\(\int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [1024]

Optimal result

Integrand size = 35, antiderivative size = 255 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {35 i a^{9/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f} \]

[Out]

-35*I*a^(9/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/c^(3/2)/f+35/2*I*a^4*(
a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^2/f+35/6*I*a^3*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^
(3/2)/c^2/f+14/3*I*a^2*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(1/2)-2/3*I*a*(a+I*a*tan(f*x+e))^(7/2)/
f/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3604, 49, 52, 65, 223, 209} \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {35 i a^{9/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[In]

Int[(a + I*a*Tan[e + f*x])^(9/2)/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

((-35*I)*a^(9/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(3/2)*f
) - (((2*I)/3)*a*(a + I*a*Tan[e + f*x])^(7/2))/(f*(c - I*c*Tan[e + f*x])^(3/2)) + (((14*I)/3)*a^2*(a + I*a*Tan
[e + f*x])^(5/2))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (((35*I)/2)*a^4*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*T
an[e + f*x]])/(c^2*f) + (((35*I)/6)*a^3*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f)

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3604

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

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{7/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {\left (7 a^2\right ) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {\left (35 a^3\right ) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 c f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (35 a^4\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (35 a^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (35 i a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{c f} \\ & = -\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (35 i a^4\right ) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{c f} \\ & = -\frac {35 i a^{9/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 9.65 (sec) , antiderivative size = 524, normalized size of antiderivative = 2.05 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) \sqrt {1-i \tan (e+f x)} (a+i a \tan (e+f x))^{7/2}}{7 \sqrt {2} c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a \left (\frac {i a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {a+i a \tan (e+f x)}{2 a}\right ) (a+i a \tan (e+f x))^{5/2} \sqrt {\frac {c-i c \tan (e+f x)}{c}}}{5 \sqrt {2} c \sqrt {c-i c \tan (e+f x)}}+2 a \left (\frac {2 i a^2 (-i+\tan (e+f x)) \sqrt {i a (-i+\tan (e+f x))}}{3 c (i+\tan (e+f x)) \sqrt {-i c (i+\tan (e+f x))}}+\frac {2 i \sqrt {2} a^3 \sqrt {\frac {c-i c \tan (e+f x)}{c}} \left (-1+\frac {a+i a \tan (e+f x)}{2 a}\right ) \left (\frac {\arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a} \sqrt {1-\frac {a+i a \tan (e+f x)}{2 a}}}+\frac {a+i a \tan (e+f x)}{2 a \left (-1+\frac {a+i a \tan (e+f x)}{2 a}\right )}\right )}{c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} \sqrt {1-\frac {a+i a \tan (e+f x)}{2 a}}}\right )\right )}{f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^(9/2)/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

((I/7)*a*Hypergeometric2F1[3/2, 7/2, 9/2, (1 + I*Tan[e + f*x])/2]*Sqrt[1 - I*Tan[e + f*x]]*(a + I*a*Tan[e + f*
x])^(7/2))/(Sqrt[2]*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (2*a*(((I/5)*a*Hypergeometric2F1[3/2, 5/2, 7/2, (a + I*a
*Tan[e + f*x])/(2*a)]*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[(c - I*c*Tan[e + f*x])/c])/(Sqrt[2]*c*Sqrt[c - I*c*Tan
[e + f*x]]) + 2*a*((((2*I)/3)*a^2*(-I + Tan[e + f*x])*Sqrt[I*a*(-I + Tan[e + f*x])])/(c*(I + Tan[e + f*x])*Sqr
t[(-I)*c*(I + Tan[e + f*x])]) + ((2*I)*Sqrt[2]*a^3*Sqrt[(c - I*c*Tan[e + f*x])/c]*(-1 + (a + I*a*Tan[e + f*x])
/(2*a))*((ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*Sqrt[a]*Sq
rt[1 - (a + I*a*Tan[e + f*x])/(2*a)]) + (a + I*a*Tan[e + f*x])/(2*a*(-1 + (a + I*a*Tan[e + f*x])/(2*a)))))/(c*
Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]]*Sqrt[1 - (a + I*a*Tan[e + f*x])/(2*a)]))))/f

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{4} \left (315 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+27 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+105 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )-3 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )-105 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -393 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-259 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{2}\left (f x +e \right )\right )+164 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \,c^{2} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(409\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{4} \left (315 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+27 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+105 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )-3 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )-105 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -393 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-259 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{2}\left (f x +e \right )\right )+164 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \,c^{2} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(409\)

[In]

int((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^4/c^2*(315*I*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x
+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2+27*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x
+e)^3+105*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^3-3*(a*c)^(
1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^4-105*I*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/
2))/(a*c)^(1/2))*a*c-393*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-315*ln((c*a*tan(f*x+e)+(a*c*(1+
tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-259*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan
(f*x+e)^2+164*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)/(tan(f*x+e)+I
)^3

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (189) = 378\).

