3.2.77 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx\) [177]

Optimal. Leaf size=326 \[ \frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \]

[Out]

(25/32+21/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(3/2)/f*2^(1/2)-(25/32+21/32*I)*arctan(1+
2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(3/2)/f*2^(1/2)+(-25/64+21/64*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))
^(1/2)+d^(1/2)*tan(f*x+e))/a^2/d^(3/2)/f*2^(1/2)+(25/64-21/64*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/
2)*tan(f*x+e))/a^2/d^(3/2)/f*2^(1/2)-25/8/a^2/d/f/(d*tan(f*x+e))^(1/2)+7/8/a^2/d/f/(d*tan(f*x+e))^(1/2)/(1+I*t
an(f*x+e))+1/4/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2

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Rubi [A]
time = 0.34, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3640, 3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

((25/16 + (21*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/16 + (
21*I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/32 - (21*I)/32)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) + ((25/32 - (21*I)/
32)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) - 25/(8*a^2*d*
f*Sqrt[d*Tan[e + f*x]]) + 7/(8*a^2*d*f*(1 + I*Tan[e + f*x])*Sqrt[d*Tan[e + f*x]]) + 1/(4*d*f*Sqrt[d*Tan[e + f*
x]]*(a + I*a*Tan[e + f*x])^2)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

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 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3640

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

Rule 3677

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

Rubi steps

\begin {align*} \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx &=\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {9 a d}{2}-\frac {5}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx}{4 a^2 d}\\ &=\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {25 a^2 d^2}{2}-\frac {21}{2} i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{8 a^4 d^2}\\ &=-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {\int \frac {-\frac {21}{2} i a^2 d^3-\frac {25}{2} a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4 d^4}\\ &=-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {21}{2} i a^2 d^4-\frac {25}{2} a^2 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^4 d^4 f}\\ &=-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d f}\\ &=-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}+-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}+-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d f}+-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d f}\\ &=-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}\\ &=\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{32}-\frac {21 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 1.16, size = 231, normalized size = 0.71 \begin {gather*} \frac {\sec ^3(e+f x) \left (23 \cos (e+f x)+41 \cos (3 (e+f x))+43 i \sin (e+f x)-(25-21 i) \cos (2 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}-(21+25 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sin ^{\frac {3}{2}}(2 (e+f x))+(21-25 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) \sqrt {\sin (2 (e+f x))} (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+43 i \sin (3 (e+f x))\right )}{32 a^2 d f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

(Sec[e + f*x]^3*(23*Cos[e + f*x] + 41*Cos[3*(e + f*x)] + (43*I)*Sin[e + f*x] - (25 - 21*I)*Cos[2*(e + f*x)]*Lo
g[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - (21 + 25*I)*Log[Cos[e + f*x]
+ Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (21 - 25*I)*ArcSin[Cos[e + f*x] - Sin[e + f*
x]]*Sqrt[Sin[2*(e + f*x)]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + (43*I)*Sin[3*(e + f*x)]))/(32*a^2*d*f*
Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^2)

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Maple [A]
time = 0.16, size = 131, normalized size = 0.40

