3.12.24 \(\int \frac {1}{(a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}} \, dx\) [1124]

Optimal. Leaf size=298 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 \sqrt {c-i d} f}+\frac {\left (2 i c^3-8 c^2 d-13 i c d^2+12 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 (c+i d)^{7/2} f}-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

1/16*(2*I*c^3-8*c^2*d-13*I*c*d^2+12*d^3)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a^3/(c+I*d)^(7/2)/f-1/8
*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^3/f/(c-I*d)^(1/2)-1/6*(c+d*tan(f*x+e))^(1/2)/(I*c-d)/f/(a+I
*a*tan(f*x+e))^3+1/24*(3*I*c-8*d)*(c+d*tan(f*x+e))^(1/2)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^2+1/16*(2*c^2+7*I*c*
d-10*d^2)*(c+d*tan(f*x+e))^(1/2)/(I*c-d)^3/f/(a^3+I*a^3*tan(f*x+e))

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Rubi [A]
time = 0.66, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3640, 3677, 3620, 3618, 65, 214} \begin {gather*} \frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 f (-d+i c)^3 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (2 i c^3-8 c^2 d-13 i c d^2+12 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 f (c+i d)^{7/2}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 f \sqrt {c-i d}}+\frac {(-8 d+3 i c) \sqrt {c+d \tan (e+f x)}}{24 a f (c+i d)^2 (a+i a \tan (e+f x))^2}-\frac {\sqrt {c+d \tan (e+f x)}}{6 f (-d+i c) (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

((-1/8*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^3*Sqrt[c - I*d]*f) + (((2*I)*c^3 - 8*c^2*d - (13
*I)*c*d^2 + 12*d^3)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(16*a^3*(c + I*d)^(7/2)*f) - Sqrt[c + d*T
an[e + f*x]]/(6*(I*c - d)*f*(a + I*a*Tan[e + f*x])^3) + (((3*I)*c - 8*d)*Sqrt[c + d*Tan[e + f*x]])/(24*a*(c +
I*d)^2*f*(a + I*a*Tan[e + f*x])^2) + ((2*c^2 + (7*I)*c*d - 10*d^2)*Sqrt[c + d*Tan[e + f*x]])/(16*(I*c - d)^3*f
*(a^3 + I*a^3*Tan[e + f*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 214

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

Rule 3618

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

Rule 3620

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

Rule 3640

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

Rule 3677

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}-\frac {\int \frac {-\frac {1}{2} a (6 i c-11 d)-\frac {5}{2} i a d \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx}{6 a^2 (i c-d)}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {-\frac {3}{2} a^2 \left (4 c^2+11 i c d-12 d^2\right )-\frac {3}{2} a^2 (3 c+8 i d) d \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{24 a^4 (c+i d)^2}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\int \frac {\frac {3}{2} a^3 (i c-2 d) \left (4 c^2+6 i c d-7 d^2\right )+\frac {3}{2} a^3 d \left (2 i c^2-7 c d-10 i d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{48 a^6 (i c-d)^3}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{16 a^3}+\frac {\left (2 c^3+8 i c^2 d-13 c d^2-12 i d^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{32 a^3 (c+i d)^3}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{16 a^3 f}-\frac {\left (2 i c^3-8 c^2 d-13 i c d^2+12 d^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{32 a^3 (c+i d)^3 f}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{8 a^3 d f}-\frac {\left (2 c^3+8 i c^2 d-13 c d^2-12 i d^3\right ) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{16 a^3 (c+i d)^3 d f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 \sqrt {c-i d} f}+\frac {\left (2 i c^3-8 c^2 d-13 i c d^2+12 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 (c+i d)^{7/2} f}-\frac {\sqrt {c+d \tan (e+f x)}}{6 (i c-d) f (a+i a \tan (e+f x))^3}+\frac {(3 i c-8 d) \sqrt {c+d \tan (e+f x)}}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2}+\frac {\left (2 c^2+7 i c d-10 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 3.42, size = 324, normalized size = 1.09 \begin {gather*} \frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (-\frac {2 \left (\sqrt {-c+i d} \left (-2 i c^3+8 c^2 d+13 i c d^2-12 d^3\right ) \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-2 i (-c-i d)^{7/2} \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )\right ) (\cos (3 e)+i \sin (3 e))}{(-c-i d)^{7/2} \sqrt {-c+i d}}+\frac {2 \cos (e+f x) (i \cos (3 f x)+\sin (3 f x)) \left (7 c^2+19 i c d-12 d^2+\left (13 c^2+40 i c d-42 d^2\right ) \cos (2 (e+f x))+i \left (9 c^2+32 i c d-38 d^2\right ) \sin (2 (e+f x))\right ) \sqrt {c+d \tan (e+f x)}}{3 (c+i d)^3}\right )}{32 f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*((-2*(Sqrt[-c + I*d]*((-2*I)*c^3 + 8*c^2*d + (13*I)*c*d^2 - 12*d^3)*
ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c - I*d]] - (2*I)*(-c - I*d)^(7/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[
-c + I*d]])*(Cos[3*e] + I*Sin[3*e]))/((-c - I*d)^(7/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[3*f*x] + Sin[3
*f*x])*(7*c^2 + (19*I)*c*d - 12*d^2 + (13*c^2 + (40*I)*c*d - 42*d^2)*Cos[2*(e + f*x)] + I*(9*c^2 + (32*I)*c*d
- 38*d^2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/(3*(c + I*d)^3)))/(32*f*(a + I*a*Tan[e + f*x])^3)

