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*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))

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Rubi [A]  time = 0.10147, antiderivative size = 159, normalized size of antiderivative = 1., 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 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 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 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 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 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 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])

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*}

Mathematica [A]  time = 0.0334661, 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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Maple [A]  time = 0.026, size = 138, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}a}{3}}+{\frac{b{x}^{3}\arctan \left ( c{x}^{2} \right ) }{3}}-{\frac{2\,bx}{3\,c}}+{\frac{b\sqrt{2}}{12\,c}\sqrt [4]{{c}^{-2}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) \left ({x}^{2}-\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) ^{-1}} \right ) }+{\frac{b\sqrt{2}}{6\,c}\sqrt [4]{{c}^{-2}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{c}^{-2}}}}}+1 \right ) }+{\frac{b\sqrt{2}}{6\,c}\sqrt [4]{{c}^{-2}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{c}^{-2}}}}}-1 \right ) } \end{align*}

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 [B]  time = 1.53196, size = 354, normalized size = 2.23 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{12} \,{\left (4 \, x^{3} \arctan \left (c x^{2}\right ) + c{\left (\frac{\frac{\sqrt{2} \log \left (\sqrt{c^{2}} x^{2} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}}} - \frac{\sqrt{2} \log \left (\sqrt{c^{2}} x^{2} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}}} + \frac{\sqrt{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{-\sqrt{c^{2}}}} + \frac{\sqrt{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{-\sqrt{c^{2}}}}}{c^{2}} - \frac{8 \, x}{c^{2}}\right )}\right )} b \end{align*}

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*((sqrt(2)*log(sqrt(c^2)*x^2 + sqrt(2)*(c^2)^(1/4)*x + 1)/(c^2)^(1/4)
- sqrt(2)*log(sqrt(c^2)*x^2 - sqrt(2)*(c^2)^(1/4)*x + 1)/(c^2)^(1/4) + sqrt(2)*log((2*sqrt(c^2)*x - sqrt(2)*s
qrt(-sqrt(c^2)) + sqrt(2)*(c^2)^(1/4))/(2*sqrt(c^2)*x + sqrt(2)*sqrt(-sqrt(c^2)) + sqrt(2)*(c^2)^(1/4)))/sqrt(
-sqrt(c^2)) + sqrt(2)*log((2*sqrt(c^2)*x - sqrt(2)*sqrt(-sqrt(c^2)) - sqrt(2)*(c^2)^(1/4))/(2*sqrt(c^2)*x + sq
rt(2)*sqrt(-sqrt(c^2)) - sqrt(2)*(c^2)^(1/4)))/sqrt(-sqrt(c^2)))/c^2 - 8*x/c^2))*b

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Fricas [B]  time = 2.80276, size = 807, normalized size = 5.08 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 28.3316, size = 1207, normalized size = 7.59 \begin{align*} \text{result too large to display} \end{align*}

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, Eq(c, 0)), (x**3*(a - b*atan((-sqrt(2)/2 - sqrt(2)*I/2)**(-2)))/3, Eq(c, -1/(x**2*(-sqrt(
2)/2 - sqrt(2)*I/2)**2))), (x**3*(a - b*atan((-sqrt(2)/2 + sqrt(2)*I/2)**(-2)))/3, Eq(c, -1/(x**2*(-sqrt(2)/2
+ sqrt(2)*I/2)**2))), (x**3*(a - b*atan((sqrt(2)/2 - sqrt(2)*I/2)**(-2)))/3, Eq(c, -1/(x**2*(sqrt(2)/2 - sqrt(
2)*I/2)**2))), (x**3*(a - b*atan((sqrt(2)/2 + sqrt(2)*I/2)**(-2)))/3, Eq(c, -1/(x**2*(sqrt(2)/2 + sqrt(2)*I/2)
**2))), (-2*(-1)**(3/4)*a*c**7*x**7*(c**(-2))**(7/4)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4
)*c**5*(c**(-2))**(7/4)) - 2*(-1)**(3/4)*a*c**5*x**3*(c**(-2))**(7/4)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/
4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) + 2*I*b*c**15*x**4*(c**(-2))**(13/2)*atan(c*x**2)/(-6*(-1)**(3/4)*c*
*7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) - 2*(-1)**(3/4)*b*c**7*x**7*(c**(-2))**(7/4)*a
tan(c*x**2)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) + 2*I*b*c**7*(c*
*(-2))**(7/2)*atan(c*x**2)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) +
4*(-1)**(3/4)*b*c**6*x**5*(c**(-2))**(7/4)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c
**(-2))**(7/4)) - 2*(-1)**(3/4)*b*c**5*x**3*(c**(-2))**(7/4)*atan(c*x**2)/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))*
*(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) + 4*(-1)**(3/4)*b*c**4*x*(c**(-2))**(7/4)/(-6*(-1)**(3/4)*c**7*x
**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) - 2*b*c**2*x**4*log(x - (-1)**(1/4)*(c**(-2))**(1/
4))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) + b*c**2*x**4*log(x**2 +
I*sqrt(c**(-2)))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)) - 2*b*c**2
*x**4*atan((-1)**(3/4)*x/(c**(-2))**(1/4))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)**(3/4)*c**5*(c*
*(-2))**(7/4)) - 2*b*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*(-1)
**(3/4)*c**5*(c**(-2))**(7/4)) + b*log(x**2 + I*sqrt(c**(-2)))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-2))**(7/4) - 6*
(-1)**(3/4)*c**5*(c**(-2))**(7/4)) - 2*b*atan((-1)**(3/4)*x/(c**(-2))**(1/4))/(-6*(-1)**(3/4)*c**7*x**4*(c**(-
2))**(7/4) - 6*(-1)**(3/4)*c**5*(c**(-2))**(7/4)), True))

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Giac [A]  time = 1.23932, size = 223, normalized size = 1.4 \begin{align*} \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} \end{align*}

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