3.1130 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=109 \[ -\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (3 c^4 d^2+6 c^2 d e-e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}+b c d^2 \log (x)-\frac{b e^2 x^2}{6 c} \]

[Out]

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

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Rubi [A]  time = 0.155636, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {270, 4976, 1251, 893} \[ -\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (3 c^4 d^2+6 c^2 d e-e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}+b c d^2 \log (x)-\frac{b e^2 x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.106088, size = 114, normalized size = 1.05 \[ \frac{1}{6} \left (-\frac{6 a d^2}{x}+12 a d e x+2 a e^2 x^3+\frac{b \left (-3 c^4 d^2-6 c^2 d e+e^2\right ) \log \left (c^2 x^2+1\right )}{c^3}+\frac{2 b \tan ^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )}{x}+6 b c d^2 \log (x)-\frac{b e^2 x^2}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 138, normalized size = 1.3 \begin{align*}{\frac{a{x}^{3}{e}^{2}}{3}}+2\,aedx-{\frac{a{d}^{2}}{x}}+{\frac{b\arctan \left ( cx \right ){x}^{3}{e}^{2}}{3}}+2\,b\arctan \left ( cx \right ) edx-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{x}}-{\frac{b{e}^{2}{x}^{2}}{6\,c}}-{\frac{cb\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}}{2}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) ed}{c}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{2}}{6\,{c}^{3}}}+cb{d}^{2}\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970993, size = 176, normalized size = 1.61 \begin{align*} \frac{1}{3} \, a e^{2} x^{3} - \frac{1}{2} \,{\left (c{\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \arctan \left (c x\right )}{x}\right )} b d^{2} + \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^{2} + 2 \, a d e x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d e}{c} - \frac{a d^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59273, size = 302, normalized size = 2.77 \begin{align*} \frac{2 \, a c^{3} e^{2} x^{4} + 6 \, b c^{4} d^{2} x \log \left (x\right ) + 12 \, a c^{3} d e x^{2} - b c^{2} e^{2} x^{3} - 6 \, a c^{3} d^{2} -{\left (3 \, b c^{4} d^{2} + 6 \, b c^{2} d e - b e^{2}\right )} x \log \left (c^{2} x^{2} + 1\right ) + 2 \,{\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2}\right )} \arctan \left (c x\right )}{6 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.5094, size = 165, normalized size = 1.51 \begin{align*} \begin{cases} - \frac{a d^{2}}{x} + 2 a d e x + \frac{a e^{2} x^{3}}{3} + b c d^{2} \log{\left (x \right )} - \frac{b c d^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{2} \operatorname{atan}{\left (c x \right )}}{x} + 2 b d e x \operatorname{atan}{\left (c x \right )} + \frac{b e^{2} x^{3} \operatorname{atan}{\left (c x \right )}}{3} - \frac{b d e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{c} - \frac{b e^{2} x^{2}}{6 c} + \frac{b e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{2}}{x} + 2 d e x + \frac{e^{2} 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)**2*(a+b*atan(c*x))/x**2,x)

[Out]

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

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Giac [A]  time = 1.08155, size = 220, normalized size = 2.02 \begin{align*} \frac{2 \, b c^{3} x^{4} \arctan \left (c x\right ) e^{2} + 2 \, a c^{3} x^{4} e^{2} + 12 \, b c^{3} d x^{2} \arctan \left (c x\right ) e - 3 \, b c^{4} d^{2} x \log \left (c^{2} x^{2} + 1\right ) + 6 \, b c^{4} d^{2} x \log \left (x\right ) + 12 \, a c^{3} d x^{2} e - 6 \, b c^{3} d^{2} \arctan \left (c x\right ) - b c^{2} x^{3} e^{2} - 6 \, b c^{2} d x e \log \left (c^{2} x^{2} + 1\right ) - 6 \, a c^{3} d^{2} + b x e^{2} \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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