3.12.32 \(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1132]

Optimal. Leaf size=127 \[ -\frac {4 i a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {2 a^2 (i c-d)}{3 d (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 127, 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 = {3623, 3610, 3618, 65, 214} \begin {gather*} \frac {4 i a^2}{f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 a^2 (-d+i c)}{3 d f (d+i c) (c+d \tan (e+f x))^{3/2}}-\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)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

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

Rule 3623

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

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Mathematica [A]
time = 5.14, size = 218, normalized size = 1.72 \begin {gather*} \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\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)^{5/2}}+\frac {2 \cos (e+f x) (\cos (2 e)-i \sin (2 e)) \left (\left (c^2+6 i c d+d^2\right ) \cos (e+f x)+6 i d^2 \sin (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 (c-i d)^2 d (c \cos (e+f x)+d \sin (e+f x))^2}\right )}{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/(c + d*Tan[e + f*x])^(5/2),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)^(5/2)*E^((2*I)*e)) + (2*Cos[e + f*x]*(Cos[2*e] - I*Sin[2*e])*((c^2 +
(6*I)*c*d + d^2)*Cos[e + f*x] + (6*I)*d^2*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*(c - I*d)^2*d*(c*Cos[e +
f*x] + d*Sin[e + f*x])^2)))/(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. 898 vs. \(2 (108 ) = 216\).
time = 0.29, size = 899, normalized size = 7.08

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {-2 i c d -c^{2}+d^{2}}{3 \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 d \left (i c^{2}-i d^{2}-2 c d \right )}{\left (c^{2}+d^{2}\right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {2 d \left (\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\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 (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\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 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\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 (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\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 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{\left (c^{2}+d^{2}\right )^{2}}\right )}{f d}\)	\(899\)
default	\(\frac {2 a^{2} \left (-\frac {-2 i c d -c^{2}+d^{2}}{3 \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 d \left (i c^{2}-i d^{2}-2 c d \right )}{\left (c^{2}+d^{2}\right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {2 d \left (\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\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 (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\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 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\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 (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\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 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{\left (c^{2}+d^{2}\right )^{2}}\right )}{f d}\)	\(899\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/f*a^2/d*(-1/3*(-2*I*c*d-c^2+d^2)/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)+2*d*(I*c^2-I*d^2-2*c*d)/(c^2+d^2)^2/(c+d*t
an(f*x+e))^(1/2)-2*d/(c^2+d^2)^2*(1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(1/2*(I*c^2*(c^2+d^2)^(1/2
)-I*d^2*(c^2+d^2)^(1/2)+I*c^3-3*I*c*d^2-2*c*d*(c^2+d^2)^(1/2)-3*c^2*d+d^3)*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*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3*I*(2*(c^2+d^2)^(
1/2)+2*c)^(1/2)*c*d^2-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^3-1/2*(I*c^2*(c^2+
d^2)^(1/2)-I*d^2*(c^2+d^2)^(1/2)+I*c^3-3*I*c*d^2-2*c*d*(c^2+d^2)^(1/2)-3*c^2*d+d^3)*(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)))+1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(1/2*(-I*c^2*(c^2+d^2)^(1/2)+I*d^2*(c^2
+d^2)^(1/2)-I*c^3+3*I*c*d^2+2*c*d*(c^2+d^2)^(1/2)+3*c^2*d-d^3)*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*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3*I*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*c*d^2-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^3+1/2*(-I*c^2*(c^2+d^2)^(1/2)+
I*d^2*(c^2+d^2)^(1/2)-I*c^3+3*I*c*d^2+2*c*d*(c^2+d^2)^(1/2)+3*c^2*d-d^3)*(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((a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^2/(d*tan(f*x + e) + c)^(5/2), 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. 929 vs. \(2 (104) = 208\).
time = 1.04, size = 929, normalized size = 7.31 \begin {gather*} \frac {3 \, {\left ({\left (c^{4} d - 4 i \, c^{3} d^{2} - 6 \, c^{2} d^{3} + 4 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} d - 2 i \, c^{3} d^{2} - 2 i \, c d^{4} - d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f\right )} \sqrt {-\frac {16 i \, a^{4}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c + {\left ({\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} f\right )} \sqrt {-\frac {16 i \, a^{4}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\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}} + 4 \, {\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 )}}{2 \, a^{2}}\right ) - 3 \, {\left ({\left (c^{4} d - 4 i \, c^{3} d^{2} - 6 \, c^{2} d^{3} + 4 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} d - 2 i \, c^{3} d^{2} - 2 i \, c d^{4} - d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f\right )} \sqrt {-\frac {16 i \, a^{4}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c + {\left ({\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f\right )} \sqrt {-\frac {16 i \, a^{4}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\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}} + 4 \, {\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 )}}{2 \, a^{2}}\right ) + 8 \, {\left (a^{2} c^{2} + 6 i \, a^{2} c d - 5 \, a^{2} d^{2} + {\left (a^{2} c^{2} + 6 i \, a^{2} c d + 7 \, a^{2} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (a^{2} c^{2} + 6 i \, a^{2} c d + 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}}}{12 \, {\left ({\left (c^{4} d - 4 i \, c^{3} d^{2} - 6 \, c^{2} d^{3} + 4 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} d - 2 i \, c^{3} d^{2} - 2 i \, c d^{4} - d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f\right )}} \end {gather*}

