3.3.35 \(\int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) [235]

Optimal. Leaf size=299 \[ -\frac {\text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {\text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \]

[Out]

1/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+1/12*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+1/6*arctan(tan(d*x+c)
^(1/3))/a/d+1/3*I*ln(1+tan(d*x+c)^(2/3))/a/d-1/6*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a/d+1/3*I*arctan(1/
3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a/d*3^(1/2)-1/24*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)
+1/24*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)-1/2*tan(d*x+c)^(1/3)/d/(a+I*a*tan(d*x+c))

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Rubi [A]
time = 0.26, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3631, 3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31} \begin {gather*} \frac {i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {\text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\text {ArcTan}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}+\frac {\text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(4/3)/(a + I*a*Tan[c + d*x]),x]

[Out]

-1/12*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)]/(a*d) + ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)]/(12*a*d) + (I*ArcT
an[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*a*d) + ArcTan[Tan[c + d*x]^(1/3)]/(6*a*d) + ((I/3)*Log[1 + Ta
n[c + d*x]^(2/3)])/(a*d) - Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)]/(8*Sqrt[3]*a*d) + Log[1 +
Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)]/(8*Sqrt[3]*a*d) - ((I/6)*Log[1 - Tan[c + d*x]^(2/3) + Tan[c +
d*x]^(4/3)])/(a*d) - Tan[c + d*x]^(1/3)/(2*d*(a + I*a*Tan[c + d*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 215

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

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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 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 3631

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

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {\int \frac {\frac {a}{3}-\frac {4}{3} i a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {(2 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{3 a}+\frac {\int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{6 a}\\ &=-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {(2 i) \text {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{3 a d}+\frac {\text {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 a d}\\ &=-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {(2 i) \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}+\frac {\text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}\\ &=-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\text {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\text {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}\\ &=\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {i \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {\text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {\text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}-\frac {\text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}+\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}\\ &=\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}-\frac {i \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}\\ &=-\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {i \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=-\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.17, size = 162, normalized size = 0.54 \begin {gather*} -\frac {e^{-2 i (c+d x)} \left (3\ 2^{2/3} e^{2 i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2 \left (1+e^{2 i (c+d x)}-5 e^{2 i (c+d x)} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt [3]{\tan (c+d x)}}{8 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(4/3)/(a + I*a*Tan[c + d*x]),x]

[Out]

-1/8*((3*2^(2/3)*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, (1 - E^(
(2*I)*(c + d*x)))/2] + 2*(1 + E^((2*I)*(c + d*x)) - 5*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/3, 1, 4/3, -((-1
+ E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))]))*Tan[c + d*x]^(1/3))/(a*d*E^((2*I)*(c + d*x)))

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Maple [A]
time = 0.18, size = 190, normalized size = 0.64

method	result	size
derivativedivides	\(\frac {\frac {5 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}-\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {5 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\)	\(190\)
default	\(\frac {\frac {5 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}-\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {5 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(5/12*I*ln(tan(d*x+c)^(1/3)+I)-1/6/(tan(d*x+c)^(1/3)+I)-1/8*I*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-
1/4*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/12*(-2*tan(d*x+c)^(1/3)-2*I)/(-I*tan(d*x+c)^(1/3)+ta
n(d*x+c)^(2/3)-1)-5/24*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+5/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^
(1/3))*3^(1/2))+1/4*I*ln(tan(d*x+c)^(1/3)-I))

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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(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),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. 492 vs. \(2 (238) = 476\).
time = 0.77, size = 492, normalized size = 1.65 \begin {gather*} -\frac {{\left (3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) - 3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) - 5 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + 5 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) - 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) + 6 \, \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{24 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/24*(3*(sqrt(3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2*I*c))*log(1/2*sqrt(3)*a*d*sqrt(
1/(a^2*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) - 3*(sqrt(3)*a*d*sqrt(1
/(a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d*x + 2*I*c))*log(-1/2*sqrt(3)*a*d*sqrt(1/(a^2*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) - 5*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x
+ 2*I*c) - I*e^(2*I*d*x + 2*I*c))*log(3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*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) + 5*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*
x + 2*I*c))*log(-3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*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) - 10*I*e^(2*I*d*x + 2*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) - 6*I*e^(2*I*d*x + 2*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) + 6*((
-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(e^(2*I*d*x + 2*I*c) + 1))*e^(-2*I*d*x - 2*I*c)/(
a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {\tan ^{\frac {4}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(4/3)/(a+I*a*tan(d*x+c)),x)

[Out]

-I*Integral(tan(c + d*x)**(4/3)/(tan(c + d*x) - I), x)/a

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Giac [A]
time = 0.69, size = 215, normalized size = 0.72 \begin {gather*} -\frac {5 \, \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 )}{24 \, a d} + \frac {\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 )}{8 \, a d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{8 \, a d} - \frac {5 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{24 \, a d} + \frac {5 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{12 \, a d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{4 \, a d} + \frac {i \, \tan \left (d x + c\right )^{\frac {1}{3}}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-5/24*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I))/(a*d) + 1/8*sqrt
(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a*d) - 1/8*I*log(tan(d*x
+ c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a*d) - 5/24*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1)/(a*d)
+ 5/12*I*log(tan(d*x + c)^(1/3) + I)/(a*d) + 1/4*I*log(tan(d*x + c)^(1/3) - I)/(a*d) + 1/2*I*tan(d*x + c)^(1/
3)/(a*d*(tan(d*x + c) - I))

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Mupad [B]
time = 5.16, size = 622, normalized size = 2.08 \begin {gather*} \ln \left (\left (a^3\,d^3\,14112{}\mathrm {i}-165888\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}+\ln \left (\left (a^3\,d^3\,14112{}\mathrm {i}-165888\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}+a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(4/3)/(a + a*tan(c + d*x)*1i),x)

[Out]

log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^(1/3)*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3) - a
^2*d^2*tan(c + d*x)^(1/3)*6120i)*(-1i/(64*a^3*d^3))^(1/3) + log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^
(1/3)*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3) - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(-125i/(1
728*a^3*d^3))^(1/3) + (log(((3^(1/2)*1i - 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i -
1)^2*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i - 1
)*(-1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(
1/2)*1i + 1)^2*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 + a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1
/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^
(1/3)*(3^(1/2)*1i - 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^
(1/3)*6120i)*(3^(1/2)*1i - 1)*(-125i/(1728*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*14112i - 41472
*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3))/2 +
a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i + 1)*(-125i/(1728*a^3*d^3))^(1/3))/2 - tan(c + d*x)^(1/3)/(2*a*
d*(tan(c + d*x)*1i + 1))

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