\(\int x^3 (a+b \arctan (c x^3)) \, dx\) [104]

Optimal result

Integrand size = 14, antiderivative size = 174 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=-\frac {3 b x}{4 c}+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )-\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac {b \arctan \left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}} \]

[Out]

-3/4*b*x/c+1/4*b*arctan(c^(1/3)*x)/c^(4/3)+1/4*x^4*(a+b*arctan(c*x^3))+1/8*b*arctan(2*c^(1/3)*x-3^(1/2))/c^(4/
3)+1/8*b*arctan(2*c^(1/3)*x+3^(1/2))/c^(4/3)-1/16*b*ln(1+c^(2/3)*x^2-c^(1/3)*x*3^(1/2))*3^(1/2)/c^(4/3)+1/16*b
*ln(1+c^(2/3)*x^2+c^(1/3)*x*3^(1/2))*3^(1/2)/c^(4/3)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4946, 327, 215, 648, 632, 210, 642, 209} \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}-\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac {b \arctan \left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac {\sqrt {3} b \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}-\frac {3 b x}{4 c} \]

[In]

Int[x^3*(a + b*ArcTan[c*x^3]),x]

[Out]

(-3*b*x)/(4*c) + (b*ArcTan[c^(1/3)*x])/(4*c^(4/3)) + (x^4*(a + b*ArcTan[c*x^3]))/4 - (b*ArcTan[Sqrt[3] - 2*c^(
1/3)*x])/(8*c^(4/3)) + (b*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/(8*c^(4/3)) - (Sqrt[3]*b*Log[1 - Sqrt[3]*c^(1/3)*x +
c^(2/3)*x^2])/(16*c^(4/3)) + (Sqrt[3]*b*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(16*c^(4/3))

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )-\frac {1}{4} (3 b c) \int \frac {x^6}{1+c^2 x^6} \, dx \\ & = -\frac {3 b x}{4 c}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )+\frac {(3 b) \int \frac {1}{1+c^2 x^6} \, dx}{4 c} \\ & = -\frac {3 b x}{4 c}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )+\frac {b \int \frac {1}{1+c^{2/3} x^2} \, dx}{4 c}+\frac {b \int \frac {1-\frac {1}{2} \sqrt {3} \sqrt [3]{c} x}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 c}+\frac {b \int \frac {1+\frac {1}{2} \sqrt {3} \sqrt [3]{c} x}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 c} \\ & = -\frac {3 b x}{4 c}+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} x^4 \left (a+b \arctan \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}{16 c^{4/3}}+\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}{16 c^{4/3}}+\frac {b \int \frac {1}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{16 c}+\frac {b \int \frac {1}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{16 c} \\ & = -\frac {3 b x}{4 c}+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} x^4 \left (a+b \arctan \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 )}{16 c^{4/3}}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}}+\frac {b \text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 \sqrt {3} c^{4/3}}-\frac {b \text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 \sqrt {3} c^{4/3}} \\ & = -\frac {3 b x}{4 c}+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^3\right )\right )-\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac {b \arctan \left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.03 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=-\frac {3 b x}{4 c}+\frac {a x^4}{4}+\frac {b \arctan \left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} b x^4 \arctan \left (c x^3\right )-\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac {b \arctan \left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}} \]

[In]

Integrate[x^3*(a + b*ArcTan[c*x^3]),x]

[Out]

(-3*b*x)/(4*c) + (a*x^4)/4 + (b*ArcTan[c^(1/3)*x])/(4*c^(4/3)) + (b*x^4*ArcTan[c*x^3])/4 - (b*ArcTan[Sqrt[3] -
2*c^(1/3)*x])/(8*c^(4/3)) + (b*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/(8*c^(4/3)) - (Sqrt[3]*b*Log[1 - Sqrt[3]*c^(1/3
)*x + c^(2/3)*x^2])/(16*c^(4/3)) + (Sqrt[3]*b*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(16*c^(4/3))