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Maple [A]
time = 0.36, size = 357, normalized size = 1.20

method result size
derivativedivides \(\frac {2 d^{4} \left (\frac {\frac {-\frac {d \left (7 i c d +2 c^{2}-10 d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 d \left (15 i c^{2} d -19 i d^{3}+3 c^{3}-31 c \,d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}-\frac {d \left (13 i c^{3} d -45 i c \,d^{3}+2 c^{4}-38 c^{2} d^{2}+18 d^{4}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (2 i c^{3}-13 i c \,d^{2}-8 c^{2} d +12 d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4} \sqrt {i d -c}}\right )}{f \,a^{3}}\) \(357\)
default \(\frac {2 d^{4} \left (\frac {\frac {-\frac {d \left (7 i c d +2 c^{2}-10 d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 d \left (15 i c^{2} d -19 i d^{3}+3 c^{3}-31 c \,d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}-\frac {d \left (13 i c^{3} d -45 i c \,d^{3}+2 c^{4}-38 c^{2} d^{2}+18 d^{4}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (2 i c^{3}-13 i c \,d^{2}-8 c^{2} d +12 d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4} \sqrt {i d -c}}\right )}{f \,a^{3}}\) \(357\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(1/16/d^4*((-1/2*d*(2*c^2+7*I*c*d-10*d^2)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(5/2)+2/3
*d*(-31*c*d^2+15*I*c^2*d-19*I*d^3+3*c^3)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(3/2)-1/2*d*(13*I*c^3*
d-45*I*c*d^3+2*c^4-38*c^2*d^2+18*d^4)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d
)^3-1/2*(2*I*c^3-8*c^2*d-13*I*c*d^2+12*d^3)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e
))^(1/2)/(-I*d-c)^(1/2)))+1/16*I/d^4/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1777 vs. \(2 (248) = 496\).
time = 1.42, size = 1777, normalized size = 5.96 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/192*(48*(I*a^3*c^3 - 3*a^3*c^2*d - 3*I*a^3*c*d^2 + a^3*d^3)*f*sqrt(1/64*I/((-I*a^6*c - a^6*d)*f^2))*e^(6*I*f
*x + 6*I*e)*log(-2*(8*((I*a^3*c + a^3*d)*f*e^(2*I*f*x + 2*I*e) + (I*a^3*c + a^3*d)*f)*sqrt(((c - I*d)*e^(2*I*f
*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I/((-I*a^6*c - a^6*d)*f^2)) - (c - I*d)*e^(2*I*f*x
 + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 48*(-I*a^3*c^3 + 3*a^3*c^2*d + 3*I*a^3*c*d^2 - a^3*d^3)*f*sqrt(1/64*I/(
(-I*a^6*c - a^6*d)*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(8*((-I*a^3*c - a^3*d)*f*e^(2*I*f*x + 2*I*e) + (-I*a^3*c -
 a^3*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I/((-I*a^6*c -
a^6*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 3*(-I*a^3*c^3 + 3*a^3*c^2*d + 3*I*a^
