3.673 \(\int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+b \tan (c+d x))} \, dx\)

Optimal. Leaf size=467 \[ \frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {\sqrt {3} a \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 d \left (a^2+b^2\right )}-\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {a \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac {a \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac {\sqrt {3} b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{a} d \left (a^2+b^2\right )}-\frac {3 b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 \sqrt [3]{a} d \left (a^2+b^2\right )}+\frac {b^{4/3} \log (a+b \tan (c+d x))}{2 \sqrt [3]{a} d \left (a^2+b^2\right )} \]

[Out]

-1/2*b*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d-1/2*b*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d-b*
arctan(tan(d*x+c)^(1/3))/(a^2+b^2)/d-3/2*b^(4/3)*ln(a^(1/3)+b^(1/3)*tan(d*x+c)^(1/3))/a^(1/3)/(a^2+b^2)/d+1/2*
a*ln(1+tan(d*x+c)^(2/3))/(a^2+b^2)/d+1/2*b^(4/3)*ln(a+b*tan(d*x+c))/a^(1/3)/(a^2+b^2)/d-1/4*a*ln(1-tan(d*x+c)^
(2/3)+tan(d*x+c)^(4/3))/(a^2+b^2)/d-b^(4/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*tan(d*x+c)^(1/3))/a^(1/3)*3^(1/2))*3
^(1/2)/a^(1/3)/(a^2+b^2)/d-1/2*a*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^(1/2)/(a^2+b^2)/d-1/4*b*ln(1-3^(
1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/(a^2+b^2)/d+1/4*b*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/
3))*3^(1/2)/(a^2+b^2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.58, antiderivative size = 467, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 16, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3574, 3538, 3476, 329, 275, 200, 31, 634, 618, 204, 628, 295, 203, 3634, 56, 617} \[ -\frac {\sqrt {3} b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{a} d \left (a^2+b^2\right )}+\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {\sqrt {3} a \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 d \left (a^2+b^2\right )}-\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {a \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac {a \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac {3 b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 \sqrt [3]{a} d \left (a^2+b^2\right )}+\frac {b^{4/3} \log (a+b \tan (c+d x))}{2 \sqrt [3]{a} d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(2*(a^2 + b^2)*d) - (b*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(2*(
a^2 + b^2)*d) - (Sqrt[3]*b^(4/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x]^(1/3))/(Sqrt[3]*a^(1/3))])/(a^(1/3)*
(a^2 + b^2)*d) - (Sqrt[3]*a*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(2*(a^2 + b^2)*d) - (b*ArcTan[Tan[c +
d*x]^(1/3)])/((a^2 + b^2)*d) - (3*b^(4/3)*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]^(1/3)])/(2*a^(1/3)*(a^2 + b^2)*d)
+ (a*Log[1 + Tan[c + d*x]^(2/3)])/(2*(a^2 + b^2)*d) - (Sqrt[3]*b*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c +
d*x]^(2/3)])/(4*(a^2 + b^2)*d) + (Sqrt[3]*b*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(4*(a^2
+ b^2)*d) + (b^(4/3)*Log[a + b*Tan[c + d*x]])/(2*a^(1/3)*(a^2 + b^2)*d) - (a*Log[1 - Tan[c + d*x]^(2/3) + Tan
[c + d*x]^(4/3)])/(4*(a^2 + b^2)*d)

Rule 31

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

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 200

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 203

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

Rule 204

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

Rule 275

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 295

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)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
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 329

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 617

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 618

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 628

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 634

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 3476

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 3538

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 3574

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

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

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

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Mathematica [C]  time = 0.33, size = 162, normalized size = 0.35 \[ \frac {30 b^2 \tan ^{\frac {2}{3}}(c+d x) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b \tan (c+d x)}{a}\right )-a \left (5 a \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )-2 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )+\log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )+12 b \tan ^{\frac {5}{3}}(c+d x) \, _2F_1\left (\frac {5}{6},1;\frac {11}{6};-\tan ^2(c+d x)\right )\right )}{20 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*b^2*Hypergeometric2F1[2/3, 1, 5/3, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(2/3) - a*(5*a*(2*Sqrt[3]*ArcTan[(1
- 2*Tan[c + d*x]^(2/3))/Sqrt[3]] - 2*Log[1 + Tan[c + d*x]^(2/3)] + Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^
(4/3)]) + 12*b*Hypergeometric2F1[5/6, 1, 11/6, -Tan[c + d*x]^2]*Tan[c + d*x]^(5/3)))/(20*a*(a^2 + b^2)*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*tan(d*x + c) + a)*tan(d*x + c)^(1/3)), x)

