\(\int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\) [247]

Optimal result

Integrand size = 26, antiderivative size = 381 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {91 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {25 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{6 \sqrt {3} a^2 d}+\frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {91 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

[Out]

91/72*I*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+91/72*I*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+91/36*I*arc
tan(tan(d*x+c)^(1/3))/a^2/d-25/18*ln(1+tan(d*x+c)^(2/3))/a^2/d+25/36*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a
^2/d+25/18*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)+91/144*I*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan
(d*x+c)^(2/3))/a^2/d*3^(1/2)-91/144*I*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)-25/12/a^2/
d/tan(d*x+c)^(4/3)+13/12/a^2/d/(1+I*tan(d*x+c))/tan(d*x+c)^(4/3)+91/12*I/a^2/d/tan(d*x+c)^(1/3)+1/4/d/tan(d*x+
c)^(4/3)/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3640, 3677, 3610, 3619, 3557, 335, 281, 206, 31, 648, 632, 210, 642, 301, 209} \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {25 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{6 \sqrt {3} a^2 d}-\frac {91 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}-\frac {25 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{18 a^2 d}+\frac {91 i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{36 a^2 d}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

[In]

Int[1/(Tan[c + d*x]^(7/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(((-91*I)/72)*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(a^2*d) + (((91*I)/72)*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(
1/3)])/(a^2*d) + (25*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(6*Sqrt[3]*a^2*d) + (((91*I)/36)*ArcTan[Tan[c
+ d*x]^(1/3)])/(a^2*d) - (25*Log[1 + Tan[c + d*x]^(2/3)])/(18*a^2*d) + (((91*I)/48)*Log[1 - Sqrt[3]*Tan[c + d
*x]^(1/3) + Tan[c + d*x]^(2/3)])/(Sqrt[3]*a^2*d) - (((91*I)/48)*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d
*x]^(2/3)])/(Sqrt[3]*a^2*d) + (25*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(36*a^2*d) - 25/(12*a^2*d*
Tan[c + d*x]^(4/3)) + 13/(12*a^2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(4/3)) + ((91*I)/12)/(a^2*d*Tan[c + d*x]^
(1/3)) + 1/(4*d*Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x])^2)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 209

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

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 301

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(
2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 +
2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))*Int[1/(r^2 + s^2*x^2), x] +
Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

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*} \text {integral}& = \frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {16 a}{3}-\frac {10}{3} i a \tan (c+d x)}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {200 a^2}{9}-\frac {182}{9} i a^2 \tan (c+d x)}{\tan ^{\frac {7}{3}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {182 i a^2}{9}-\frac {200}{9} a^2 \tan (c+d x)}{\tan ^{\frac {4}{3}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {200 a^2}{9}+\frac {182}{9} i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(91 i) \int \tan ^{\frac {2}{3}}(c+d x) \, dx}{36 a^2}-\frac {25 \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{9 a^2} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(91 i) \text {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{9 a^2 d} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(91 i) \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d} \\ & = -\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(91 i) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {(91 i) \text {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {(91 i) \text {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a^2 d} \\ & = \frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(91 i) \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac {(91 i) \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {(91 i) \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d}-\frac {(91 i) \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d} \\ & = \frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {91 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(91 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {(91 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {25 \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{36 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{12 a^2 d} \\ & = -\frac {91 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {91 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {25 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{6 a^2 d} \\ & = -\frac {91 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {25 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{6 \sqrt {3} a^2 d}+\frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {91 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.52 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\frac {13 a}{3 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))}+\frac {\frac {182 i a^2 \left (1+\frac {1}{6} i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-\frac {1}{6} i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-\frac {1}{6} (-1)^{5/6} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+\frac {1}{6} \sqrt [6]{-1} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-\frac {1}{6} \sqrt [6]{-1} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+\frac {1}{6} (-1)^{5/6} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}\right )}{3 d \sqrt [3]{\tan (c+d x)}}-\frac {50 a^2 \left (1+\frac {2}{3} \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}-\frac {2}{3} \sqrt [3]{-1} \log \left (1-e^{-\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}+\frac {2}{3} (-1)^{2/3} \log \left (1-e^{\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}\right )}{3 d \tan ^{\frac {4}{3}}(c+d x)}}{2 a^2}}{4 a^2} \]

