3.1126 \(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=92 \[ \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {4 i a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3542, 3537, 63, 208} \[ \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {4 i a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{3/2}} \]

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

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 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 3542

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

Rubi steps

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

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Mathematica [B]  time = 4.15, size = 189, normalized size = 2.05 \[ \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (\frac {2 (c+i d) (\cos (2 e)-i \sin (2 e)) \cos (e+f x) \sqrt {c+d \tan (e+f x)}}{d (c-i d) (c \cos (e+f x)+d \sin (e+f x))}-\frac {4 i e^{-2 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2}}\right )}{f (\cos (f x)+i \sin (f x))^2} \]

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)*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^((2*I)*e)) + (2*(c + I*d)*Cos[e + f*x]*(Cos[2*e] - I*Sin[2*e
])*Sqrt[c + d*Tan[e + f*x]])/((c - I*d)*d*(c*Cos[e + f*x] + d*Sin[e + f*x]))))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [B]  time = 0.53, size = 586, normalized size = 6.37 \[ \frac {{\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} 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 {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \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}} + {\left (4 \, a^{2} c - 4 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 )}}{2 \, a^{2}}\right ) - {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} 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 {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \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}} + {\left (4 \, a^{2} c - 4 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 )}}{2 \, a^{2}}\right ) + {\left (8 \, a^{2} c + 8 i \, a^{2} d + {\left (8 \, a^{2} c + 8 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}}}{{\left (4 \, c^{2} d - 8 i \, c d^{2} - 4 \, d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (c^{2} d + d^{3}\right )} f} \]

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]

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

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giac [B]  time = 0.97, size = 209, normalized size = 2.27 \[ \frac {16 \, a^{2} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c f - d f\right )} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, a^{2} c + 2 i \, a^{2} d}{{\left (c d f - i \, d^{2} f\right )} \sqrt {d \tan \left (f x + e\right ) + c}} \]

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]

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

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maple [B]  time = 0.28, size = 2534, normalized size = 27.54 \[ \text {result too large to display} \]

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]

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

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

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)

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mupad [B]  time = 6.38, size = 142, normalized size = 1.54 \[ \frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,4{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}+\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \]

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]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \frac {\tan ^{2}{\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx + \int \left (- \frac {1}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx\right ) \]

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(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) + I
ntegral(-2*I*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) + Integra
l(-1/(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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