3.1108 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=131 \[ -\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^2 (d+i c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

[Out]

-4*I*a^2*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+4*a^2*(I*c+d)*(c+d*tan(f*x+e))^(1/2)/f+
4/3*I*a^2*(c+d*tan(f*x+e))^(3/2)/f-2/5*a^2*(c+d*tan(f*x+e))^(5/2)/d/f

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3543, 3528, 3537, 63, 208} \[ -\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^2 (d+i c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (4*a^2*(I*c + d)*Sqrt[c + d*T
an[e + f*x]])/f + (((4*I)/3)*a^2*(c + d*Tan[e + f*x])^(3/2))/f - (2*a^2*(c + d*Tan[e + f*x])^(5/2))/(5*d*f)

Rule 63

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 208

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

Rule 3528

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 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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 3543

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 (c+d \tan (e+f x))^{3/2} \, dx &=-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2} \, dx\\ &=\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \sqrt {c+d \tan (e+f x)} \left (2 a^2 (c-i d)+2 a^2 (i c+d) \tan (e+f x)\right ) \, dx\\ &=\frac {4 a^2 (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \frac {2 a^2 (c-i d)^2+2 i a^2 (c-i d)^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {4 a^2 (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {\left (4 i a^4 (c-i d)^4\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-4 a^4 (c-i d)^4+2 a^2 (c-i d)^2 x\right ) \sqrt {c-\frac {i d x}{2 a^2 (c-i d)^2}}} \, dx,x,2 i a^2 (c-i d)^2 \tan (e+f x)\right )}{f}\\ &=\frac {4 a^2 (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left (16 a^6 (c-i d)^6\right ) \operatorname {Subst}\left (\int \frac {1}{-4 a^4 (c-i d)^4-\frac {4 i a^4 c (c-i d)^4}{d}+\frac {4 i a^4 (c-i d)^4 x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac {4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 a^2 (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 4.47, size = 221, normalized size = 1.69 \[ \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\frac {(\cos (2 e)-i \sin (2 e)) \sec ^2(e+f x) \sqrt {c+d \tan (e+f x)} \left (\left (3 c^2-40 i c d-33 d^2\right ) \cos (2 (e+f x))+3 c^2+2 d (3 c-5 i d) \sin (2 (e+f x))-40 i c d-27 d^2\right )}{15 d}-4 i e^{-2 i e} (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[e + f*x] + I*Sin[e + f*x])^2*(((-4*I)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)
)))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^((2*I)*e) - (Sec[e + f*x]^2*(Cos[2*e] - I*Sin[2*e])*(3*c^2 -
(40*I)*c*d - 27*d^2 + (3*c^2 - (40*I)*c*d - 33*d^2)*Cos[2*(e + f*x)] + 2*(3*c - (5*I)*d)*d*Sin[2*(e + f*x)])*S
qrt[c + d*Tan[e + f*x]])/(15*d)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

________________________________________________________________________________________

fricas [B]  time = 0.60, size = 650, normalized size = 4.96 \[ -\frac {15 \, {\left (d f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 \, a^{4} c^{3} - 48 i \, a^{4} c^{2} d - 48 \, a^{4} c d^{2} + 16 i \, a^{4} d^{3}}{f^{2}}} \log \left (-\frac {{\left (4 i \, a^{2} c^{2} + 4 \, a^{2} c d + {\left (f e^{\left (2 i \, f x + 2 i \, e\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 {16 \, a^{4} c^{3} - 48 i \, a^{4} c^{2} d - 48 \, a^{4} c d^{2} + 16 i \, a^{4} d^{3}}{f^{2}}} + {\left (4 i \, a^{2} c^{2} + 8 \, a^{2} c d - 4 i \, a^{2} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (-i \, a^{2} c - a^{2} d\right )}}\right ) - 15 \, {\left (d f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 \, a^{4} c^{3} - 48 i \, a^{4} c^{2} d - 48 \, a^{4} c d^{2} + 16 i \, a^{4} d^{3}}{f^{2}}} \log \left (-\frac {{\left (4 i \, a^{2} c^{2} + 4 \, a^{2} c d - {\left (f e^{\left (2 i \, f x + 2 i \, e\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 {16 \, a^{4} c^{3} - 48 i \, a^{4} c^{2} d - 48 \, a^{4} c d^{2} + 16 i \, a^{4} d^{3}}{f^{2}}} + {\left (4 i \, a^{2} c^{2} + 8 \, a^{2} c d - 4 i \, a^{2} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (-i \, a^{2} c - a^{2} d\right )}}\right ) + 8 \, {\left (3 \, a^{2} c^{2} - 34 i \, a^{2} c d - 23 \, a^{2} d^{2} + {\left (3 \, a^{2} c^{2} - 46 i \, a^{2} c d - 43 \, a^{2} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (3 \, a^{2} c^{2} - 40 i \, a^{2} c d - 27 \, a^{2} 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}}}{60 \, {\left (d f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*
a^4*c*d^2 + 16*I*a^4*d^3)/f^2)*log(-1/2*(4*I*a^2*c^2 + 4*a^2*c*d + (f*e^(2*I*f*x + 2*I*e) + 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(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*a^4*c*d^2 +
16*I*a^4*d^3)/f^2) + (4*I*a^2*c^2 + 8*a^2*c*d - 4*I*a^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a^
2*c - a^2*d)) - 15*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^3 - 48*I*a^4*c^
2*d - 48*a^4*c*d^2 + 16*I*a^4*d^3)/f^2)*log(-1/2*(4*I*a^2*c^2 + 4*a^2*c*d - (f*e^(2*I*f*x + 2*I*e) + 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(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*a^
4*c*d^2 + 16*I*a^4*d^3)/f^2) + (4*I*a^2*c^2 + 8*a^2*c*d - 4*I*a^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*
e)/(-I*a^2*c - a^2*d)) + 8*(3*a^2*c^2 - 34*I*a^2*c*d - 23*a^2*d^2 + (3*a^2*c^2 - 46*I*a^2*c*d - 43*a^2*d^2)*e^
(4*I*f*x + 4*I*e) + 2*(3*a^2*c^2 - 40*I*a^2*c*d - 27*a^2*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)))/(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)