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {11 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{4}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{4} \sqrt {i d}}\right )}{f \,a^{2}}\)	\(131\)
default	\(\frac {2 d^{3} \left (-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {11 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{4}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{4} \sqrt {i d}}\right )}{f \,a^{2}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*d^3*(-1/d^4/(d*tan(f*x+e))^(1/2)-1/8/d^4*((-9/2*(d*tan(f*x+e))^(3/2)+11/2*I*d*(d*tan(f*x+e))^(1/2))/(I
*d*tan(f*x+e)+d)^2+23/2/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/8/d^4/(I*d)^(1/2)*arctan((d*
tan(f*x+e))^(1/2)/(I*d)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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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. 730 vs. \(2 (253) = 506\).
time = 0.40, size = 730, normalized size = 2.24 \begin {gather*} \frac {4 \, {\left (a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i}{16 \, a^{4} d^{3} f^{2}}} \log \left (-2 \, {\left (4 \, {\left (i \, a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{16 \, a^{4} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 4 \, {\left (a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i}{16 \, a^{4} d^{3} f^{2}}} \log \left (-2 \, {\left (4 \, {\left (-i \, a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{16 \, a^{4} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + 4 \, {\left (a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {529 i}{64 \, a^{4} d^{3} f^{2}}} \log \left (\frac {{\left (8 \, {\left (a^{2} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {529 i}{64 \, a^{4} d^{3} f^{2}}} + 23\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} d f}\right ) - 4 \, {\left (a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {529 i}{64 \, a^{4} d^{3} f^{2}}} \log \left (-\frac {{\left (8 \, {\left (a^{2} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {529 i}{64 \, a^{4} d^{3} f^{2}}} - 23\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} d f}\right ) + \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-42 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 33 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 10 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{16 \, {\left (a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/16*(4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(1/16*I/(a^4*d^3*f^2))*log(-2*(4*(
I*a^2*d^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(1/16*I/(a^4*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 4*(a^2*d^2*f*e^(6*I*f*x + 6*I*
e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(1/16*I/(a^4*d^3*f^2))*log(-2*(4*(-I*a^2*d^2*f*e^(2*I*f*x + 2*I*e) - I
*a^2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I/(a^4*d^3*f^2)) + I*d*
e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))
*sqrt(-529/64*I/(a^4*d^3*f^2))*log(1/8*(8*(a^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*
e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-529/64*I/(a^4*d^3*f^2)) + 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) - 4*(
a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-529/64*I/(a^4*d^3*f^2))*log(-1/8*(8*(a^2*
d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-529/
64*I/(a^4*d^3*f^2)) - 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) + sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x
+ 2*I*e) + 1))*(-42*I*e^(6*I*f*x + 6*I*e) - 33*I*e^(4*I*f*x + 4*I*e) + 10*I*e^(2*I*f*x + 2*I*e) + I))/(a^2*d^2
*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} - \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

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

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Giac [A]
time = 0.67, size = 221, normalized size = 0.68 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {23 i \, \sqrt {2} \arctan \left (-\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {16}{\sqrt {d \tan \left (f x + e\right )} a^{2} f} + \frac {9 \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 11 i \, \sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/8*(2*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(
a^2*sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - 23*I*sqrt(2)*arctan(-8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3
/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*sqrt(d)*f*(I*d/sqrt(d^2) + 1)) + 16/(sqrt(d*tan(f*x + e))*a^2*f) + (9
*sqrt(d*tan(f*x + e))*d*tan(f*x + e) - 11*I*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^2*a^2*f))/d

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Mupad [B]
time = 5.96, size = 185, normalized size = 0.57 \begin {gather*} 2\,\mathrm {atanh}\left (8\,a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{64\,a^4\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{64\,a^4\,d^3\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {529{}\mathrm {i}}{256\,a^4\,d^3\,f^2}}}{23}\right )\,\sqrt {-\frac {529{}\mathrm {i}}{256\,a^4\,d^3\,f^2}}-\frac {\frac {2\,d}{a^2\,f}-\frac {25\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,a^2\,f}+\frac {d\,\mathrm {tan}\left (e+f\,x\right )\,43{}\mathrm {i}}{8\,a^2\,f}}{d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x)*1i)^2),x)

[Out]

2*atanh(8*a^2*d*f*(d*tan(e + f*x))^(1/2)*(1i/(64*a^4*d^3*f^2))^(1/2))*(1i/(64*a^4*d^3*f^2))^(1/2) + 2*atanh((1
6*a^2*d*f*(d*tan(e + f*x))^(1/2)*(-529i/(256*a^4*d^3*f^2))^(1/2))/23)*(-529i/(256*a^4*d^3*f^2))^(1/2) - ((2*d)
/(a^2*f) - (25*d*tan(e + f*x)^2)/(8*a^2*f) + (d*tan(e + f*x)*43i)/(8*a^2*f))/(d*(d*tan(e + f*x))^(3/2)*2i - (d
*tan(e + f*x))^(5/2) + d^2*(d*tan(e + f*x))^(1/2))

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