3*c*d^2 - a^3*d^3)*f*sqrt(-(-4*I*c^6 + 32*c^5*d + 116*I*c^4*d^2 - 256*c^3*d^3 - 361*I*c^2*d^4 + 312*c*d^5 + 14
4*I*d^6)/((-I*a^6*c^7 + 7*a^6*c^6*d + 21*I*a^6*c^5*d^2 - 35*a^6*c^4*d^3 - 35*I*a^6*c^3*d^4 + 21*a^6*c^2*d^5 +
7*I*a^6*c*d^6 - a^6*d^7)*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*(2*I*c^4 - 10*c^3*d - 21*I*c^2*d^2 + 25*c*d^3 + 12
*I*d^4 + ((a^3*c^4 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f*e^(2*I*f*x + 2*I*e) + (a^3*c^4
 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/
(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-4*I*c^6 + 32*c^5*d + 116*I*c^4*d^2 - 256*c^3*d^3 - 361*I*c^2*d^4 + 312*c*d^
5 + 144*I*d^6)/((-I*a^6*c^7 + 7*a^6*c^6*d + 21*I*a^6*c^5*d^2 - 35*a^6*c^4*d^3 - 35*I*a^6*c^3*d^4 + 21*a^6*c^2*
d^5 + 7*I*a^6*c*d^6 - a^6*d^7)*f^2)) + (2*I*c^4 - 8*c^3*d - 13*I*c^2*d^2 + 12*c*d^3)*e^(2*I*f*x + 2*I*e))*e^(-
2*I*f*x - 2*I*e)/((a^3*c^4 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f)) + 3*(I*a^3*c^3 - 3*a
^3*c^2*d - 3*I*a^3*c*d^2 + a^3*d^3)*f*sqrt(-(-4*I*c^6 + 32*c^5*d + 116*I*c^4*d^2 - 256*c^3*d^3 - 361*I*c^2*d^4
 + 312*c*d^5 + 144*I*d^6)/((-I*a^6*c^7 + 7*a^6*c^6*d + 21*I*a^6*c^5*d^2 - 35*a^6*c^4*d^3 - 35*I*a^6*c^3*d^4 +
21*a^6*c^2*d^5 + 7*I*a^6*c*d^6 - a^6*d^7)*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*(2*I*c^4 - 10*c^3*d - 21*I*c^2*d^
2 + 25*c*d^3 + 12*I*d^4 - ((a^3*c^4 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f*e^(2*I*f*x +
2*I*e) + (a^3*c^4 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2
*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-4*I*c^6 + 32*c^5*d + 116*I*c^4*d^2 - 256*c^3*d^3 - 361*I*c
^2*d^4 + 312*c*d^5 + 144*I*d^6)/((-I*a^6*c^7 + 7*a^6*c^6*d + 21*I*a^6*c^5*d^2 - 35*a^6*c^4*d^3 - 35*I*a^6*c^3*
d^4 + 21*a^6*c^2*d^5 + 7*I*a^6*c*d^6 - a^6*d^7)*f^2)) + (2*I*c^4 - 8*c^3*d - 13*I*c^2*d^2 + 12*c*d^3)*e^(2*I*f
*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((a^3*c^4 + 4*I*a^3*c^3*d - 6*a^3*c^2*d^2 - 4*I*a^3*c*d^3 + a^3*d^4)*f)) + 2
*(2*c^2 + 4*I*c*d - 2*d^2 + (11*c^2 + 36*I*c*d - 40*d^2)*e^(6*I*f*x + 6*I*e) + (18*c^2 + 55*I*c*d - 52*d^2)*e^
(4*I*f*x + 4*I*e) + (9*c^2 + 23*I*c*d - 14*d^2)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c +
 I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/((-I*a^3*c^3 + 3*a^3*c^2*d + 3*I*a^3*c*d^2 - a^3*d^3)*f
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {1}{\sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + i \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

I*Integral(1/(sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3 - 3*I*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2 - 3*sqrt
(c + d*tan(e + f*x))*tan(e + f*x) + I*sqrt(c + d*tan(e + f*x))), x)/a**3

