3.15 \(\int x^3 (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=112 \[ \frac{a b x}{2 c^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 x^2}{12 c^2}-\frac{b^2 \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 x \tan ^{-1}(c x)}{2 c^3} \]

[Out]

(a*b*x)/(2*c^3) + (b^2*x^2)/(12*c^2) + (b^2*x*ArcTan[c*x])/(2*c^3) - (b*x^3*(a + b*ArcTan[c*x]))/(6*c) - (a +
b*ArcTan[c*x])^2/(4*c^4) + (x^4*(a + b*ArcTan[c*x])^2)/4 - (b^2*Log[1 + c^2*x^2])/(3*c^4)

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Rubi [A]  time = 0.208729, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4852, 4916, 266, 43, 4846, 260, 4884} \[ \frac{a b x}{2 c^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 x^2}{12 c^2}-\frac{b^2 \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 x \tan ^{-1}(c x)}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTan[c*x])^2,x]

[Out]

(a*b*x)/(2*c^3) + (b^2*x^2)/(12*c^2) + (b^2*x*ArcTan[c*x])/(2*c^3) - (b*x^3*(a + b*ArcTan[c*x]))/(6*c) - (a +
b*ArcTan[c*x])^2/(4*c^4) + (x^4*(a + b*ArcTan[c*x])^2)/4 - (b^2*Log[1 + c^2*x^2])/(3*c^4)

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 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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 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]

Rubi steps

\begin{align*} \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} (b c) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac{b \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{6} b^2 \int \frac{x^3}{1+c^2 x^2} \, dx+\frac{b \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac{b \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}\\ &=\frac{a b x}{2 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{12} b^2 \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )+\frac{b^2 \int \tan ^{-1}(c x) \, dx}{2 c^3}\\ &=\frac{a b x}{2 c^3}+\frac{b^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{12} b^2 \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac{b^2 \int \frac{x}{1+c^2 x^2} \, dx}{2 c^2}\\ &=\frac{a b x}{2 c^3}+\frac{b^2 x^2}{12 c^2}+\frac{b^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b^2 \log \left (1+c^2 x^2\right )}{3 c^4}\\ \end{align*}

Mathematica [A]  time = 0.0794119, size = 111, normalized size = 0.99 \[ \frac{c x \left (3 a^2 c^3 x^3-2 a b c^2 x^2+6 a b+b^2 c x\right )-2 b \tan ^{-1}(c x) \left (a \left (3-3 c^4 x^4\right )+b c x \left (c^2 x^2-3\right )\right )-4 b^2 \log \left (c^2 x^2+1\right )+3 b^2 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^2}{12 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTan[c*x])^2,x]

[Out]

(c*x*(6*a*b + b^2*c*x - 2*a*b*c^2*x^2 + 3*a^2*c^3*x^3) - 2*b*(b*c*x*(-3 + c^2*x^2) + a*(3 - 3*c^4*x^4))*ArcTan
[c*x] + 3*b^2*(-1 + c^4*x^4)*ArcTan[c*x]^2 - 4*b^2*Log[1 + c^2*x^2])/(12*c^4)

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Maple [A]  time = 0.016, size = 135, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}{x}^{4}}{4}}+{\frac{{b}^{2}{x}^{4} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{4}}-{\frac{{b}^{2}\arctan \left ( cx \right ){x}^{3}}{6\,c}}+{\frac{{b}^{2}x\arctan \left ( cx \right ) }{2\,{c}^{3}}}-{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{4\,{c}^{4}}}+{\frac{{b}^{2}{x}^{2}}{12\,{c}^{2}}}-{\frac{{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,{c}^{4}}}+{\frac{{x}^{4}ab\arctan \left ( cx \right ) }{2}}-{\frac{ab{x}^{3}}{6\,c}}+{\frac{xab}{2\,{c}^{3}}}-{\frac{ab\arctan \left ( cx \right ) }{2\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x))^2,x)

[Out]

1/4*a^2*x^4+1/4*b^2*x^4*arctan(c*x)^2-1/6/c*b^2*arctan(c*x)*x^3+1/2*b^2*x*arctan(c*x)/c^3-1/4/c^4*b^2*arctan(c
*x)^2+1/12*b^2*x^2/c^2-1/3*b^2*ln(c^2*x^2+1)/c^4+1/2*x^4*a*b*arctan(c*x)-1/6*a*b*x^3/c+1/2*a*b*x/c^3-1/2/c^4*a
*b*arctan(c*x)

