3.110 \(\int x (a+b \tan ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=165 \[ \frac {1}{2} x^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {\sqrt {3} b \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {\sqrt {3} b \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}+\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{4 c^{2/3}} \]

[Out]

-1/2*b*arctan(c^(1/3)*x)/c^(2/3)+1/2*x^2*(a+b*arctan(c*x^3))-1/4*b*arctan(2*c^(1/3)*x-3^(1/2))/c^(2/3)-1/4*b*a
rctan(2*c^(1/3)*x+3^(1/2))/c^(2/3)-1/8*b*ln(1+c^(2/3)*x^2-c^(1/3)*x*3^(1/2))*3^(1/2)/c^(2/3)+1/8*b*ln(1+c^(2/3
)*x^2+c^(1/3)*x*3^(1/2))*3^(1/2)/c^(2/3)

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Rubi [A]  time = 0.39, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5033, 295, 634, 618, 204, 628, 203} \[ \frac {1}{2} x^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {\sqrt {3} b \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {\sqrt {3} b \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}+\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{4 c^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x^3]),x]

[Out]

-(b*ArcTan[c^(1/3)*x])/(2*c^(2/3)) + (x^2*(a + b*ArcTan[c*x^3]))/2 + (b*ArcTan[Sqrt[3] - 2*c^(1/3)*x])/(4*c^(2
/3)) - (b*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/(4*c^(2/3)) - (Sqrt[3]*b*Log[1 - Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(8
*c^(2/3)) + (Sqrt[3]*b*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3))

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 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 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 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 170, normalized size = 1.03 \[ \frac {a x^2}{2}-\frac {\sqrt {3} b \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {\sqrt {3} b \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}+\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{4 c^{2/3}}+\frac {1}{2} b x^2 \tan ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTan[c*x^3]),x]

[Out]

(a*x^2)/2 - (b*ArcTan[c^(1/3)*x])/(2*c^(2/3)) + (b*x^2*ArcTan[c*x^3])/2 + (b*ArcTan[Sqrt[3] - 2*c^(1/3)*x])/(4
*c^(2/3)) - (b*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/(4*c^(2/3)) - (Sqrt[3]*b*Log[1 - Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2
])/(8*c^(2/3)) + (Sqrt[3]*b*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3))

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fricas [B]  time = 0.47, size = 408, normalized size = 2.47 \[ \frac {1}{2} \, b x^{2} \arctan \left (c x^{3}\right ) + \frac {1}{2} \, a x^{2} + \frac {1}{8} \, \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} \log \left (b^{10} x^{2} + \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {5}{6}} b^{5} c^{3} x + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {2}{3}} b^{6} c^{2}\right ) - \frac {1}{8} \, \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} \log \left (b^{10} x^{2} - \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {5}{6}} b^{5} c^{3} x + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {2}{3}} b^{6} c^{2}\right ) + \frac {1}{2} \, \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} \arctan \left (-\frac {2 \, \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} b^{5} c x + \sqrt {3} b^{6} - 2 \, \sqrt {b^{10} x^{2} + \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {5}{6}} b^{5} c^{3} x + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {2}{3}} b^{6} c^{2}} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} c}{b^{6}}\right ) + \frac {1}{2} \, \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} \arctan \left (-\frac {2 \, \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} b^{5} c x - \sqrt {3} b^{6} - 2 \, \sqrt {b^{10} x^{2} - \sqrt {3} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {5}{6}} b^{5} c^{3} x + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {2}{3}} b^{6} c^{2}} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} c}{b^{6}}\right ) + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} \arctan \left (-\frac {\left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} b^{5} c x - \sqrt {b^{10} x^{2} + \left (\frac {b^{6}}{c^{4}}\right )^{\frac {2}{3}} b^{6} c^{2}} \left (\frac {b^{6}}{c^{4}}\right )^{\frac {1}{6}} c}{b^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x^3)),x, algorithm="fricas")

[Out]

