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

Optimal. Leaf size=76 \[ \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {a b x}{c}+\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {b^2 x \tan ^{-1}(c x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4852, 4916, 4846, 260, 4884} \[ \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {a b x}{c}+\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {b^2 x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x])^2,x]

[Out]

-((a*b*x)/c) - (b^2*x*ArcTan[c*x])/c + (a + b*ArcTan[c*x])^2/(2*c^2) + (x^2*(a + b*ArcTan[c*x])^2)/2 + (b^2*Lo
g[1 + c^2*x^2])/(2*c^2)

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 4852

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

Rule 4884

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

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 75, normalized size = 0.99 \[ \frac {2 b \tan ^{-1}(c x) \left (a c^2 x^2+a-b c x\right )+a c x (a c x-2 b)+b^2 \log \left (c^2 x^2+1\right )+b^2 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTan[c*x])^2,x]

[Out]

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

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fricas [A]  time = 0.43, size = 83, normalized size = 1.09 \[ \frac {a^{2} c^{2} x^{2} - 2 \, a b c x + {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x\right )^{2} + b^{2} \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (a b c^{2} x^{2} - b^{2} c x + a b\right )} \arctan \left (c x\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [A]  time = 0.02, size = 97, normalized size = 1.28 \[ \frac {a^{2} x^{2}}{2}+\frac {x^{2} b^{2} \arctan \left (c x \right )^{2}}{2}+\frac {b^{2} \arctan \left (c x \right )^{2}}{2 c^{2}}-\frac {b^{2} x \arctan \left (c x \right )}{c}+\frac {b^{2} \ln \left (c^{2} x^{2}+1\right )}{2 c^{2}}+a b \,x^{2} \arctan \left (c x \right )+\frac {a b \arctan \left (c x \right )}{c^{2}}-\frac {a b x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 104, normalized size = 1.37 \[ \frac {1}{2} \, b^{2} x^{2} \arctan \left (c x\right )^{2} + \frac {1}{2} \, a^{2} x^{2} + {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} a b - \frac {1}{2} \, {\left (2 \, c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )} \arctan \left (c x\right ) + \frac {\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right )}{c^{2}}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.41, size = 88, normalized size = 1.16 \[ \frac {\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2}+\frac {b^2\,\ln \left (c^2\,x^2+1\right )}{2}-c\,\left (x\,\mathrm {atan}\left (c\,x\right )\,b^2+a\,x\,b\right )+a\,b\,\mathrm {atan}\left (c\,x\right )}{c^2}+\frac {a^2\,x^2}{2}+\frac {b^2\,x^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2}+a\,b\,x^2\,\mathrm {atan}\left (c\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atan(c*x))^2,x)

[Out]

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

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sympy [A]  time = 0.73, size = 107, normalized size = 1.41 \[ \begin {cases} \frac {a^{2} x^{2}}{2} + a b x^{2} \operatorname {atan}{\left (c x \right )} - \frac {a b x}{c} + \frac {a b \operatorname {atan}{\left (c x \right )}}{c^{2}} + \frac {b^{2} x^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2} - \frac {b^{2} x \operatorname {atan}{\left (c x \right )}}{c} + \frac {b^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2}} + \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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