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 669 vs. \(2 (248) = 496\).
time = 0.83, size = 669, normalized size = 2.24 \begin {gather*} -\frac {2 \, {\left (2 \, c^{3} + 8 i \, c^{2} d - 13 \, c d^{2} - 12 i \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{-16 \, {\left (i \, a^{3} c^{3} f - 3 \, a^{3} c^{2} d f - 3 i \, a^{3} c d^{2} f + a^{3} d^{3} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (-6 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} d + 12 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} d - 6 i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{4} d + 21 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c d^{2} - 60 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} d^{2} + 39 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{3} d^{2} + 30 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} d^{3} - 124 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{3} + 114 i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d^{3} + 76 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{4} - 135 \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{4} - 54 i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{5}\right )}}{-96 \, {\left (i \, a^{3} c^{3} f - 3 \, a^{3} c^{2} d f - 3 i \, a^{3} c d^{2} f + a^{3} d^{3} f\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}} + \frac {i \, \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{4 \, a^{3} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-2*(2*c^3 + 8*I*c^2*d - 13*c*d^2 - 12*I*d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan
(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(
-2*c + 2*sqrt(c^2 + d^2))))/((-16*I*a^3*c^3*f + 48*a^3*c^2*d*f + 48*I*a^3*c*d^2*f - 16*a^3*d^3*f)*sqrt(-2*c +
2*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + 2*(-6*I*(d*tan(f*x + e) + c)^(5/2)*c^2*d + 12*I*(d*tan(f
*x + e) + c)^(3/2)*c^3*d - 6*I*sqrt(d*tan(f*x + e) + c)*c^4*d + 21*(d*tan(f*x + e) + c)^(5/2)*c*d^2 - 60*(d*ta
n(f*x + e) + c)^(3/2)*c^2*d^2 + 39*sqrt(d*tan(f*x + e) + c)*c^3*d^2 + 30*I*(d*tan(f*x + e) + c)^(5/2)*d^3 - 12
4*I*(d*tan(f*x + e) + c)^(3/2)*c*d^3 + 114*I*sqrt(d*tan(f*x + e) + c)*c^2*d^3 + 76*(d*tan(f*x + e) + c)^(3/2)*
d^4 - 135*sqrt(d*tan(f*x + e) + c)*c*d^4 - 54*I*sqrt(d*tan(f*x + e) + c)*d^5)/((-96*I*a^3*c^3*f + 288*a^3*c^2*
d*f + 288*I*a^3*c*d^2*f - 96*a^3*d^3*f)*(d*tan(f*x + e) - I*d)^3) + 1/4*I*arctan(2*(sqrt(d*tan(f*x + e) + c)*c
 - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^
2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a^3*sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt
(c^2 + d^2)) + 1))