Time = 0.26 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.69 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{2} f\right )}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (-i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2} f\right )}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 4 \, {\left (8 i \, a^{4} e^{\left (7 i \, f x + 7 i \, e\right )} - 56 i \, a^{4} e^{\left (5 i \, f x + 5 i \, e\right )} - 175 i \, a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} - 105 i \, a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/12*(105*sqrt(a^9/(c^3*f^2))*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*log(4*(2*(a^4*e^(3*I*f*x + 3*I*e) + a^4*e^(I
*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^9/(c^3*f^2))*(I*c^2*
f*e^(2*I*f*x + 2*I*e) - I*c^2*f))/(a^4*e^(2*I*f*x + 2*I*e) + a^4)) - 105*sqrt(a^9/(c^3*f^2))*(c^2*f*e^(2*I*f*x
+ 2*I*e) + c^2*f)*log(4*(2*(a^4*e^(3*I*f*x + 3*I*e) + a^4*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*
sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^9/(c^3*f^2))*(-I*c^2*f*e^(2*I*f*x + 2*I*e) + I*c^2*f))/(a^4*e^(2*I*
f*x + 2*I*e) + a^4)) - 4*(8*I*a^4*e^(7*I*f*x + 7*I*e) - 56*I*a^4*e^(5*I*f*x + 5*I*e) - 175*I*a^4*e^(3*I*f*x +
3*I*e) - 105*I*a^4*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2*
f*e^(2*I*f*x + 2*I*e) + c^2*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(f*x+e))**(9/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 915 vs. \(2 (189) = 378\).

Time = 0.44 (sec) , antiderivative size = 915, normalized size of antiderivative = 3.59 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-6*(210*(a^4*cos(4*f*x + 4*e) + 2*a^4*cos(2*f*x + 2*e) + I*a^4*sin(4*f*x + 4*e) + 2*I*a^4*sin(2*f*x + 2*e) + a
^4)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 1) + 210*(a^4*cos(4*f*x + 4*e) + 2*a^4*cos(2*f*x + 2*e) + I*a^4*sin(4*f*x + 4*e) + 2*I*a^4*sin(2*f*
x + 2*e) + a^4)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))) + 1) + 4*(8*a^4*cos(4*f*x + 4*e) + 16*a^4*cos(2*f*x + 2*e) + 8*I*a^4*sin(4*f*x + 4*e) +
16*I*a^4*sin(2*f*x + 2*e) - 31*a^4)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*(24*a^4*cos(4*f*
x + 4*e) + 48*a^4*cos(2*f*x + 2*e) + 24*I*a^4*sin(4*f*x + 4*e) + 48*I*a^4*sin(2*f*x + 2*e) + 35*a^4)*cos(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 105*(I*a^4*cos(4*f*x + 4*e) + 2*I*a^4*cos(2*f*x + 2*e) - a^4*sin
(4*f*x + 4*e) - 2*a^4*sin(2*f*x + 2*e) + I*a^4)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + s
in(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 1) + 105*(-I*a^4*cos(4*f*x + 4*e) - 2*I*a^4*cos(2*f*x + 2*e) + a^4*sin(4*f*x + 4*e) + 2*a^4*sin(2*f*x + 2*e
) - I*a^4)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 4*(8*I*a^4*cos(4*f*x + 4*e) +
16*I*a^4*cos(2*f*x + 2*e) - 8*a^4*sin(4*f*x + 4*e) - 16*a^4*sin(2*f*x + 2*e) - 31*I*a^4)*sin(3/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*(-24*I*a^4*cos(4*f*x + 4*e) - 48*I*a^4*cos(2*f*x + 2*e) + 24*a^4*sin(4*f
*x + 4*e) + 48*a^4*sin(2*f*x + 2*e) - 35*I*a^4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*
sqrt(c)/((-72*I*c^2*cos(4*f*x + 4*e) - 144*I*c^2*cos(2*f*x + 2*e) + 72*c^2*sin(4*f*x + 4*e) + 144*c^2*sin(2*f*
x + 2*e) - 72*I*c^2)*f)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {9}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(9/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(9/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

[In]

int((a + a*tan(e + f*x)*1i)^(9/2)/(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(9/2)/(c - c*tan(e + f*x)*1i)^(3/2), x)