[In]

Integrate[1/(Tan[c + d*x]^(7/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

1/(4*d*Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x])^2) + ((13*a)/(3*d*Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x]))
+ ((((182*I)/3)*a^2*(1 + (I/6)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) - (I/6)*Log[1 + I*(Tan
[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) - ((-1)^(5/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^((I/6)*Pi)]*(Tan[c +
d*x]^2)^(1/6))/6 + ((-1)^(1/6)*Log[1 - E^((I/6)*Pi)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6))/6 - ((-1)
^(1/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^(((5*I)/6)*Pi)]*(Tan[c + d*x]^2)^(1/6))/6 + ((-1)^(5/6)*Log[1 - E^(((5
*I)/6)*Pi)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6))/6))/(d*Tan[c + d*x]^(1/3)) - (50*a^2*(1 + (2*Log[1
+ (Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(2/3))/3 - (2*(-1)^(1/3)*Log[1 - (Tan[c + d*x]^2)^(1/3)/E^((I/3)*Pi
)]*(Tan[c + d*x]^2)^(2/3))/3 + (2*(-1)^(2/3)*Log[1 - E^((I/3)*Pi)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(2/
3))/3))/(3*d*Tan[c + d*x]^(4/3)))/(2*a^2))/(4*a^2)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {-\frac {3}{4 \tan \left (d x +c \right )^{\frac {4}{3}}}+\frac {6 i}{\tan \left (d x +c \right )^{\frac {1}{3}}}+\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {19 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}-\frac {191 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {-76 i \tan \left (d x +c \right )-116 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+122 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+40}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}+\frac {191 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {191 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}-\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\)	\(244\)
default	\(\frac {-\frac {3}{4 \tan \left (d x +c \right )^{\frac {4}{3}}}+\frac {6 i}{\tan \left (d x +c \right )^{\frac {1}{3}}}+\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {19 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}-\frac {191 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {-76 i \tan \left (d x +c \right )-116 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+122 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+40}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}+\frac {191 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {191 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}-\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\)	\(244\)

[In]

int(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-3/4/tan(d*x+c)^(4/3)+6*I/tan(d*x+c)^(1/3)+1/16*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/8*I*3^(1/
2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))+19/36*I/(tan(d*x+c)^(1/3)+I)-1/36/(tan(d*x+c)^(1/3)+I)^2-191/72
*ln(tan(d*x+c)^(1/3)+I)-1/72*(-76*I*tan(d*x+c)-116*tan(d*x+c)^(2/3)+122*I*tan(d*x+c)^(1/3)+40)/(-I*tan(d*x+c)^
(1/3)+tan(d*x+c)^(2/3)-1)^2+191/144*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-191/72*I*3^(1/2)*arctanh(1/3*(-
I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/8*ln(tan(d*x+c)^(1/3)-I))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 836 vs. \(2 (298) = 596\).

Time = 0.26 (sec) , antiderivative size = 836, normalized size of antiderivative = 2.19 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/144*(9*(sqrt(3)*(-I*a^2*d*e^(8*I*d*x + 8*I*c) + 2*I*a^2*d*e^(6*I*d*x + 6*I*c) - I*a^2*d*e^(4*I*d*x + 4*I*c)
)*sqrt(1/(a^4*d^2)) - e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(1/2*sqrt(3)*a^2*d
*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 9*(sqrt(3)*(I*a
^2*d*e^(8*I*d*x + 8*I*c) - 2*I*a^2*d*e^(6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - e^
(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(-1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((
-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 191*(3*sqrt(1/3)*(I*a^2*d*e^(8*I*d*x +
8*I*c) - 2*I*a^2*d*e^(6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - e^(8*I*d*x + 8*I*c)
+ 2*e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x +
2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 191*(3*sqrt(1/3)*(-I*a^2*d*e^(8*I*d*x + 8*I*c) + 2*I*a
^2*d*e^(6*I*d*x + 6*I*c) - I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x
+ 6*I*c) - e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e
^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 382*(e^(8*I*d*x + 8*I*c) - 2*e^(6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*
c))*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) + 18*(e^(8*I*d*x + 8*I*c) - 2*e^(6
*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - I)
+ 3*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(293*e^(8*I*d*x + 8*I*c) - 110*e^(6*I*d*x
+ 6*I*c) - 368*e^(4*I*d*x + 4*I*c) + 38*e^(2*I*d*x + 2*I*c) + 3))/(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*
d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(1/tan(d*x+c)**(7/3)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 1.07 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {191 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} - \frac {i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} + \frac {\log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} + \frac {191 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} - \frac {191 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} - \frac {\log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} - \frac {3 \, {\left (-8 i \, \tan \left (d x + c\right ) + 1\right )}}{4 \, a^{2} d \tan \left (d x + c\right )^{\frac {4}{3}}} + \frac {19 i \, \tan \left (d x + c\right )^{\frac {5}{3}} + 22 \, \tan \left (d x + c\right )^{\frac {2}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]