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Maxima [A]  time = 1.51971, size = 184, normalized size = 1.64 \begin{align*} \frac{1}{4} \, b^{2} x^{4} \arctan \left (c x\right )^{2} + \frac{1}{4} \, a^{2} x^{4} + \frac{1}{6} \,{\left (3 \, x^{4} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a b - \frac{1}{12} \,{\left (2 \, c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )} \arctan \left (c x\right ) - \frac{c^{2} x^{2} + 3 \, \arctan \left (c x\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arctan(c*x)^2 + 1/4*a^2*x^4 + 1/6*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5)
)*a*b - 1/12*(2*c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5)*arctan(c*x) - (c^2*x^2 + 3*arctan(c*x)^2 - 4*log(c
^2*x^2 + 1))/c^4)*b^2

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Fricas [A]  time = 2.40703, size = 266, normalized size = 2.38 \begin{align*} \frac{3 \, a^{2} c^{4} x^{4} - 2 \, a b c^{3} x^{3} + b^{2} c^{2} x^{2} + 6 \, a b c x + 3 \,{\left (b^{2} c^{4} x^{4} - b^{2}\right )} \arctan \left (c x\right )^{2} - 4 \, b^{2} \log \left (c^{2} x^{2} + 1\right ) + 2 \,{\left (3 \, a b c^{4} x^{4} - b^{2} c^{3} x^{3} + 3 \, b^{2} c x - 3 \, a b\right )} \arctan \left (c x\right )}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

1/12*(3*a^2*c^4*x^4 - 2*a*b*c^3*x^3 + b^2*c^2*x^2 + 6*a*b*c*x + 3*(b^2*c^4*x^4 - b^2)*arctan(c*x)^2 - 4*b^2*lo
g(c^2*x^2 + 1) + 2*(3*a*b*c^4*x^4 - b^2*c^3*x^3 + 3*b^2*c*x - 3*a*b)*arctan(c*x))/c^4

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Sympy [A]  time = 2.08856, size = 155, normalized size = 1.38 \begin{align*} \begin{cases} \frac{a^{2} x^{4}}{4} + \frac{a b x^{4} \operatorname{atan}{\left (c x \right )}}{2} - \frac{a b x^{3}}{6 c} + \frac{a b x}{2 c^{3}} - \frac{a b \operatorname{atan}{\left (c x \right )}}{2 c^{4}} + \frac{b^{2} x^{4} \operatorname{atan}^{2}{\left (c x \right )}}{4} - \frac{b^{2} x^{3} \operatorname{atan}{\left (c x \right )}}{6 c} + \frac{b^{2} x^{2}}{12 c^{2}} + \frac{b^{2} x \operatorname{atan}{\left (c x \right )}}{2 c^{3}} - \frac{b^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{3 c^{4}} - \frac{b^{2} \operatorname{atan}^{2}{\left (c x \right )}}{4 c^{4}} & \text{for}\: c \neq 0 \\\frac{a^{2} x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x))**2,x)

[Out]

Piecewise((a**2*x**4/4 + a*b*x**4*atan(c*x)/2 - a*b*x**3/(6*c) + a*b*x/(2*c**3) - a*b*atan(c*x)/(2*c**4) + b**
2*x**4*atan(c*x)**2/4 - b**2*x**3*atan(c*x)/(6*c) + b**2*x**2/(12*c**2) + b**2*x*atan(c*x)/(2*c**3) - b**2*log
(x**2 + c**(-2))/(3*c**4) - b**2*atan(c*x)**2/(4*c**4), Ne(c, 0)), (a**2*x**4/4, True))

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Giac [A]  time = 1.18741, size = 181, normalized size = 1.62 \begin{align*} \frac{3 \, b^{2} c^{4} x^{4} \arctan \left (c x\right )^{2} + 6 \, a b c^{4} x^{4} \arctan \left (c x\right ) + 3 \, a^{2} c^{4} x^{4} - 2 \, b^{2} c^{3} x^{3} \arctan \left (c x\right ) - 2 \, a b c^{3} x^{3} + b^{2} c^{2} x^{2} + 6 \, b^{2} c x \arctan \left (c x\right ) + 6 \, a b c x - 3 \, b^{2} \arctan \left (c x\right )^{2} - 6 \, a b \arctan \left (c x\right ) - 4 \, b^{2} \log \left (c^{2} x^{2} + 1\right )}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))^2,x, algorithm="giac")

[Out]

1/12*(3*b^2*c^4*x^4*arctan(c*x)^2 + 6*a*b*c^4*x^4*arctan(c*x) + 3*a^2*c^4*x^4 - 2*b^2*c^3*x^3*arctan(c*x) - 2*
a*b*c^3*x^3 + b^2*c^2*x^2 + 6*b^2*c*x*arctan(c*x) + 6*a*b*c*x - 3*b^2*arctan(c*x)^2 - 6*a*b*arctan(c*x) - 4*b^
2*log(c^2*x^2 + 1))/c^4