3.12.25 \(\int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1125]

Optimal. Leaf size=139 \[ -\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f} \]

[Out]

-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f+4*a^3*c*(c+d*tan(f*x+e))^(1/2)/d^2/(I*c
+d)/f+2*(c+I*d)*(a^3+I*a^3*tan(f*x+e))/(c-I*d)/d/f/(c+d*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3634, 3673, 3618, 65, 214} \begin {gather*} \frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 f (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f) + (2*(c + I*d)*(a^3 + I*a^3*T
an[e + f*x]))/((c - I*d)*d*f*Sqrt[c + d*Tan[e + f*x]]) + (4*a^3*c*Sqrt[c + d*Tan[e + f*x]])/(d^2*(I*c + d)*f)

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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] /;
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] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3673

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {(a+i a \tan (e+f x)) \left (-a^2 (c+2 i d)+i a^2 c \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{d (i c+d)}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f}-\frac {2 \int \frac {-2 i a^3 d+2 a^3 d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{d (i c+d)}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f}+\frac {\left (8 a^6 d\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {x}{2 a^3}} \left (4 a^6 d^2-2 i a^3 d x\right )} \, dx,x,2 a^3 d \tan (e+f x)\right )}{(c-i d) f}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f}+\frac {\left (32 a^9 d\right ) \text {Subst}\left (\int \frac {1}{4 i a^6 c d+4 a^6 d^2-4 i a^6 d x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(c-i d) f}\\ &=-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f}\\ \end {align*}

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Mathematica [A]
time = 4.37, size = 219, normalized size = 1.58 \begin {gather*} \frac {a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (-\frac {8 i e^{-3 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2}}+\frac {2 (\cos (3 e)-i \sin (3 e)) \left (\left (-2 i c^2+c d+i d^2\right ) \cos (e+f x)+(-i c-d) d \sin (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{(c-i d) d^2 (c \cos (e+f x)+d \sin (e+f x))}\right )}{f (\cos (f x)+i \sin (f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I
)*(e + f*x)))]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*E^((3*I)*e)) + (2*(Cos[3*e] - I*Sin[3*e])*(((-2*I)*c^2 + c*d +
I*d^2)*Cos[e + f*x] + ((-I)*c - d)*d*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/((c - I*d)*d^2*(c*Cos[e + f*x] +
d*Sin[e + f*x]))))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1055 vs. \(2 (122 ) = 244\).
time = 0.29, size = 1056, normalized size = 7.60

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-i \sqrt {c +d \tan \left (f x +e \right )}-\frac {i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}}{\left (c^{2}+d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}-\frac {4 d^{2} \left (\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i c^{2} \sqrt {c^{2}+d^{2}}-2 i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}-2 i c \,d^{2}-4 c d \sqrt {c^{2}+d^{2}}-4 c^{2} d -\frac {\left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}+\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i c^{2} \sqrt {c^{2}+d^{2}}-2 i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}-2 i c \,d^{2}-4 c d \sqrt {c^{2}+d^{2}}-4 c^{2} d +\frac {\left (-i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}+\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}\right )}{c^{2}+d^{2}}\right )}{f \,d^{2}}\)	\(1056\)
default	\(\frac {2 a^{3} \left (-i \sqrt {c +d \tan \left (f x +e \right )}-\frac {i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}}{\left (c^{2}+d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}-\frac {4 d^{2} \left (\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i c^{2} \sqrt {c^{2}+d^{2}}-2 i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}-2 i c \,d^{2}-4 c d \sqrt {c^{2}+d^{2}}-4 c^{2} d -\frac {\left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}+\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i c^{2} \sqrt {c^{2}+d^{2}}-2 i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}-2 i c \,d^{2}-4 c d \sqrt {c^{2}+d^{2}}-4 c^{2} d +\frac {\left (-i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{2}+\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}\right )}{c^{2}+d^{2}}\right )}{f \,d^{2}}\)	\(1056\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^3/d^2*(-I*(c+d*tan(f*x+e))^(1/2)-(I*c^3-3*I*c*d^2-3*c^2*d+d^3)/(c^2+d^2)/(c+d*tan(f*x+e))^(1/2)-4*d^2/(c
^2+d^2)*(1/4/((c^2+d^2)^(1/2)+c)/(c^2+d^2)^(1/2)*(1/2*(I*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+I*(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*d-2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)+(c^2+d^2)^(1/2))+2*(2*I*(c^2+d^2)^(1/2)*c^2-2*I*(c^2+d^2)^(1/2)*d^2+2*I*c^3-2*I*c*d^2-4*c*d*(c^2+d^2)^(1/2)
-4*c^2*d-1/2*(I*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-I*(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*d^2-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d-2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d)
*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1
/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+1/4/((c^2+d^2)^(1/2)+c)/(c^2+d^2)^(1/2)*(1/2*(-I*(c^2+d^2)^(1/
2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2+(c^
2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d)*ln(d*tan(f*x+e)+c-(c+d*tan(f
*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(2*I*(c^2+d^2)^(1/2)*c^2-2*I*(c^2+d^2)^(1/2)*d^2
+2*I*c^3-2*I*c*d^2-4*c*d*(c^2+d^2)^(1/2)-4*c^2*d+1/2*(-I*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-I*(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2+(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*d+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(
(2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 620 vs. \(2 (119) = 238\).
time = 1.24, size = 620, normalized size = 4.46 \begin {gather*} \frac {\sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) + 16 \, {\left (-i \, a^{3} c^{2} + a^{3} c d + {\left (-i \, a^{3} c^{2} + i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(64*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((c^2*d^2 - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I
*e) + (c^2*d^2 + d^4)*f)*log(1/4*(8*a^3*c + sqrt(64*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((I*c^2
+ 2*c*d - I*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*c^2 + 2*c*d - I*d^2)*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)) + 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - sqrt(64
*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((c^2*d^2 - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^2*d
^2 + d^4)*f)*log(1/4*(8*a^3*c + sqrt(64*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((-I*c^2 - 2*c*d + I
*d^2)*f*e^(2*I*f*x + 2*I*e) + (-I*c^2 - 2*c*d + I*d^2)*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)) + 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) + 16*(-I*a^3*c^2 +
a^3*c*d + (-I*a^3*c^2 + I*a^3*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)))/((c^2*d^2 - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^2*d^2 + d^4)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 247 vs. \(2 (119) = 238\).
time = 0.78, size = 247, normalized size = 1.78 \begin {gather*} \frac {16 \, a^{3} \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 )}{{\left (-i \, c f - d 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 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3}}{d^{2} f} + \frac {2 \, {\left (-i \, a^{3} c^{2} + 2 \, a^{3} c d + i \, a^{3} d^{2}\right )}}{{\left (c d^{2} f - i \, d^{3} f\right )} \sqrt {d \tan \left (f x + e\right ) + c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.59, size = 182, normalized size = 1.31 \begin {gather*} -\frac {a^3\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,8{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\left (a^3\,c^2+a^3\,c\,d\,2{}\mathrm {i}-a^3\,d^2\right )\,2{}\mathrm {i}}{d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^4*f^2 + 2*d^4*f^2 + 4*c^2*d^2*f^2))/(2*f*(d*1i - c)^(3/2)*(c^3*f +
d^3*f*1i + c*d^2*f + c^2*d*f*1i)))*8i)/(f*(d*1i - c)^(3/2)) - (a^3*(c + d*tan(e + f*x))^(1/2)*2i)/(d^2*f) - ((
a^3*c^2 - a^3*d^2 + a^3*c*d*2i)*2i)/(d^2*f*(c - d*1i)*(c + d*tan(e + f*x))^(1/2))

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