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {a \,x^{4}}{4}+b \left (\frac {x^{4} \arctan \left (c \,x^{3}\right )}{4}-\frac {3 c \left (\frac {x}{c^{2}}-\frac {\frac {\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{12}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6}-\frac {\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{12}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\right )}{3}}{c^{2}}\right )}{4}\right )\)	\(155\)
parts	\(\frac {a \,x^{4}}{4}+b \left (\frac {x^{4} \arctan \left (c \,x^{3}\right )}{4}-\frac {3 c \left (\frac {x}{c^{2}}-\frac {\frac {\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{12}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6}-\frac {\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{12}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6}+\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} \arctan \left (\frac {x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\right )}{3}}{c^{2}}\right )}{4}\right )\)	\(155\)

[In]

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

[Out]

1/4*a*x^4+b*(1/4*x^4*arctan(c*x^3)-3/4*c*(1/c^2*x-(1/12*3^(1/2)*(1/c^2)^(1/6)*ln(x^2+3^(1/2)*(1/c^2)^(1/6)*x+(
1/c^2)^(1/3))+1/6*(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(1/6)+3^(1/2))-1/12*3^(1/2)*(1/c^2)^(1/6)*ln(x^2-3^(1/2)*(1
/c^2)^(1/6)*x+(1/c^2)^(1/3))+1/6*(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(1/6)-3^(1/2))+1/3*(1/c^2)^(1/6)*arctan(x/(1
/c^2)^(1/6)))/c^2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (126) = 252\).

Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.55 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\frac {4 \, b c x^{4} \arctan \left (c x^{3}\right ) + 4 \, a c x^{4} + {\left (\sqrt {-3} c + c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x + \frac {1}{2} \, {\left (\sqrt {-3} c + c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) - {\left (\sqrt {-3} c + c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x - \frac {1}{2} \, {\left (\sqrt {-3} c + c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) + {\left (\sqrt {-3} c - c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x + \frac {1}{2} \, {\left (\sqrt {-3} c - c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) - {\left (\sqrt {-3} c - c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x - \frac {1}{2} \, {\left (\sqrt {-3} c - c\right )} \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) + 2 \, c \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x + c \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) - 2 \, c \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}} \log \left (b x - c \left (-\frac {b^{6}}{c^{8}}\right )^{\frac {1}{6}}\right ) - 12 \, b x}{16 \, c} \]

[In]

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

[Out]

1/16*(4*b*c*x^4*arctan(c*x^3) + 4*a*c*x^4 + (sqrt(-3)*c + c)*(-b^6/c^8)^(1/6)*log(b*x + 1/2*(sqrt(-3)*c + c)*(
-b^6/c^8)^(1/6)) - (sqrt(-3)*c + c)*(-b^6/c^8)^(1/6)*log(b*x - 1/2*(sqrt(-3)*c + c)*(-b^6/c^8)^(1/6)) + (sqrt(
-3)*c - c)*(-b^6/c^8)^(1/6)*log(b*x + 1/2*(sqrt(-3)*c - c)*(-b^6/c^8)^(1/6)) - (sqrt(-3)*c - c)*(-b^6/c^8)^(1/
6)*log(b*x - 1/2*(sqrt(-3)*c - c)*(-b^6/c^8)^(1/6)) + 2*c*(-b^6/c^8)^(1/6)*log(b*x + c*(-b^6/c^8)^(1/6)) - 2*c
*(-b^6/c^8)^(1/6)*log(b*x - c*(-b^6/c^8)^(1/6)) - 12*b*x)/c

Sympy [A] (verification not implemented)

Time = 24.30 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.47 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atan}{\left (c x^{3} \right )}}{4} - \frac {3 b x}{4 c} - \frac {3 b \sqrt [6]{- \frac {1}{c^{2}}} \log {\left (4 x^{2} - 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{16 c} + \frac {3 b \sqrt [6]{- \frac {1}{c^{2}}} \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{16 c} + \frac {\sqrt {3} b \sqrt [6]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} - \frac {\sqrt {3}}{3} \right )}}{8 c} + \frac {\sqrt {3} b \sqrt [6]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3}}{3} \right )}}{8 c} + \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{4 c^{2} \sqrt [3]{- \frac {1}{c^{2}}}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*atan(c*x**3)/4 - 3*b*x/(4*c) - 3*b*(-1/c**2)**(1/6)*log(4*x**2 - 4*x*(-1/c**2)**(
1/6) + 4*(-1/c**2)**(1/3))/(16*c) + 3*b*(-1/c**2)**(1/6)*log(4*x**2 + 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3
))/(16*c) + sqrt(3)*b*(-1/c**2)**(1/6)*atan(2*sqrt(3)*x/(3*(-1/c**2)**(1/6)) - sqrt(3)/3)/(8*c) + sqrt(3)*b*(-
1/c**2)**(1/6)*atan(2*sqrt(3)*x/(3*(-1/c**2)**(1/6)) + sqrt(3)/3)/(8*c) + b*atan(c*x**3)/(4*c**2*(-1/c**2)**(1
/3)), Ne(c, 0)), (a*x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.85 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{16} \, {\left (4 \, x^{4} \arctan \left (c x^{3}\right ) + c {\left (\frac {\frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} + \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {1}{3}}} - \frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} - \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {1}{3}}} + \frac {4 \, \arctan \left (c^{\frac {1}{3}} x\right )}{c^{\frac {1}{3}}} + \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x + \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {1}{3}}} + \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x - \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {1}{3}}}}{c^{2}} - \frac {12 \, x}{c^{2}}\right )}\right )} b \]

