3.12.2 \(\int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx\) [1102]

Optimal. Leaf size=100 \[ -\frac {4 i a^2 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 100, 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 = {3624, 3609, 3618, 65, 214} \begin {gather*} -\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*Sqrt[c + d*Tan[e + f
*x]])/f - (2*a^2*(c + d*Tan[e + f*x])^(3/2))/(3*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 3609

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

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 3624

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

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx &=-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)} \, dx\\ &=\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \frac {2 a^2 (c-i d)+2 a^2 (i c+d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {\left (4 i a^4 (c-i d)^2\right ) \text {Subst}\left (\int \frac {1}{\left (4 a^4 (i c+d)^2+2 a^2 (c-i d) x\right ) \sqrt {c+\frac {d x}{2 a^2 (i c+d)}}} \, dx,x,2 a^2 (i c+d) \tan (e+f x)\right )}{f}\\ &=\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}-\frac {\left (16 a^6 (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {4 a^4 c (c-i d) (i c+d)}{d}+4 a^4 (i c+d)^2+\frac {4 a^4 (c-i d) (i c+d) x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac {4 i a^2 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\\ \end {align*}

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Mathematica [A]
time = 2.98, size = 155, normalized size = 1.55 \begin {gather*} -\frac {2 a^2 e^{-2 i e} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (6 i \sqrt {c-i d} d \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 )+\sqrt {c+d \tan (e+f x)} (c-6 i d+d \tan (e+f x))\right )}{3 d f (\cos (f x)+i \sin (f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((6*I)*Sqrt[c - I*d]*d*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e
+ f*x))))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(c - (6*I)*d + d*Tan[e + f*x])
))/(3*d*E^((2*I)*e)*f*(Cos[f*x] + I*Sin[f*x])^2)

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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. 703 vs. \(2 (84 ) = 168\).
time = 0.25, size = 704, normalized size = 7.04

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i d \sqrt {c +d \tan \left (f x +e \right )}-2 d \left (\frac {\frac {\left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i \sqrt {c^{2}+d^{2}}\, c +2 i c^{2}+2 i d^{2}+\frac {\left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \sqrt {c^{2}+d^{2}}+4 c}+\frac {\frac {\left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i \sqrt {c^{2}+d^{2}}\, c +2 i c^{2}+2 i d^{2}-\frac {\left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \sqrt {c^{2}+d^{2}}+4 c}\right )\right )}{f d}\)	\(704\)
default	\(\frac {2 a^{2} \left (-\frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i d \sqrt {c +d \tan \left (f x +e \right )}-2 d \left (\frac {\frac {\left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i \sqrt {c^{2}+d^{2}}\, c +2 i c^{2}+2 i d^{2}+\frac {\left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \sqrt {c^{2}+d^{2}}+4 c}+\frac {\frac {\left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 i \sqrt {c^{2}+d^{2}}\, c +2 i c^{2}+2 i d^{2}-\frac {\left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 \sqrt {c^{2}+d^{2}}+4 c}\right )\right )}{f d}\)	\(704\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^2/d*(-1/3*(c+d*tan(f*x+e))^(3/2)+2*I*d*(c+d*tan(f*x+e))^(1/2)-2*d*(1/(4*(c^2+d^2)^(1/2)+4*c)*(1/2*(-I*(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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*I*c^2+2*I*d^2+1/2*(-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*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)+4*c)*(1/2*(
I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+(2*(c^2+d^2)^(1/2)+2*c)^(1/2
)*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*I*c^2+2*I*d^2-1/2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c
+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^2*sqrt(d*tan(f*x + e) + c), 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. 413 vs. \(2 (83) = 166\).
time = 1.04, size = 413, normalized size = 4.13 \begin {gather*} \frac {3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 {a^{4} c - i \, a^{4} d}{f^{2}}} + {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2}}\right ) - 3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 {a^{4} c - i \, a^{4} d}{f^{2}}} + {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2}}\right ) - 2 \, {\left (a^{2} c - 5 i \, a^{2} d + {\left (a^{2} c - 7 i \, a^{2} d\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}}}{3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^4*c - I*a^4*d)/f^2)*log(2*(a^2*c + (I*f*e^(2*I*f*x + 2*I*e) +
I*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(-(a^4*c - I*a^4*d)/f^2) +
(a^2*c - I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - 3*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^
4*c - I*a^4*d)/f^2)*log(2*(a^2*c + (-I*f*e^(2*I*f*x + 2*I*e) - I*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(-(a^4*c - I*a^4*d)/f^2) + (a^2*c - I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*
f*x - 2*I*e)/a^2) - 2*(a^2*c - 5*I*a^2*d + (a^2*c - 7*I*a^2*d)*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)))/(d*f*e^(2*I*f*x + 2*I*e) + d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-2*I*sqrt(c + d*tan(e + f*x))*tan(e +
f*x), x) + Integral(-sqrt(c + d*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. 227 vs. \(2 (83) = 166\).
time = 0.57, size = 227, normalized size = 2.27 \begin {gather*} -\frac {8 \, {\left (-i \, a^{2} c - a^{2} d\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 )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} d^{2} f^{2} - 6 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} d^{3} f^{2}\right )}}{3 \, d^{3} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 7.07, size = 90, normalized size = 0.90 \begin {gather*} \frac {a^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,4{}\mathrm {i}}{f}-\frac {2\,a^2\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d\,f}-\frac {2\,\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atanh}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d+c\,1{}\mathrm {i}}}\right )\,\sqrt {d+c\,1{}\mathrm {i}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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