3.1117 \(\int (d+e x^2) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=68 \[ d x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (3 c^2 d-e\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{b e x^2}{6 c} \]

[Out]

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

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Rubi [A]  time = 0.0718078, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4912, 1593, 444, 43} \[ d x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (3 c^2 d-e\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{b e x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4912

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[u/(1 + c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x]
&& (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.0100485, size = 85, normalized size = 1.25 \[ a d x+\frac{1}{3} a e x^3-\frac{b d \log \left (c^2 x^2+1\right )}{2 c}+\frac{b e \log \left (c^2 x^2+1\right )}{6 c^3}+b d x \tan ^{-1}(c x)-\frac{b e x^2}{6 c}+\frac{1}{3} b e x^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 76, normalized size = 1.1 \begin{align*}{\frac{ae{x}^{3}}{3}}+adx+{\frac{be{x}^{3}\arctan \left ( cx \right ) }{3}}+b\arctan \left ( cx \right ) dx-{\frac{be{x}^{2}}{6\,c}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) d}{2\,c}}+{\frac{be\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05791, size = 108, normalized size = 1.59 \begin{align*} \frac{1}{3} \, a e x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b e + a d x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69946, size = 181, normalized size = 2.66 \begin{align*} \frac{2 \, a c^{3} e x^{3} + 6 \, a c^{3} d x - b c^{2} e x^{2} + 2 \,{\left (b c^{3} e x^{3} + 3 \, b c^{3} d x\right )} \arctan \left (c x\right ) -{\left (3 \, b c^{2} d - b e\right )} \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.03202, size = 94, normalized size = 1.38 \begin{align*} \begin{cases} a d x + \frac{a e x^{3}}{3} + b d x \operatorname{atan}{\left (c x \right )} + \frac{b e x^{3} \operatorname{atan}{\left (c x \right )}}{3} - \frac{b d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} - \frac{b e x^{2}}{6 c} + \frac{b e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} & \text{for}\: c \neq 0 \\a \left (d x + \frac{e x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10571, size = 127, normalized size = 1.87 \begin{align*} \frac{2 \, b c^{3} x^{3} \arctan \left (c x\right ) e + 2 \, a c^{3} x^{3} e + 6 \, b c^{3} d x \arctan \left (c x\right ) + 6 \, a c^{3} d x - b c^{2} x^{2} e - 3 \, b c^{2} d \log \left (c^{2} x^{2} + 1\right ) + b e \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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