[In]

integrate(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

191/144*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I))/(a^2*d) - 1/
16*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a^2*d) + 1/16*lo
g(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) + 191/144*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/
3) - 1)/(a^2*d) - 191/72*log(tan(d*x + c)^(1/3) + I)/(a^2*d) - 1/8*log(tan(d*x + c)^(1/3) - I)/(a^2*d) - 3/4*(
-8*I*tan(d*x + c) + 1)/(a^2*d*tan(d*x + c)^(4/3)) + 1/12*(19*I*tan(d*x + c)^(5/3) + 22*tan(d*x + c)^(2/3))/(a^
2*d*(tan(d*x + c) - I)^2)

Mupad [B] (verification not implemented)

Time = 6.14 (sec) , antiderivative size = 652, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(1/(tan(c + d*x)^(7/3)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

(191*log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - (36481*((a^6*d^3*891794176i)/3 - (893867776*a^8*d^4*tan(c + d
*x)^(1/3)*(-1/(a^6*d^3))^(1/3))/3)*(-1/(a^6*d^3))^(2/3))/5184)*(-1/(a^6*d^3))^(1/3))/72 + (3i/(4*a^2*d) + (9*t
an(c + d*x))/(2*a^2*d) + (tan(c + d*x)^2*157i)/(12*a^2*d) - (91*tan(c + d*x)^3)/(12*a^2*d))/(2*tan(c + d*x)^(7
/3) - tan(c + d*x)^(4/3)*1i + tan(c + d*x)^(10/3)*1i) + log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - ((a^6*d^3*
891794176i)/3 - 112318464*a^8*d^4*tan(c + d*x)^(1/3)*(-1/(512*a^6*d^3))^(1/3))*(-1/(512*a^6*d^3))^(2/3))*(-1/(
512*a^6*d^3))^(1/3) + log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i)/2 - 1/2)^2*((a^6*d^3*891794176
i)/3 - 112318464*a^8*d^4*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(-1/(512*a^6*d^3))^(1/3))*(-1/(512*a^6*d^3)
)^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-1/(512*a^6*d^3))^(1/3) - log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/
2)*1i)/2 + 1/2)^2*((a^6*d^3*891794176i)/3 + 112318464*a^8*d^4*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*(-1/(5
12*a^6*d^3))^(1/3))*(-1/(512*a^6*d^3))^(2/3))*((3^(1/2)*1i)/2 + 1/2)*(-1/(512*a^6*d^3))^(1/3) + (191*log((1459
2400*a^2*d*tan(c + d*x)^(1/3))/3 - (36481*(3^(1/2)*1i - 1)^2*((a^6*d^3*891794176i)/3 - (446933888*a^8*d^4*tan(
c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1/(a^6*d^3))^(1/3))/3)*(-1/(a^6*d^3))^(2/3))/20736)*(3^(1/2)*1i - 1)*(-1/(a^
6*d^3))^(1/3))/144 - (191*log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - (36481*(3^(1/2)*1i + 1)^2*((a^6*d^3*8917
94176i)/3 + (446933888*a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(-1/(a^6*d^3))^(1/3))/3)*(-1/(a^6*d^3))^(2/
3))/20736)*(3^(1/2)*1i + 1)*(-1/(a^6*d^3))^(1/3))/144