)^2*a^3 - (9*I + 9)*(I*a*tan(d*x + c) + a)*a^4 + (2*I + 2)*a^5)