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maple [A]  time = 0.23, size = 526, normalized size = 1.13 \[ -\frac {b \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {b \ln \left (\tan ^{\frac {2}{3}}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{2 d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {3 \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right ) \sqrt {3}\, b}{4 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right ) a}{4 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}\, a}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) b}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right ) \sqrt {3}\, b}{4 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right ) a}{4 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}\, a}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) b}{2 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{d \left (3 a^{2}+3 b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*b/(a^2+b^2)/(1/b*a)^(1/3)*ln(tan(d*x+c)^(1/3)+(1/b*a)^(1/3))+1/2/d*b/(a^2+b^2)/(1/b*a)^(1/3)*ln(tan(d*x+c
)^(2/3)-(1/b*a)^(1/3)*tan(d*x+c)^(1/3)+(1/b*a)^(2/3))+1/d*b/(a^2+b^2)*3^(1/2)/(1/b*a)^(1/3)*arctan(1/3*3^(1/2)
*(2/(1/b*a)^(1/3)*tan(d*x+c)^(1/3)-1))+3/4/d/(3*a^2+3*b^2)*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(
1/2)*b-3/4/d/(3*a^2+3*b^2)*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*a-3/2/d/(3*a^2+3*b^2)*arctan(3^(1/2
)+2*tan(d*x+c)^(1/3))*3^(1/2)*a-3/2/d/(3*a^2+3*b^2)*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))*b-3/4/d/(3*a^2+3*b^2)*l
n(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)*b-3/4/d/(3*a^2+3*b^2)*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan
(d*x+c)^(2/3))*a+3/2/d/(3*a^2+3*b^2)*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))*3^(1/2)*a-3/2/d/(3*a^2+3*b^2)*arctan(
-3^(1/2)+2*tan(d*x+c)^(1/3))*b+3/2/d/(3*a^2+3*b^2)*a*ln(1+tan(d*x+c)^(2/3))-3/d/(3*a^2+3*b^2)*b*arctan(tan(d*x
+c)^(1/3))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [B]  time = 11.67, size = 2050, normalized size = 4.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c + d*x)^(1/3) + 1i)*1i)/(2*(a*d*1i - b*d)) + log(tan(c + d*x)^(1/3)*1i + 1)/(2*(a*d - b*d*1i)) + sym
sum(log((6561*b^7*tan(c + d*x)^(1/3))/d^8 - root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b
^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)^2*(root(32*a^2*b^2*d^4*z^4 +
16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*
z + 1, z, k)*(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 -
4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)^2*((6561*(20*a^2*b^10*d^3 + 44*a^4*b^8*d^3 + 28*a^6*b^6*d^
3 + 4*a^8*b^4*d^3))/d^6 - root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^
3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)*((6561*tan(c + d*x)^(1/3)*(64*b^13*d^6 + 120*a
^2*b^11*d^6 + 96*a^4*b^9*d^6 + 80*a^6*b^7*d^6 + 32*a^8*b^5*d^6 - 8*a^10*b^3*d^6))/d^8 + (6561*root(32*a^2*b^2*
d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2
