\(\int (c+a^2 c x^2) \arctan (a x) \, dx\) [152]

Optimal result

Integrand size = 15, antiderivative size = 50 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{6} a c x^2+c x \arctan (a x)+\frac {1}{3} a^2 c x^3 \arctan (a x)-\frac {c \log \left (1+a^2 x^2\right )}{3 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4998, 4930, 266} \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {1}{3} c x \left (a^2 x^2+1\right ) \arctan (a x)-\frac {c \left (a^2 x^2+1\right )}{6 a}-\frac {c \log \left (a^2 x^2+1\right )}{3 a}+\frac {2}{3} c x \arctan (a x) \]

[In]

Int[(c + a^2*c*x^2)*ArcTan[a*x],x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

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

Rule 4998

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(-b)*((d + e*x^2)^q/(2*c
*q*(2*q + 1))), x] + (Dist[2*d*(q/(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[x*(d
+ e*x^2)^q*((a + b*ArcTan[c*x])/(2*q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {c \left (1+a^2 x^2\right )}{6 a}+\frac {1}{3} c x \left (1+a^2 x^2\right ) \arctan (a x)+\frac {1}{3} (2 c) \int \arctan (a x) \, dx \\ & = -\frac {c \left (1+a^2 x^2\right )}{6 a}+\frac {2}{3} c x \arctan (a x)+\frac {1}{3} c x \left (1+a^2 x^2\right ) \arctan (a x)-\frac {1}{3} (2 a c) \int \frac {x}{1+a^2 x^2} \, dx \\ & = -\frac {c \left (1+a^2 x^2\right )}{6 a}+\frac {2}{3} c x \arctan (a x)+\frac {1}{3} c x \left (1+a^2 x^2\right ) \arctan (a x)-\frac {c \log \left (1+a^2 x^2\right )}{3 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{6} a c x^2+c x \arctan (a x)+\frac {1}{3} a^2 c x^3 \arctan (a x)-\frac {c \log \left (1+a^2 x^2\right )}{3 a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92

method	result	size
parts	\(\frac {a^{2} c \,x^{3} \arctan \left (a x \right )}{3}+c x \arctan \left (a x \right )-\frac {c a \left (\frac {x^{2}}{2}+\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{2}}\right )}{3}\)	\(46\)
derivativedivides	\(\frac {\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}+a c x \arctan \left (a x \right )-\frac {c \left (\frac {a^{2} x^{2}}{2}+\ln \left (a^{2} x^{2}+1\right )\right )}{3}}{a}\)	\(49\)
default	\(\frac {\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}+a c x \arctan \left (a x \right )-\frac {c \left (\frac {a^{2} x^{2}}{2}+\ln \left (a^{2} x^{2}+1\right )\right )}{3}}{a}\)	\(49\)
parallelrisch	\(-\frac {-2 c \arctan \left (a x \right ) a^{3} x^{3}+a^{2} c \,x^{2}-6 a c x \arctan \left (a x \right )+2 c \ln \left (a^{2} x^{2}+1\right )}{6 a}\)	\(50\)
risch	\(-\frac {i c x \left (a^{2} x^{2}+3\right ) \ln \left (i a x +1\right )}{6}+\frac {i c \,a^{2} x^{3} \ln \left (-i a x +1\right )}{6}-\frac {a c \,x^{2}}{6}+\frac {i c x \ln \left (-i a x +1\right )}{2}-\frac {c \ln \left (-a^{2} x^{2}-1\right )}{3 a}\)	\(79\)
meijerg	\(\frac {c \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4 a}+\frac {c \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4 a}\)	\(102\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {a^{2} c x^{2} - 2 \, {\left (a^{3} c x^{3} + 3 \, a c x\right )} \arctan \left (a x\right ) + 2 \, c \log \left (a^{2} x^{2} + 1\right )}{6 \, a} \]

[In]

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

[Out]

-1/6*(a^2*c*x^2 - 2*(a^3*c*x^3 + 3*a*c*x)*arctan(a*x) + 2*c*log(a^2*x^2 + 1))/a

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\begin {cases} \frac {a^{2} c x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {a c x^{2}}{6} + c x \operatorname {atan}{\left (a x \right )} - \frac {c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*c*x**3*atan(a*x)/3 - a*c*x**2/6 + c*x*atan(a*x) - c*log(x**2 + a**(-2))/(3*a), Ne(a, 0)), (0,
True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{6} \, {\left (c x^{2} + \frac {2 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a + \frac {1}{3} \, {\left (a^{2} c x^{3} + 3 \, c x\right )} \arctan \left (a x\right ) \]

[In]

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

[Out]

-1/6*(c*x^2 + 2*c*log(a^2*x^2 + 1)/a^2)*a + 1/3*(a^2*c*x^3 + 3*c*x)*arctan(a*x)

Giac [F]

\[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right ) \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c\,\left (2\,\ln \left (a^2\,x^2+1\right )+a^2\,x^2-2\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-6\,a\,x\,\mathrm {atan}\left (a\,x\right )\right )}{6\,a} \]

[In]

int(atan(a*x)*(c + a^2*c*x^2),x)

[Out]

-(c*(2*log(a^2*x^2 + 1) + a^2*x^2 - 2*a^3*x^3*atan(a*x) - 6*a*x*atan(a*x)))/(6*a)