\(\int x^2 (a+b \arctan (c x^2)) \, dx\) [69]

Optimal result

Integrand size = 14, antiderivative size = 125 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {2 b x}{3 c}+\frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {b \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} x}{1+c x^2}\right )}{3 \sqrt {2} c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {2 b x}{3 c}+\frac {a x^3}{3}+\frac {1}{3} b x^3 \arctan \left (c x^2\right )+\frac {b \arctan \left (\frac {-\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}+\frac {b \arctan \left (\frac {\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}} \] Input:

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

Output:

(-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*Sqr 
t[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))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.35, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5361, 843, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \int \frac {x^4}{c^2 x^4+1}dx\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\int \frac {1}{c^2 x^4+1}dx}{c^2}\right )\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \int \frac {c x^2+1}{c^2 x^4+1}dx}{c^2}\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )}{c^2}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} \sqrt {c} x\right )^2-1}d\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{-\left (\sqrt {2} \sqrt {c} x+1\right )^2-1}d\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 \sqrt {2} c}+\frac {\int \frac {\sqrt {2} \sqrt {c} x+1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{3} x^3 \left (a+b \arctan \left (c x^2\right )\right )-\frac {2}{3} b c \left (\frac {x}{c^2}-\frac {\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}-\frac {\log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}\right )}{c^2}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[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 843

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] - Simp[ 
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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

rule 5361

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {a \,x^{3}}{3}+b \left (\frac {x^{3} \arctan \left (c \,x^{2}\right )}{3}-\frac {2 c \left (\frac {x}{c^{2}}-\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2}}\right )}{3}\right )\)	\(118\)
parts	\(\frac {a \,x^{3}}{3}+b \left (\frac {x^{3} \arctan \left (c \,x^{2}\right )}{3}-\frac {2 c \left (\frac {x}{c^{2}}-\frac {\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2}}\right )}{3}\right )\)	\(118\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {4 \, b c x^{3} \arctan \left (c x^{2}\right ) + 4 \, a c x^{3} - 8 \, b x + \frac {2 \, \sqrt {2} b \arctan \left (\sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} b \arctan \left (\sqrt {2} \sqrt {c} x - 1\right )}{\sqrt {c}} + \frac {\sqrt {2} b \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} b \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}}{12 \, c} \] Input:

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

Output:

1/12*(4*b*c*x^3*arctan(c*x^2) + 4*a*c*x^3 - 8*b*x + 2*sqrt(2)*b*arctan(sqr 
t(2)*sqrt(c)*x + 1)/sqrt(c) + 2*sqrt(2)*b*arctan(sqrt(2)*sqrt(c)*x - 1)/sq 
rt(c) + sqrt(2)*b*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/sqrt(c) - sqrt(2)*b*l 
og(c*x^2 - sqrt(2)*sqrt(c)*x + 1)/sqrt(c))/c
 

Sympy [A] (verification not implemented)

Time = 7.06 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.14 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {atan}{\left (c x^{2} \right )}}{3} + \frac {b \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (c x^{2} \right )}}{3} - \frac {2 b x}{3 c} - \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x - \sqrt [4]{- \frac {1}{c^{2}}} \right )}}{3 c} + \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{6 c} + \frac {b \sqrt [4]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {x}{\sqrt [4]{- \frac {1}{c^{2}}}} \right )}}{3 c} & \text {for}\: c \neq 0 \\\frac {a x^{3}}{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \arctan \left (c x^{2}\right ) - c {\left (\frac {8 \, x}{c^{2}} - \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}}{c^{2}}\right )}\right )} b \] Input:

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

Output:

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))/sqrt(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
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.32 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{12} \, b c^{5} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{6} \sqrt {{\left | c \right |}}}\right )} + \frac {b c x^{3} \arctan \left (c x^{2}\right ) + a c x^{3} - 2 \, b x}{3 \, c} \] Input:

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

Output:

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*sqrt(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
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.50 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a\,x^3}{3}+\frac {b\,x^3\,\mathrm {atan}\left (c\,x^2\right )}{3}-\frac {2\,b\,x}{3\,c}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )\,1{}\mathrm {i}}{3\,c^{3/2}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )}{3\,c^{3/2}} \] Input:

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

Output:

(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))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94 \[ \int x^2 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {-4 \sqrt {c}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {c}\, \sqrt {2}-2 c x}{\sqrt {c}\, \sqrt {2}}\right ) b -2 \sqrt {c}\, \sqrt {2}\, \mathit {atan} \left (c \,x^{2}\right ) b +4 \mathit {atan} \left (c \,x^{2}\right ) b \,c^{2} x^{3}-\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b +4 a \,c^{2} x^{3}-8 b c x}{12 c^{2}} \] Input:

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

Output:

( - 4*sqrt(c)*sqrt(2)*atan((sqrt(c)*sqrt(2) - 2*c*x)/(sqrt(c)*sqrt(2)))*b 
- 2*sqrt(c)*sqrt(2)*atan(c*x**2)*b + 4*atan(c*x**2)*b*c**2*x**3 - sqrt(c)* 
sqrt(2)*log( - sqrt(c)*sqrt(2)*x + c*x**2 + 1)*b + sqrt(c)*sqrt(2)*log(sqr 
t(c)*sqrt(2)*x + c*x**2 + 1)*b + 4*a*c**2*x**3 - 8*b*c*x)/(12*c**2)