3.12.27 \(\int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1127]

Optimal. Leaf size=76 \[ -\frac {2 i a \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}-\frac {2 a}{(i c+d) f \sqrt {c+d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3610, 3618, 65, 214} \begin {gather*} -\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {2 i a \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])/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((-2*I)*a*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f) - (2*a)/((I*c + d)*f*Sqrt[c + d
*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 3610

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

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]

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(158\) vs. \(2(76)=152\).
time = 2.62, size = 158, normalized size = 2.08 \begin {gather*} \frac {2 i a e^{-i e} (\cos (e)+i \sin (e)) \left (-\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 \cos (e+f x)+d \sin (e+f x))+\sqrt {c-i d} \cos (e+f x) \sqrt {c+d \tan (e+f x)}\right )}{(c-i d)^{3/2} f (c \cos (e+f x)+d \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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. 1014 vs. \(2 (64 ) = 128\).
time = 0.31, size = 1015, normalized size = 13.36

method	result	size
derivativedivides	\(\frac {a \left (\frac {\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}}}{2 \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}}}{2 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}}{c^{2}+d^{2}}-\frac {2 \left (-i c +d \right )}{\left (c^{2}+d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}\right )}{f}\)	\(1015\)
default	\(\frac {a \left (\frac {\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}}}{2 \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}}}{2 \left (\sqrt {c^{2}+d^{2}}+c \right ) \sqrt {c^{2}+d^{2}}}}{c^{2}+d^{2}}-\frac {2 \left (-i c +d \right )}{\left (c^{2}+d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}\right )}{f}\)	\(1015\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((-(2*c*d^4)/((c^2-d^2)^2>0)',
see `assume

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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. 550 vs. \(2 (62) = 124\).
time = 1.00, size = 550, normalized size = 7.24 \begin {gather*} \frac {{\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\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}} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - {\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\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}} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) + 8 \, {\left (i \, a e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a\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} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- \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 )}}\right )\, dx + \int \frac {\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 )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*(Integral(-I/(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)/(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. 192 vs. \(2 (62) = 124\).
time = 0.64, size = 192, normalized size = 2.53 \begin {gather*} -2 \, a {\left (\frac {1}{{\left (i \, c f + d f\right )} \sqrt {d \tan \left (f x + e\right ) + c}} - \frac {2 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 )}{{\left (c f - i \, d f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 14.43, size = 2500, normalized size = 32.89 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2) - ((((16*c*d^2*(c + d*tan(e + f*x))^(1/2)*((4*(-a^4*d^2*f^4*(3*c^2 - d
^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) - (32*a*d^2*(c^2*1i - d^2*1i))/(f*
(c^2 + d^2)))*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3)
)^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/2))/(f^2*(c^2 + d^2)^2))*((4*(-a^4*d^2*f^4*(3*c^2
- d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4)*(((96*a^4*c^2*d^4*f^4 - 16
*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f
^4 + 3*c^4*d^2*f^4))^(1/2))/4 + (log((a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2) - ((((16*c*d^2*(c + d*tan(e + f*x))^(1
/2)*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) -
(32*a*d^2*(c^2*1i - d^2*1i))/(f*(c^2 + d^2)))*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*
a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/2))/(f^2*(c^2 +
d^2)^2))*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(
1/2))/4)*(-((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f
^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2))/4 - log(((((16*c*d^2*(c + d*tan(e + f*x))^(1/2
)*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) + (3
2*a*d^2*(c^2*1i - d^2*1i))/(f*(c^2 + d^2)))*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*
c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/2))/(f^2*(c^2 + d^2
)^2))*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))
/4 + (a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2))*(((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) -
4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4))^(1/2) - log(((
((16*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2
*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) + (32*a*d^2*(c^2*1i - d^2*1i))/(f*(c^2 + d^2)))*(-(4*(-a^4*d^2*f^4*(3*c^2 - d
^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c +
d*tan(e + f*x))^(1/2))/(f^2*(c^2 + d^2)^2))*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2
*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2))*(-((96*a^4*c^2*d^4*f^4 - 16*a^
4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d
^4*f^4 + 48*c^4*d^2*f^4))^(1/2) + (log(((((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) -
4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*((c + d*tan(e + f
*x))^(1/2)*(16*a^2*d^10*f^3 + 32*a^2*c^2*d^8*f^3 - 32*a^2*c^6*d^4*f^3 - 16*a^2*c^8*d^2*f^3) - ((((96*a^4*c^2*d
^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 +
3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(((((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) -
4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(c + d*tan(e + f*
x))^(1/2)*(64*c*d^12*f^5 + 320*c^3*d^10*f^5 + 640*c^5*d^8*f^5 + 640*c^7*d^6*f^5 + 320*c^9*d^4*f^5 + 64*c^11*d^
2*f^5))/4 + 256*a*c^3*d^9*f^4 + 384*a*c^5*d^7*f^4 + 256*a*c^7*d^5*f^4 + 64*a*c^9*d^3*f^4 + 64*a*c*d^11*f^4))/4
))/4 + 8*a^3*d^9*f^2 + 24*a^3*c^2*d^7*f^2 + 24*a^3*c^4*d^5*f^2 + 8*a^3*c^6*d^3*f^2)*(((96*a^4*c^2*d^4*f^4 - 16
*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f
^4 + 3*c^4*d^2*f^4))^(1/2))/4 + (log(((-((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4
*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*((c + d*tan(e + f*
x))^(1/2)*(16*a^2*d^10*f^3 + 32*a^2*c^2*d^8*f^3 - 32*a^2*c^6*d^4*f^3 - 16*a^2*c^8*d^2*f^3) - ((-((96*a^4*c^2*d
^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 +
3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(((-((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) +
4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(c + d*tan(e + f
*x))^(1/2)*(64*c*d^12*f^5 + 320*c^3*d^10*f^5 + 640*c^5*d^8*f^5 + 640*c^7*d^6*f^5 + 320*c^9*d^4*f^5 + 64*c^11*d
^2*f^5))/4 + 256*a*c^3*d^9*f^4 + 384*a*c^5*d^7*f^4 + 256*a*c^7*d^5*f^4 + 64*a*c^9*d^3*f^4 + 64*a*c*d^11*f^4))/
4))/4 + 8*a^3*d^9*f^2 + 24*a^3*c^2*d^7*f^2 + 24*a^3*c^4*d^5*f^2 + 8*a^3*c^6*d^3*f^2)*(-((96*a^4*c^2*d^4*f^4 -
16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4
*f^4 + 3*c^4*d^2*f^4))^(1/2))/4 - log(8*a^3*d^9...

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