________________________________________________________________________________________

giac [B]  time = 1.16, size = 288, normalized size = 2.20 \[ \frac {2 \, {\left (8 i \, a^{2} c^{2} + 16 \, a^{2} c d - 8 i \, a^{2} d^{2}\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {6 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} d^{4} f^{4} - 20 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} d^{5} f^{4} - 60 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} c d^{5} f^{4} - 60 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} d^{6} f^{4}}{15 \, d^{5} f^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(8*I*a^2*c^2 + 16*a^2*c*d - 8*I*a^2*d^2)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f
*x + e) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8
*c + 8*sqrt(c^2 + d^2))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 1/15*(6*(d*tan
(f*x + e) + c)^(5/2)*a^2*d^4*f^4 - 20*I*(d*tan(f*x + e) + c)^(3/2)*a^2*d^5*f^4 - 60*I*sqrt(d*tan(f*x + e) + c)
*a^2*c*d^5*f^4 - 60*sqrt(d*tan(f*x + e) + c)*a^2*d^6*f^4)/(d^5*f^5)

________________________________________________________________________________________

maple [B]  time = 0.23, size = 2484, normalized size = 18.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/f*a^2*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*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))*c+2*I/f*a^2*d^2/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*
(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-2*I/f*a^2*d^2/(c^2+d^2)^
(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))*c+4/3*I*a^2*(c+d*tan(f*x+e))^(3/2)/f+2*I/f*a^2/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1
/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-2*I/f*a
^2/(c^2+d^2)^(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))*c^3+I/f*a^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e)
)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^3+I/f*a^2*d^2/(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*
c+4/f*a^2*d*(c+d*tan(f*x+e))^(1/2)+2/f*a^2*d/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2
)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-I/f*a^2/(2*(c^2+d^2)^(1/2)+2*c
)^(1/2)/(c^2+d^2)^(1/2)*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)
)*c^3+4*I/f*a^2*c*(c+d*tan(f*x+e))^(1/2)-1/f*a^2*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*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))*c^2-2/f*a^2*d/(c^2+d^2)^(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))*c^2+1/f*a^2*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(
1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^2-2/5*a^2*(c+d*tan(f*x+e))^(5/2)/d/f-1/f*a^2*d^3/(2*(c^2+d^2
)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*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/f*a^2*d^3/(c^2+d^2)^(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/f*a^2*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*
x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c+I/f*a^2/(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^2-4/f*a^2*d/(2*(
c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2
*c)^(1/2))*c+I/f*a^2*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))+1/f*a^2*d^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))
^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))-2*I/f*a^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a
rctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+2*I/f*a^2*d^
2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(
1/2)-2*c)^(1/2))+4/f*a^2*d/(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))*c+2*I/f*a^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))*c^2+2/f*a^2*d^3/(c^2+d^2)^(1/2)/(2*(c^2+d
^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(
1/2))-2/f*a^2*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c-I/f*a^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c^2-I/f*a^2*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1
/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))-2*I/f*a^2*d^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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 10.32, size = 196, normalized size = 1.50 \[ -\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{3\,d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{3\,d\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {2\,a^2\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}+\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,{\left (-d-c\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,\left (c^2\,1{}\mathrm {i}+2\,c\,d-d^2\,1{}\mathrm {i}\right )}\right )\,{\left (-d-c\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{f} \]

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))^(3/2),x)

[Out]

(4i^(1/2)*a^2*atan((4i^(1/2)*(- c*1i - d)^(3/2)*(c + d*tan(e + f*x))^(1/2)*1i)/(2*(2*c*d + c^2*1i - d^2*1i)))*
(- c*1i - d)^(3/2)*2i)/f - (c - d*1i)*((2*a^2*(c - d*1i))/(d*f) - (2*a^2*(c + d*1i))/(d*f))*(c + d*tan(e + f*x
))^(1/2) - (2*a^2*(c + d*tan(e + f*x))^(5/2))/(5*d*f) - ((2*a^2*(c - d*1i))/(3*d*f) - (2*a^2*(c + d*1i))/(3*d*
f))*(c + d*tan(e + f*x))^(3/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________