1/2*b*x^2*arctan(c*x^3) + 1/2*a*x^2 + 1/8*sqrt(3)*(b^6/c^4)^(1/6)*log(b^10*x^2 + sqrt(3)*(b^6/c^4)^(5/6)*b^5*c
^3*x + (b^6/c^4)^(2/3)*b^6*c^2) - 1/8*sqrt(3)*(b^6/c^4)^(1/6)*log(b^10*x^2 - sqrt(3)*(b^6/c^4)^(5/6)*b^5*c^3*x
+ (b^6/c^4)^(2/3)*b^6*c^2) + 1/2*(b^6/c^4)^(1/6)*arctan(-(2*(b^6/c^4)^(1/6)*b^5*c*x + sqrt(3)*b^6 - 2*sqrt(b^
10*x^2 + sqrt(3)*(b^6/c^4)^(5/6)*b^5*c^3*x + (b^6/c^4)^(2/3)*b^6*c^2)*(b^6/c^4)^(1/6)*c)/b^6) + 1/2*(b^6/c^4)^
(1/6)*arctan(-(2*(b^6/c^4)^(1/6)*b^5*c*x - sqrt(3)*b^6 - 2*sqrt(b^10*x^2 - sqrt(3)*(b^6/c^4)^(5/6)*b^5*c^3*x +
(b^6/c^4)^(2/3)*b^6*c^2)*(b^6/c^4)^(1/6)*c)/b^6) + (b^6/c^4)^(1/6)*arctan(-((b^6/c^4)^(1/6)*b^5*c*x - sqrt(b^
10*x^2 + (b^6/c^4)^(2/3)*b^6*c^2)*(b^6/c^4)^(1/6)*c)/b^6)

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giac [A]  time = 2.11, size = 157, normalized size = 0.95 \[ \frac {1}{8} \, b c^{5} {\left (\frac {\sqrt {3} \log \left (x^{2} + \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{4} {\left | c \right |}^{\frac {5}{3}}} - \frac {\sqrt {3} {\left | c \right |}^{\frac {1}{3}} \log \left (x^{2} - \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{6}} - \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left ({\left (2 \, x + \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{6}} - \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left ({\left (2 \, x - \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{6}} - \frac {4 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left (x {\left | c \right |}^{\frac {1}{3}}\right )}{c^{6}}\right )} + \frac {1}{2} \, b x^{2} \arctan \left (c x^{3}\right ) + \frac {1}{2} \, a x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x^3)),x, algorithm="giac")

[Out]

1/8*b*c^5*(sqrt(3)*log(x^2 + sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/(c^4*abs(c)^(5/3)) - sqrt(3)*abs(c)^(1/3
)*log(x^2 - sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^6 - 2*abs(c)^(1/3)*arctan((2*x + sqrt(3)/abs(c)^(1/3))*
abs(c)^(1/3))/c^6 - 2*abs(c)^(1/3)*arctan((2*x - sqrt(3)/abs(c)^(1/3))*abs(c)^(1/3))/c^6 - 4*abs(c)^(1/3)*arct
an(x*abs(c)^(1/3))/c^6) + 1/2*b*x^2*arctan(c*x^3) + 1/2*a*x^2

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maple [A]  time = 0.11, size = 154, normalized size = 0.93 \[ \frac {a \,x^{2}}{2}+\frac {b \,x^{2} \arctan \left (c \,x^{3}\right )}{2}-\frac {b \arctan \left (\frac {x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\frac {b c \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{8}-\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\frac {b c \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{8}-\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctan(c*x^3)),x)

[Out]

1/2*a*x^2+1/2*b*x^2*arctan(c*x^3)-1/2*b/c/(1/c^2)^(1/6)*arctan(x/(1/c^2)^(1/6))-1/8*b*c*3^(1/2)*(1/c^2)^(5/6)*
ln(x^2-3^(1/2)*(1/c^2)^(1/6)*x+(1/c^2)^(1/3))-1/4*b/c/(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(1/6)-3^(1/2))+1/8*b*c*
3^(1/2)*(1/c^2)^(5/6)*ln(x^2+3^(1/2)*(1/c^2)^(1/6)*x+(1/c^2)^(1/3))-1/4*b/c/(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(
1/6)+3^(1/2))

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maxima [A]  time = 0.43, size = 137, normalized size = 0.83 \[ \frac {1}{2} \, a x^{2} + \frac {1}{8} \, {\left (4 \, x^{2} \arctan \left (c x^{3}\right ) + c {\left (\frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} + \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} - \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {4 \, \arctan \left (c^{\frac {1}{3}} x\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x + \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x - \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x^3)),x, algorithm="maxima")

[Out]

