3.1109 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=98 \[ \frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 i a (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3528, 3537, 63, 208} \[ \frac {2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d+i c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 i a (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 2.16, size = 111, normalized size = 1.13 \[ \frac {2 a \left ((4 i c+i d \tan (e+f x)+3 d) \sqrt {c+d \tan (e+f x)}-3 i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 506, normalized size = 5.16 \[ -\frac {3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} \log \left (\frac {{\left (2 i \, a c^{2} + 2 \, a c d + {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} + {\left (2 i \, a c^{2} + 4 \, a c d - 2 i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) - 3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} \log \left (\frac {{\left (2 i \, a c^{2} + 2 \, a c d - {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} + {\left (2 i \, a c^{2} + 4 \, a c d - 2 i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) + 16 \, {\left (-2 i \, a c - a d + 2 \, {\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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]

-1/12*(3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2)*log(
(2*I*a*c^2 + 2*a*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(-(4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2) + (2*I*a*c^2 + 4*a*c*d - 2
*I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*a*c + a*d)) - 3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^
2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2)*log((2*I*a*c^2 + 2*a*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(-(4*a^2*c^3 - 12*I*a^2*c^2*d
- 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2) + (2*I*a*c^2 + 4*a*c*d - 2*I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I
*e)/(I*a*c + a*d)) + 16*(-2*I*a*c - a*d + 2*(-I*a*c - a*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)))/(f*e^(2*I*f*x + 2*I*e) + f)

________________________________________________________________________________________

giac [B]  time = 0.48, size = 241, normalized size = 2.46 \[ \frac {2 \, {\left (4 i \, a c^{2} + 8 \, a c d - 4 i \, a d^{2}\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {-2 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a f^{2} - 6 i \, \sqrt {d \tan \left (f x + e\right ) + c} a c f^{2} - 6 \, \sqrt {d \tan \left (f x + e\right ) + c} a d f^{2}}{3 \, f^{3}} \]

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

________________________________________________________________________________________

maple [B]  time = 0.21, size = 2398, normalized size = 24.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

mupad [B]  time = 16.32, size = 2869, normalized size = 29.28 \[ \text {result too large to display} \]

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^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(((-a^4*d^2
*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*c^2*1i + a*d^2*1i - f*(((-a^4*d^2*f
^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2
*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (a^3*d^2*(c^2 - d^2)*(c^2*1i + d^2*1i)^2*8i
)/f^3)*((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - (a^2*c^3)/(4*f^2) + (3*a^2*c*d^2
)/(4*f^2))^(1/2) - log(((((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((1
6*c*d^2*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*c^2*1i + a*d^2*1
i + f*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^
(1/2)))/f + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (a^3*d^2*(c^2 - d^2)*(c^
2*1i + d^2*1i)^2*8i)/f^3)*(((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) - a^2*c^3*f^2 + 3*a^2*
c*d^2*f^2)/(4*f^4))^(1/2) - log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)
^(1/2)*((16*c*d^2*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*c^2*1
i + a*d^2*1i + f*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*ta
n(e + f*x))^(1/2)))/f + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (a^3*d^2*(c^
2 - d^2)*(c^2*1i + d^2*1i)^2*8i)/f^3)*(-((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) + a^2*c^3
*f^2 - 3*a^2*c*d^2*f^2)/(4*f^4))^(1/2) + log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*
d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(
1/2)*(a*c^2*1i + a*d^2*1i - f*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1
/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 -
(a^3*d^2*(c^2 - d^2)*(c^2*1i + d^2*1i)^2*8i)/f^3)*((3*a^2*c*d^2)/(4*f^2) - (a^2*c^3)/(4*f^2) - (6*a^4*c^2*d^4
*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2) - log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) +
a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a
^2*c*d^2*f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d + c*f*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*
c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c
^2*d^2))/f^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*(-((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^
(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/(4*f^4))^(1/2) - log(((((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*
f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*
f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d + c*f*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)
/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f
^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*(((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) - a^2
*c^3*f^2 + 3*a^2*c*d^2*f^2)/(4*f^4))^(1/2) + log(((((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2
*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(
1/2)*(a*d^3 + a*c^2*d - c*f*(((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)
*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (1
6*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - (a^2*c^
3)/(4*f^2) + (3*a^2*c*d^2)/(4*f^2))^(1/2) + log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2
*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^
(1/2)*(a*d^3 + a*c^2*d - c*f*(-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/
2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 -
(16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*((3*a^2*c*d^2)/(4*f^2) - (a^2*c^3)/(4*f^2) - (6*a^4*c^2*d^4*f^4 - a^4*d^6*f^
4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2) + (a*(c + d*tan(e + f*x))^(3/2)*2i)/(3*f) + (a*c*(c + d*tan(e + f*
x))^(1/2)*2i)/f + (2*a*d*(c + d*tan(e + f*x))^(1/2))/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*(Integral(-I*c*sqrt(c + d*tan(e + f*x)), x) + Integral(c*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integ
ral(d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x))

________________________________________________________________________________________