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

Optimal. Leaf size=174 \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac{\sqrt{3} b \log \left (c^{2/3} x^2-\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{\sqrt{3} b \log \left (c^{2/3} x^2+\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}-\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac{b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt{3}\right )}{8 c^{4/3}}-\frac{3 b x}{4 c} \]

[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))

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Rubi [A]  time = 0.324721, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5033, 321, 209, 634, 618, 204, 628, 203} \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac{\sqrt{3} b \log \left (c^{2/3} x^2-\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{\sqrt{3} b \log \left (c^{2/3} x^2+\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}-\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac{b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt{3}\right )}{8 c^{4/3}}-\frac{3 b x}{4 c} \]

Antiderivative was successfully verified.

[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 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]

Rule 321

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 209

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

Rubi steps

\begin{align*} \int x^3 \left (a+b \tan ^{-1}\left (c x^3\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}\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 \tan ^{-1}\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 \tan ^{-1}\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 \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac{1}{4} x^4 \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}{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 \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac{1}{4} x^4 \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 )}{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 \operatorname{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 \operatorname{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 \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac{b \tan ^{-1}\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]  time = 0.0428507, size = 179, normalized size = 1.03 \[ \frac{a x^4}{4}-\frac{\sqrt{3} b \log \left (c^{2/3} x^2-\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{\sqrt{3} b \log \left (c^{2/3} x^2+\sqrt{3} \sqrt [3]{c} x+1\right )}{16 c^{4/3}}+\frac{b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}-\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac{b \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt{3}\right )}{8 c^{4/3}}+\frac{1}{4} b x^4 \tan ^{-1}\left (c x^3\right )-\frac{3 b x}{4 c} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.092, size = 165, normalized size = 1. \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}\arctan \left ( c{x}^{3} \right ) }{4}}-{\frac{3\,bx}{4\,c}}+{\frac{b\sqrt{3}}{16\,c}\sqrt [6]{{c}^{-2}}\ln \left ({x}^{2}+\sqrt{3}\sqrt [6]{{c}^{-2}}x+\sqrt [3]{{c}^{-2}} \right ) }+{\frac{b}{8\,c}\sqrt [6]{{c}^{-2}}\arctan \left ( 2\,{\frac{x}{\sqrt [6]{{c}^{-2}}}}+\sqrt{3} \right ) }-{\frac{b\sqrt{3}}{16\,c}\sqrt [6]{{c}^{-2}}\ln \left ({x}^{2}-\sqrt{3}\sqrt [6]{{c}^{-2}}x+\sqrt [3]{{c}^{-2}} \right ) }+{\frac{b}{8\,c}\sqrt [6]{{c}^{-2}}\arctan \left ( 2\,{\frac{x}{\sqrt [6]{{c}^{-2}}}}-\sqrt{3} \right ) }+{\frac{b}{4\,c}\sqrt [6]{{c}^{-2}}\arctan \left ({x{\frac{1}{\sqrt [6]{{c}^{-2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.53178, size = 396, normalized size = 2.28 \begin{align*} \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 ({\left (c^{2}\right )}^{\frac{1}{3}} x^{2} + \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{6}}} - \frac{\sqrt{3} \log \left ({\left (c^{2}\right )}^{\frac{1}{3}} x^{2} - \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{6}}} + \frac{\log \left (\frac{2 \,{\left (c^{2}\right )}^{\frac{1}{3}} x + \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} - \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}{2 \,{\left (c^{2}\right )}^{\frac{1}{3}} x + \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} + \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}\right )}{\sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}} + \frac{\log \left (\frac{2 \,{\left (c^{2}\right )}^{\frac{1}{3}} x - \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} - \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}{2 \,{\left (c^{2}\right )}^{\frac{1}{3}} x - \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}} + \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}\right )}{\sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}} + \frac{2 \, \log \left (\frac{{\left (c^{2}\right )}^{\frac{1}{3}} x - \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}{{\left (c^{2}\right )}^{\frac{1}{3}} x + \sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}\right )}{\sqrt{-{\left (c^{2}\right )}^{\frac{1}{3}}}}}{c^{2}} - \frac{12 \, x}{c^{2}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)^(1/3)*x^2 + sqrt(3)*(c^2)^(1/6)*x + 1)/(c^2)^(1/
6) - sqrt(3)*log((c^2)^(1/3)*x^2 - sqrt(3)*(c^2)^(1/6)*x + 1)/(c^2)^(1/6) + log((2*(c^2)^(1/3)*x + sqrt(3)*(c^
2)^(1/6) - sqrt(-(c^2)^(1/3)))/(2*(c^2)^(1/3)*x + sqrt(3)*(c^2)^(1/6) + sqrt(-(c^2)^(1/3))))/sqrt(-(c^2)^(1/3)
) + log((2*(c^2)^(1/3)*x - sqrt(3)*(c^2)^(1/6) - sqrt(-(c^2)^(1/3)))/(2*(c^2)^(1/3)*x - sqrt(3)*(c^2)^(1/6) +
sqrt(-(c^2)^(1/3))))/sqrt(-(c^2)^(1/3)) + 2*log(((c^2)^(1/3)*x - sqrt(-(c^2)^(1/3)))/((c^2)^(1/3)*x + sqrt(-(c
^2)^(1/3))))/sqrt(-(c^2)^(1/3)))/c^2 - 12*x/c^2))*b