+ 4*a*d*z + 1, z, k)^2*(64*a*b^14*d^6 + 192*a^3*b^12*d^6 + 128*a^5*b^10*d^6 - 128*a^7*b^8*d^6 - 192*a^9*b^6*d
^6 - 64*a^11*b^4*d^6))/d^6)) - (6561*tan(c + d*x)^(1/3)*(40*a*b^9*d^3 - 7*a^3*b^7*d^3 + 2*a^5*b^5*d^3 + a^7*b^
3*d^3))/d^8) + (6561*(3*a*b^8 - a^3*b^6))/d^6))*root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16
*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k), k, 1, 4) + log((6561*b^
7*tan(c + d*x)^(1/3))/d^8 - ((((419904*a*b^4*(a^2 - b^2)*(a^2 + b^2)^4*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3) + (5
2488*b^3*tan(c + d*x)^(1/3)*(a^2 + b^2)^2*(a^6 - 8*b^6 + a^2*b^4 - 6*a^4*b^2))/d^2)*(-b^4/(a*d^3*(a^2 + b^2)^3
))^(1/3) + (26244*a^2*b^4*(a^2 + 5*b^2)*(a^2 + b^2)^2)/d^3)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3) - (6561*a*b^3*t
an(c + d*x)^(1/3)*(a^6 + 40*b^6 - 7*a^2*b^4 + 2*a^4*b^2))/d^5)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(1/3) - (6561*a*b^
6*(a^2 - 3*b^2))/d^6)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3))*(-b^4/(a^7*d^3 + a*b^6*d^3 + 3*a^3*b^4*d^3 + 3*a^5*b
^2*d^3))^(1/3) - log((6561*b^7*tan(c + d*x)^(1/3))/d^8 - ((3^(1/2)*1i)/2 - 1/2)*(((3^(1/2)*1i)/2 + 1/2)*((((3^
(1/2)*1i)/2 + 1/2)*((52488*b^3*tan(c + d*x)^(1/3)*(a^2 + b^2)^2*(a^6 - 8*b^6 + a^2*b^4 - 6*a^4*b^2))/d^2 + 419
904*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 - b^2)*(a^2 + b^2)^4*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3))*(-b^4/(a*d^3*(a
^2 + b^2)^3))^(1/3) - (26244*a^2*b^4*(a^2 + 5*b^2)*(a^2 + b^2)^2)/d^3)*((3^(1/2)*1i)/2 - 1/2)*(-b^4/(a*d^3*(a^
2 + b^2)^3))^(2/3) + (6561*a*b^3*tan(c + d*x)^(1/3)*(a^6 + 40*b^6 - 7*a^2*b^4 + 2*a^4*b^2))/d^5)*(-b^4/(a*d^3*
(a^2 + b^2)^3))^(1/3) - (6561*a*b^6*(a^2 - 3*b^2))/d^6)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3))*((3^(1/2)*1i)/2 +
1/2)*(-b^4/(a^7*d^3 + a*b^6*d^3 + 3*a^3*b^4*d^3 + 3*a^5*b^2*d^3))^(1/3) + log((6561*b^7*tan(c + d*x)^(1/3))/d^
8 - ((3^(1/2)*1i)/2 + 1/2)*(((3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 - 1/2)*((52488*b^3*tan(c + d*x)^(1/3)*(a^
2 + b^2)^2*(a^6 - 8*b^6 + a^2*b^4 - 6*a^4*b^2))/d^2 - 419904*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 - b^2)*(a^2 + b
^2)^4*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3))*(-b^4/(a*d^3*(a^2 + b^2)^3))^(1/3) + (26244*a^2*b^4*(a^2 + 5*b^2)*(a
^2 + b^2)^2)/d^3)*((3^(1/2)*1i)/2 + 1/2)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3) + (6561*a*b^3*tan(c + d*x)^(1/3)*(
a^6 + 40*b^6 - 7*a^2*b^4 + 2*a^4*b^2))/d^5)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(1/3) + (6561*a*b^6*(a^2 - 3*b^2))/d^
6)*(-b^4/(a*d^3*(a^2 + b^2)^3))^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-b^4/(a^7*d^3 + a*b^6*d^3 + 3*a^3*b^4*d^3 + 3*a
^5*b^2*d^3))^(1/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right ) \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(1/3)/(a+b*tan(d*x+c)),x)

[Out]

Integral(1/((a + b*tan(c + d*x))*tan(c + d*x)**(1/3)), x)

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