3.69 \(\int x^2 (a+b \tan ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=159 \[ \frac {1}{3} x^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac {b \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}-\frac {b \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{3 \sqrt {2} c^{3/2}}-\frac {2 b x}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5033, 321, 211, 1165, 628, 1162, 617, 204} \[ \frac {1}{3} x^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac {b \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}-\frac {b \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{3 \sqrt {2} c^{3/2}}-\frac {2 b x}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTan[c*x^2]),x]

[Out]

(-2*b*x)/(3*c) + (x^3*(a + b*ArcTan[c*x^2]))/3 - (b*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(3*Sqrt[2]*c^(3/2)) + (b*Ar
cTan[1 + Sqrt[2]*Sqrt[c]*x])/(3*Sqrt[2]*c^(3/2)) - (b*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2])/(6*Sqrt[2]*c^(3/2))
+ (b*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(6*Sqrt[2]*c^(3/2))

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 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 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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

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Mathematica [A]  time = 0.04, size = 177, normalized size = 1.11 \[ \frac {a x^3}{3}-\frac {b \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \tan ^{-1}\left (\frac {2 \sqrt {c} x-\sqrt {2}}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \tan ^{-1}\left (\frac {2 \sqrt {c} x+\sqrt {2}}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}+\frac {1}{3} b x^3 \tan ^{-1}\left (c x^2\right )-\frac {2 b x}{3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTan[c*x^2]),x]

[Out]

(-2*b*x)/(3*c) + (a*x^3)/3 + (b*x^3*ArcTan[c*x^2])/3 + (b*ArcTan[(-Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(3*Sqrt[2]
*c^(3/2)) + (b*ArcTan[(Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(3*Sqrt[2]*c^(3/2)) - (b*Log[1 - Sqrt[2]*Sqrt[c]*x + c
*x^2])/(6*Sqrt[2]*c^(3/2)) + (b*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(6*Sqrt[2]*c^(3/2))

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fricas [B]  time = 0.44, size = 337, normalized size = 2.12 \[ \frac {4 \, b c x^{3} \arctan \left (c x^{2}\right ) + 4 \, a c x^{3} - 4 \, \sqrt {2} c \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} b c^{5} x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {3}{4}} - \sqrt {2} \sqrt {b^{2} x^{2} + \sqrt {2} b c x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} + c^{2} \sqrt {\frac {b^{4}}{c^{6}}}} c^{5} \left (\frac {b^{4}}{c^{6}}\right )^{\frac {3}{4}} + b^{4}}{b^{4}}\right ) - 4 \, \sqrt {2} c \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} b c^{5} x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {3}{4}} - \sqrt {2} \sqrt {b^{2} x^{2} - \sqrt {2} b c x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} + c^{2} \sqrt {\frac {b^{4}}{c^{6}}}} c^{5} \left (\frac {b^{4}}{c^{6}}\right )^{\frac {3}{4}} - b^{4}}{b^{4}}\right ) + \sqrt {2} c \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b^{2} x^{2} + \sqrt {2} b c x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} + c^{2} \sqrt {\frac {b^{4}}{c^{6}}}\right ) - \sqrt {2} c \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} \log \left (b^{2} x^{2} - \sqrt {2} b c x \left (\frac {b^{4}}{c^{6}}\right )^{\frac {1}{4}} + c^{2} \sqrt {\frac {b^{4}}{c^{6}}}\right ) - 8 \, b x}{12 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x^2)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.07, size = 165, normalized size = 1.04 \[ \frac {1}{12} \, b c^{5} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}}\right )} + \frac {b c x^{3} \arctan \left (c x^{2}\right ) + a c x^{3} - 2 \, b x}{3 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x^2)),x, algorithm="giac")

[Out]

1/12*b*c^5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/(c^6*sqrt(abs(c))) + 2*sqr
t(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/(c^6*sqrt(abs(c))) + sqrt(2)*log(x^2 + sqrt
(2)*x/sqrt(abs(c)) + 1/abs(c))/(c^6*sqrt(abs(c))) - sqrt(2)*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/(c^6*
sqrt(abs(c)))) + 1/3*(b*c*x^3*arctan(c*x^2) + a*c*x^3 - 2*b*x)/c

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maple [A]  time = 0.03, size = 138, normalized size = 0.87 \[ \frac {x^{3} a}{3}+\frac {b \,x^{3} \arctan \left (c \,x^{2}\right )}{3}-\frac {2 b x}{3 c}+\frac {b \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{12 c}+\frac {b \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{6 c}+\frac {b \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{6 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctan(c*x^2)),x)

[Out]

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

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maxima [A]  time = 0.43, size = 145, normalized size = 0.91 \[ \frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \arctan \left (c x^{2}\right ) - c {\left (\frac {8 \, x}{c^{2}} - \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}}{c^{2}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x^2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.39, size = 62, normalized size = 0.39 \[ \frac {a\,x^3}{3}+\frac {b\,x^3\,\mathrm {atan}\left (c\,x^2\right )}{3}-\frac {2\,b\,x}{3\,c}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )\,1{}\mathrm {i}}{3\,c^{3/2}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )}{3\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atan(c*x^2)),x)

[Out]

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

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sympy [A]  time = 17.68, size = 173, normalized size = 1.09 \[ \begin {cases} \frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {atan}{\left (c x^{2} \right )}}{3} + \frac {\left (-1\right )^{\frac {3}{4}} b \left (\frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (c x^{2} \right )}}{3} - \frac {2 b x}{3 c} - \frac {\sqrt [4]{-1} b \sqrt [4]{\frac {1}{c^{2}}} \log {\left (x - \sqrt [4]{-1} \sqrt [4]{\frac {1}{c^{2}}} \right )}}{3 c} + \frac {\sqrt [4]{-1} b \sqrt [4]{\frac {1}{c^{2}}} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{6 c} - \frac {\sqrt [4]{-1} b \sqrt [4]{\frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} x}{\sqrt [4]{\frac {1}{c^{2}}}} \right )}}{3 c} & \text {for}\: c \neq 0 \\\frac {a x^{3}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atan(c*x**2)),x)

[Out]

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

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