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))^(5/2),x, algorithm="fricas")

[Out]

1/12*(3*((c^4*d - 4*I*c^3*d^2 - 6*c^2*d^3 + 4*I*c*d^4 + d^5)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4*d - 2*I*c^3*d^2 -
2*I*c*d^4 - d^5)*f*e^(2*I*f*x + 2*I*e) + (c^4*d + 2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((I*c^5 + 5*c^4*d - 10*I*
c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/2*(4*a^2*c + ((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*e^(2*I
*f*x + 2*I*e) + (I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f)*sqrt(-16*I*a^4/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^
2*d^3 + 5*I*c*d^4 + d^5)*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 - I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - 3*((c^4*d - 4*I*c^3*d^2 - 6*c^2*d^3 + 4*I*c*
d^4 + d^5)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4*d - 2*I*c^3*d^2 - 2*I*c*d^4 - d^5)*f*e^(2*I*f*x + 2*I*e) + (c^4*d +
2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(
1/2*(4*a^2*c + ((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d
^3)*f)*sqrt(-16*I*a^4/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*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 - 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^2 + 6*I*a^2*c*d - 5*a^2*d^2 + (a^2*c^2 + 6*I*a^2*c*d + 7*a^2*d^2)*e^(4*I*f*x + 4*I*
e) + 2*(a^2*c^2 + 6*I*a^2*c*d + 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)))/((c^4*d - 4*I*c^3*d^2 - 6*c^2*d^3 + 4*I*c*d^4 + d^5)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4
*d - 2*I*c^3*d^2 - 2*I*c*d^4 - d^5)*f*e^(2*I*f*x + 2*I*e) + (c^4*d + 2*c^2*d^3 + d^5)*f)

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

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))**(5/2),x)

[Out]

-a**2*(Integral(tan(e + f*x)**2/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) +
d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(-2*I*tan(e + f*x)/(c**2*sqrt(c + d*tan(e + f*x)
) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integra
l(-1/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*
x))*tan(e + f*x)**2), 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. 250 vs. \(2 (104) = 208\).
time = 0.90, size = 250, normalized size = 1.97 \begin {gather*} \frac {8 \, a^{2} \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 (-i \, c^{2} f - 2 \, c d f + i \, d^{2} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (-i \, a^{2} c^{2} + 6 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} d - i \, a^{2} d^{2}\right )}}{-3 \, {\left (-i \, c^{2} d f - 2 \, c d^{2} f + i \, d^{3} f\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \end {gather*}

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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 8.90, size = 221, normalized size = 1.74 \begin {gather*} \frac {\frac {a^2\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,4{}\mathrm {i}}{f\,{\left (c-d\,1{}\mathrm {i}\right )}^2}+\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{3\,d\,f\,\left (c-d\,1{}\mathrm {i}\right )}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^8\,f^2+8\,c^6\,d^2\,f^2+12\,c^4\,d^4\,f^2+8\,c^2\,d^6\,f^2+2\,d^8\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}\,\left (f\,c^6+2{}\mathrm {i}\,f\,c^5\,d+f\,c^4\,d^2+4{}\mathrm {i}\,f\,c^3\,d^3-f\,c^2\,d^4+2{}\mathrm {i}\,f\,c\,d^5-f\,d^6\right )}\right )\,4{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}} \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))^(5/2),x)

[Out]

((a^2*(c + d*tan(e + f*x))*4i)/(f*(c - d*1i)^2) + (2*a^2*(c + d*1i))/(3*d*f*(c - d*1i)))/(c + d*tan(e + f*x))^
(3/2) + (a^2*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^8*f^2 + 2*d^8*f^2 + 8*c^2*d^6*f^2 + 12*c^4*d^4*f^2 + 8*c^6*
d^2*f^2))/(2*f*(d*1i - c)^(5/2)*(c^6*f - d^6*f - c^2*d^4*f + c^3*d^3*f*4i + c^4*d^2*f + c*d^5*f*2i + c^5*d*f*2
i)))*4i)/(f*(d*1i - c)^(5/2))

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