1/2*a*x^2 + 1/8*(4*x^2*arctan(c*x^3) + c*(sqrt(3)*log(c^(2/3)*x^2 + sqrt(3)*c^(1/3)*x + 1)/c^(5/3) - sqrt(3)*l
og(c^(2/3)*x^2 - sqrt(3)*c^(1/3)*x + 1)/c^(5/3) - 4*arctan(c^(1/3)*x)/c^(5/3) - 2*arctan((2*c^(2/3)*x + sqrt(3
)*c^(1/3))/c^(1/3))/c^(5/3) - 2*arctan((2*c^(2/3)*x - sqrt(3)*c^(1/3))/c^(1/3))/c^(5/3)))*b

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mupad [B]  time = 0.69, size = 113, normalized size = 0.68 \[ \frac {a\,x^2}{2}+\frac {b\,\left (\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )+\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )+2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\right )}{4\,c^{2/3}}+\frac {b\,x^2\,\mathrm {atan}\left (c\,x^3\right )}{2}-\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )-\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\right )\,1{}\mathrm {i}}{4\,c^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atan(c*x^3)),x)

[Out]

(a*x^2)/2 + (b*(atan((-1)^(2/3)*c^(1/3)*x) + atan(((-1)^(2/3)*c^(1/3)*x*(3^(1/2)*1i - 1))/2) + 2*atan(((-1)^(2
/3)*c^(1/3)*x*(3^(1/2)*1i + 1))/2)))/(4*c^(2/3)) + (b*x^2*atan(c*x^3))/2 - (3^(1/2)*b*(atan((-1)^(2/3)*c^(1/3)
*x) - atan(((-1)^(2/3)*c^(1/3)*x*(3^(1/2)*1i - 1))/2))*1i)/(4*c^(2/3))

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sympy [A]  time = 39.66, size = 303, normalized size = 1.84 \[ \begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {atan}{\left (c x^{3} \right )}}{2} + \frac {3 \left (-1\right )^{\frac {5}{6}} b \log {\left (4 x^{2} - 4 \sqrt [6]{-1} x \sqrt [6]{\frac {1}{c^{2}}} + 4 \sqrt [3]{-1} \sqrt [3]{\frac {1}{c^{2}}} \right )}}{8 c \sqrt [6]{\frac {1}{c^{2}}}} - \frac {3 \left (-1\right )^{\frac {5}{6}} b \log {\left (4 x^{2} + 4 \sqrt [6]{-1} x \sqrt [6]{\frac {1}{c^{2}}} + 4 \sqrt [3]{-1} \sqrt [3]{\frac {1}{c^{2}}} \right )}}{8 c \sqrt [6]{\frac {1}{c^{2}}}} - \frac {\left (-1\right )^{\frac {5}{6}} \sqrt {3} b \operatorname {atan}{\left (\frac {2 \left (-1\right )^{\frac {5}{6}} \sqrt {3} x}{3 \sqrt [6]{\frac {1}{c^{2}}}} - \frac {\sqrt {3}}{3} \right )}}{4 c \sqrt [6]{\frac {1}{c^{2}}}} - \frac {\left (-1\right )^{\frac {5}{6}} \sqrt {3} b \operatorname {atan}{\left (\frac {2 \left (-1\right )^{\frac {5}{6}} \sqrt {3} x}{3 \sqrt [6]{\frac {1}{c^{2}}}} + \frac {\sqrt {3}}{3} \right )}}{4 c \sqrt [6]{\frac {1}{c^{2}}}} - \frac {\sqrt [3]{-1} b \operatorname {atan}{\left (c x^{3} \right )}}{2 c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atan(c*x**3)),x)

[Out]

Piecewise((a*x**2/2 + b*x**2*atan(c*x**3)/2 + 3*(-1)**(5/6)*b*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) +
4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*c*(c**(-2))**(1/6)) - 3*(-1)**(5/6)*b*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2)
)**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*c*(c**(-2))**(1/6)) - (-1)**(5/6)*sqrt(3)*b*atan(2*(-1)**(5/6)*s
qrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(4*c*(c**(-2))**(1/6)) - (-1)**(5/6)*sqrt(3)*b*atan(2*(-1)**(5/6)*s
qrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(4*c*(c**(-2))**(1/6)) - (-1)**(1/3)*b*atan(c*x**3)/(2*c**2*(c**(-2
))**(2/3)), Ne(c, 0)), (a*x**2/2, True))

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