[In]

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

[Out]

1/4*a*x^4 + 1/16*(4*x^4*arctan(c*x^3) + c*((sqrt(3)*log(c^(2/3)*x^2 + sqrt(3)*c^(1/3)*x + 1)/c^(1/3) - sqrt(3)
*log(c^(2/3)*x^2 - sqrt(3)*c^(1/3)*x + 1)/c^(1/3) + 4*arctan(c^(1/3)*x)/c^(1/3) + 2*arctan((2*c^(2/3)*x + sqrt
(3)*c^(1/3))/c^(1/3))/c^(1/3) + 2*arctan((2*c^(2/3)*x - sqrt(3)*c^(1/3))/c^(1/3))/c^(1/3))/c^2 - 12*x/c^2))*b

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\frac {1}{16} \, b c^{7} {\left (\frac {\sqrt {3} \log \left (x^{2} + \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{8} {\left | c \right |}^{\frac {1}{3}}} - \frac {\sqrt {3} \log \left (x^{2} - \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{8} {\left | c \right |}^{\frac {1}{3}}} + \frac {2 \, \arctan \left ({\left (2 \, x + \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{8} {\left | c \right |}^{\frac {1}{3}}} + \frac {2 \, \arctan \left ({\left (2 \, x - \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{8} {\left | c \right |}^{\frac {1}{3}}} + \frac {4 \, \arctan \left (x {\left | c \right |}^{\frac {1}{3}}\right )}{c^{8} {\left | c \right |}^{\frac {1}{3}}}\right )} + \frac {b c x^{4} \arctan \left (c x^{3}\right ) + a c x^{4} - 3 \, b x}{4 \, c} \]

[In]

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

[Out]

1/16*b*c^7*(sqrt(3)*log(x^2 + sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/(c^8*abs(c)^(1/3)) - sqrt(3)*log(x^2 -
sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/(c^8*abs(c)^(1/3)) + 2*arctan((2*x + sqrt(3)/abs(c)^(1/3))*abs(c)^(1/
3))/(c^8*abs(c)^(1/3)) + 2*arctan((2*x - sqrt(3)/abs(c)^(1/3))*abs(c)^(1/3))/(c^8*abs(c)^(1/3)) + 4*arctan(x*a
bs(c)^(1/3))/(c^8*abs(c)^(1/3))) + 1/4*(b*c*x^4*arctan(c*x^3) + a*c*x^4 - 3*b*x)/c

Mupad [B] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int x^3 \left (a+b \arctan \left (c x^3\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,\left (\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )-\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )+2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\right )}{8\,c^{4/3}}+\frac {b\,x^4\,\mathrm {atan}\left (c\,x^3\right )}{4}-\frac {3\,b\,x}{4\,c}-\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )+\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )\right )\,1{}\mathrm {i}}{8\,c^{4/3}} \]

[In]

int(x^3*(a + b*atan(c*x^3)),x)

[Out]

(a*x^4)/4 - (b*(atan((-1)^(2/3)*c^(1/3)*x) - atan((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)))/(8*c^(4/3)) + (b*x^4*atan(c*x^3))/4 - (3*b*x)/(4*c) - (3^(1/2)*b*(atan((c^(1/3)*x*(3
^(1/2)*1i + 1))/2) + atan((-1)^(2/3)*c^(1/3)*x))*1i)/(8*c^(4/3))