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Mupad [B]
time = 11.14, size = 2500, normalized size = 8.39 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a^3*d^14*f*240i - ((-(140*d^11 - c*d^10*140i + 35*c^2*d^9 - c^3*d^8*245i - 280*c^4*d^7 + c^5*d^6*168i + 56
*c^6*d^5 - c^7*d^4*8i - a^6*c^6*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((649*c^2*d^14)/1024 - (9*d^16)/64 + (85*c^4*d
^12)/1024 + (119*c^6*d^10)/1024 + (15*c^8*d^8)/1024 - (c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^12*f^4 + a^12*d^
12*f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d
^2*f^4) - (((69*c*d^15)/128 - (55*c^3*d^13)/512 + (57*c^5*d^11)/256 + (61*c^7*d^9)/512 + (5*c^9*d^7)/128 + (c^
11*d^5)/128)*1i)/(a^12*c^12*f^4 + a^12*d^12*f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*
f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^4)) + ((((1225*c^2*d^15)/4 - 35*d^17 + (35*c^4*d^13)/4 + (427*c^
6*d^11)/4 + (197*c^8*d^9)/4 + 12*c^10*d^7 + 2*c^12*d^5)*1i)/(a^6*c^12*f^2 + a^6*d^12*f^2 + 6*a^6*c^2*d^10*f^2
+ 15*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2) + (175*c*d^16 - (735*c^3*
d^14)/4 + (203*c^5*d^12)/4 + (83*c^7*d^10)/4 + (85*c^9*d^8)/4 + 12*c^11*d^6 + 2*c^13*d^4)/(a^6*c^12*f^2 + a^6*
d^12*f^2 + 6*a^6*c^2*d^10*f^2 + 15*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*
f^2))^2)^(1/2)*4i + a^6*d^6*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((649*c^2*d^14)/1024 - (9*d^16)/64 + (85*c^4*d^12)
/1024 + (119*c^6*d^10)/1024 + (15*c^8*d^8)/1024 - (c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^12*f^4 + a^12*d^12*f
^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f
^4) - (((69*c*d^15)/128 - (55*c^3*d^13)/512 + (57*c^5*d^11)/256 + (61*c^7*d^9)/512 + (5*c^9*d^7)/128 + (c^11*d
^5)/128)*1i)/(a^12*c^12*f^4 + a^12*d^12*f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4
+ 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^4)) + ((((1225*c^2*d^15)/4 - 35*d^17 + (35*c^4*d^13)/4 + (427*c^6*d^
11)/4 + (197*c^8*d^9)/4 + 12*c^10*d^7 + 2*c^12*d^5)*1i)/(a^6*c^12*f^2 + a^6*d^12*f^2 + 6*a^6*c^2*d^10*f^2 + 15
*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2) + (175*c*d^16 - (735*c^3*d^14
)/4 + (203*c^5*d^12)/4 + (83*c^7*d^10)/4 + (85*c^9*d^8)/4 + 12*c^11*d^6 + 2*c^13*d^4)/(a^6*c^12*f^2 + a^6*d^12
*f^2 + 6*a^6*c^2*d^10*f^2 + 15*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2)
)^2)^(1/2)*4i + 24*a^6*c*d^5*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((649*c^2*d^14)/1024 - (9*d^16)/64 + (85*c^4*d^12
)/1024 + (119*c^6*d^10)/1024 + (15*c^8*d^8)/1024 - (c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^12*f^4 + a^12*d^12*
f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*
f^4) - (((69*c*d^15)/128 - (55*c^3*d^13)/512 + (57*c^5*d^11)/256 + (61*c^7*d^9)/512 + (5*c^9*d^7)/128 + (c^11*
d^5)/128)*1i)/(a^12*c^12*f^4 + a^12*d^12*f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4
 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^4)) + ((((1225*c^2*d^15)/4 - 35*d^17 + (35*c^4*d^13)/4 + (427*c^6*d
^11)/4 + (197*c^8*d^9)/4 + 12*c^10*d^7 + 2*c^12*d^5)*1i)/(a^6*c^12*f^2 + a^6*d^12*f^2 + 6*a^6*c^2*d^10*f^2 + 1
5*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2) + (175*c*d^16 - (735*c^3*d^1
4)/4 + (203*c^5*d^12)/4 + (83*c^7*d^10)/4 + (85*c^9*d^8)/4 + 12*c^11*d^6 + 2*c^13*d^4)/(a^6*c^12*f^2 + a^6*d^1
2*f^2 + 6*a^6*c^2*d^10*f^2 + 15*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2
))^2)^(1/2) + 24*a^6*c^5*d*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((649*c^2*d^14)/1024 - (9*d^16)/64 + (85*c^4*d^12)/
1024 + (119*c^6*d^10)/1024 + (15*c^8*d^8)/1024 - (c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^12*f^4 + a^12*d^12*f^
4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^
4) - (((69*c*d^15)/128 - (55*c^3*d^13)/512 + (57*c^5*d^11)/256 + (61*c^7*d^9)/512 + (5*c^9*d^7)/128 + (c^11*d^
5)/128)*1i)/(a^12*c^12*f^4 + a^12*d^12*f^4 + 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 +
 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^4)) + ((((1225*c^2*d^15)/4 - 35*d^17 + (35*c^4*d^13)/4 + (427*c^6*d^1
1)/4 + (197*c^8*d^9)/4 + 12*c^10*d^7 + 2*c^12*d^5)*1i)/(a^6*c^12*f^2 + a^6*d^12*f^2 + 6*a^6*c^2*d^10*f^2 + 15*
a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2) + (175*c*d^16 - (735*c^3*d^14)
/4 + (203*c^5*d^12)/4 + (83*c^7*d^10)/4 + (85*c^9*d^8)/4 + 12*c^11*d^6 + 2*c^13*d^4)/(a^6*c^12*f^2 + a^6*d^12*
f^2 + 6*a^6*c^2*d^10*f^2 + 15*a^6*c^4*d^8*f^2 + 20*a^6*c^6*d^6*f^2 + 15*a^6*c^8*d^4*f^2 + 6*a^6*c^10*d^2*f^2))
^2)^(1/2) - a^6*c^2*d^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((649*c^2*d^14)/1024 - (9*d^16)/64 + (85*c^4*d^12)/102
4 + (119*c^6*d^10)/1024 + (15*c^8*d^8)/1024 - (c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^12*f^4 + a^12*d^12*f^4 +
 6*a^12*c^2*d^10*f^4 + 15*a^12*c^4*d^8*f^4 + 20*a^12*c^6*d^6*f^4 + 15*a^12*c^8*d^4*f^4 + 6*a^12*c^10*d^2*f^4)
- (((69*c*d^15)/128 - (55*c^3*d^13)/512 + (57*c^5*d^11)/256 + (61*c^7*d^9)/512 + (5*c^9*d^7)/128 + (c^11*d^5)/
128)*1i)/(a^12*c^12*f^4 + a^12*d^12*f^4 + 6*a^1...

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