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Fricas [B]  time = 3.14076, size = 948, normalized size = 5.45 \begin{align*} \frac{4 \, b c x^{4} \arctan \left (c x^{3}\right ) + 4 \, a c x^{4} + \sqrt{3} c \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} \log \left (b^{2} x^{2} + \sqrt{3} b c x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} + c^{2} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{3}}\right ) - \sqrt{3} c \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} \log \left (b^{2} x^{2} - \sqrt{3} b c x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} + c^{2} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{3}}\right ) - 4 \, c \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} \arctan \left (-\frac{2 \, b c^{7} x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}} - 2 \, \sqrt{b^{2} x^{2} + \sqrt{3} b c x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} + c^{2} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{3}}} c^{7} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}} + \sqrt{3} b^{6}}{b^{6}}\right ) - 4 \, c \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} \arctan \left (-\frac{2 \, b c^{7} x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}} - 2 \, \sqrt{b^{2} x^{2} - \sqrt{3} b c x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} + c^{2} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{3}}} c^{7} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}} - \sqrt{3} b^{6}}{b^{6}}\right ) - 8 \, c \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{6}} \arctan \left (-\frac{b c^{7} x \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}} - \sqrt{b^{2} x^{2} + c^{2} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{1}{3}}} c^{7} \left (\frac{b^{6}}{c^{8}}\right )^{\frac{5}{6}}}{b^{6}}\right ) - 12 \, b x}{16 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*(b^6/c^8)^(1/6)*log(b^2*x^2 + sqrt(3)*b*c*x*(b^6/c^8)^(1
/6) + c^2*(b^6/c^8)^(1/3)) - sqrt(3)*c*(b^6/c^8)^(1/6)*log(b^2*x^2 - sqrt(3)*b*c*x*(b^6/c^8)^(1/6) + c^2*(b^6/
c^8)^(1/3)) - 4*c*(b^6/c^8)^(1/6)*arctan(-(2*b*c^7*x*(b^6/c^8)^(5/6) - 2*sqrt(b^2*x^2 + sqrt(3)*b*c*x*(b^6/c^8
)^(1/6) + c^2*(b^6/c^8)^(1/3))*c^7*(b^6/c^8)^(5/6) + sqrt(3)*b^6)/b^6) - 4*c*(b^6/c^8)^(1/6)*arctan(-(2*b*c^7*
x*(b^6/c^8)^(5/6) - 2*sqrt(b^2*x^2 - sqrt(3)*b*c*x*(b^6/c^8)^(1/6) + c^2*(b^6/c^8)^(1/3))*c^7*(b^6/c^8)^(5/6)
- sqrt(3)*b^6)/b^6) - 8*c*(b^6/c^8)^(1/6)*arctan(-(b*c^7*x*(b^6/c^8)^(5/6) - sqrt(b^2*x^2 + c^2*(b^6/c^8)^(1/3
))*c^7*(b^6/c^8)^(5/6))/b^6) - 12*b*x)/c

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Sympy [A]  time = 111.885, size = 1287, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*(a - oo*I*b)/4, Eq(c, -I/x**3)), (x**4*(a + oo*I*b)/4, Eq(c, I/x**3)), (a*x**4/4, Eq(c, 0)), (
4*I*a*c**42*x**10*(c**(-2))**(73/2)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) + 4*I*a
*c**40*x**4*(c**(-2))**(73/2)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) - 2*(-1)**(2/
3)*sqrt(3)*b*c**71*x**6*(c**(-2))**(155/3)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(16*
I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) - 2*(-1)**(2/3)*sqrt(3)*b*c**69*(c**(-2))**(155
/3)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c
**40*(c**(-2))**(73/2)) + 4*I*b*c**42*x**10*(c**(-2))**(73/2)*atan(c*x**3)/(16*I*c**42*x**6*(c**(-2))**(73/2)
+ 16*I*c**40*(c**(-2))**(73/2)) - 12*I*b*c**41*x**7*(c**(-2))**(73/2)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*
I*c**40*(c**(-2))**(73/2)) + 4*I*b*c**40*x**4*(c**(-2))**(73/2)*atan(c*x**3)/(16*I*c**42*x**6*(c**(-2))**(73/2
) + 16*I*c**40*(c**(-2))**(73/2)) - 12*I*b*c**39*x*(c**(-2))**(73/2)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I
*c**40*(c**(-2))**(73/2)) - 3*(-1)**(2/3)*b*c**33*x**6*(c**(-2))**(98/3)*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2)
)**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2))
+ 3*(-1)**(2/3)*b*c**33*x**6*(c**(-2))**(98/3)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(
c**(-2))**(1/3))/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) - 2*(-1)**(2/3)*sqrt(3)*b*
c**33*x**6*(c**(-2))**(98/3)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(16*I*c**42*x**6*(
c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) - 3*(-1)**(2/3)*b*c**31*(c**(-2))**(98/3)*log(4*x**2 - 4*(-1)
**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(
c**(-2))**(73/2)) + 3*(-1)**(2/3)*b*c**31*(c**(-2))**(98/3)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*
(-1)**(1/3)*(c**(-2))**(1/3))/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) - 2*(-1)**(2/
3)*sqrt(3)*b*c**31*(c**(-2))**(98/3)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(16*I*c**4
2*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) + 4*(-1)**(1/6)*b*c**14*x**6*(c**(-2))**(139/6)*atan(
c*x**3)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)) + 4*(-1)**(1/6)*b*(c**(-2))**(103/6
)*atan(c*x**3)/(16*I*c**42*x**6*(c**(-2))**(73/2) + 16*I*c**40*(c**(-2))**(73/2)), True))

________________________________________________________________________________________

Giac [A]  time = 1.23519, size = 225, normalized size = 1.29 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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