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

Optimal. Leaf size=50 \[ \frac {1}{3} a^2 c x^3 \tan ^{-1}(a x)-\frac {c \log \left (a^2 x^2+1\right )}{3 a}-\frac {1}{6} a c x^2+c x \tan ^{-1}(a x) \]

[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

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4878, 4846, 260} \[ -\frac {c \left (a^2 x^2+1\right )}{6 a}-\frac {c \log \left (a^2 x^2+1\right )}{3 a}+\frac {1}{3} c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)+\frac {2}{3} c x \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)*ArcTan[a*x],x]

[Out]

-(c*(1 + a^2*x^2))/(6*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 260

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 4846

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

Rule 4878

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

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Mathematica [A]  time = 0.01, size = 50, normalized size = 1.00 \[ \frac {1}{3} a^2 c x^3 \tan ^{-1}(a x)-\frac {c \log \left (a^2 x^2+1\right )}{3 a}-\frac {1}{6} a c x^2+c x \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.01, size = 47, normalized size = 0.94 \[ -\frac {a^{2} c x^{2} - 2 \, {\left (a^{3} c x^{3} + 3 \, a c x\right )} \arctan \left (a x\right ) + 2 \, c \log \left (a^{2} x^{2} + 1\right )}{6 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.02, size = 45, normalized size = 0.90 \[ -\frac {a \,x^{2} c}{6}+c x \arctan \left (a x \right )+\frac {a^{2} c \,x^{3} \arctan \left (a x \right )}{3}-\frac {c \ln \left (a^{2} x^{2}+1\right )}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*x^2*c+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

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maxima [A]  time = 0.33, size = 45, normalized size = 0.90 \[ -\frac {1}{6} \, {\left (c x^{2} + \frac {2 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a + \frac {1}{3} \, {\left (a^{2} c x^{3} + 3 \, c x\right )} \arctan \left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.16, size = 46, normalized size = 0.92 \[ -\frac {c\,\left (2\,\ln \left (a^2\,x^2+1\right )+a^2\,x^2-2\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-6\,a\,x\,\mathrm {atan}\left (a\,x\right )\right )}{6\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.57, size = 48, normalized size = 0.96 \[ \begin {cases} \frac {a^{2} c x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {a c x^{2}}{6} + c x \operatorname {atan}{\left (a x \right )} - \frac {c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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