3.281 \(\int \frac{x^2 (d+e x)}{a+c x^2} \, dx\)

Optimal. Leaf size=61 \[ -\frac{\sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{c^{3/2}}-\frac{a e \log \left (a+c x^2\right )}{2 c^2}+\frac{d x}{c}+\frac{e x^2}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0441019, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {801, 635, 205, 260} \[ -\frac{\sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{c^{3/2}}-\frac{a e \log \left (a+c x^2\right )}{2 c^2}+\frac{d x}{c}+\frac{e x^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x))/(a + c*x^2),x]

[Out]

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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]

Rubi steps

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

Mathematica [A]  time = 0.0209669, size = 56, normalized size = 0.92 \[ \frac{-2 \sqrt{a} \sqrt{c} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )-a e \log \left (a+c x^2\right )+c x (2 d+e x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*(2*d + e*x) - 2*Sqrt[a]*Sqrt[c]*d*ArcTan[(Sqrt[c]*x)/Sqrt[a]] - a*e*Log[a + c*x^2])/(2*c^2)

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Maple [A]  time = 0.004, size = 53, normalized size = 0.9 \begin{align*}{\frac{e{x}^{2}}{2\,c}}+{\frac{dx}{c}}-{\frac{ae\ln \left ( c{x}^{2}+a \right ) }{2\,{c}^{2}}}-{\frac{ad}{c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54537, size = 285, normalized size = 4.67 \begin{align*} \left [\frac{c e x^{2} + c d \sqrt{-\frac{a}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{a}{c}} - a}{c x^{2} + a}\right ) + 2 \, c d x - a e \log \left (c x^{2} + a\right )}{2 \, c^{2}}, \frac{c e x^{2} - 2 \, c d \sqrt{\frac{a}{c}} \arctan \left (\frac{c x \sqrt{\frac{a}{c}}}{a}\right ) + 2 \, c d x - a e \log \left (c x^{2} + a\right )}{2 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(c*e*x^2 + c*d*sqrt(-a/c)*log((c*x^2 - 2*c*x*sqrt(-a/c) - a)/(c*x^2 + a)) + 2*c*d*x - a*e*log(c*x^2 + a))
/c^2, 1/2*(c*e*x^2 - 2*c*d*sqrt(a/c)*arctan(c*x*sqrt(a/c)/a) + 2*c*d*x - a*e*log(c*x^2 + a))/c^2]

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Sympy [B]  time = 0.591244, size = 151, normalized size = 2.48 \begin{align*} \left (- \frac{a e}{2 c^{2}} - \frac{d \sqrt{- a c^{5}}}{2 c^{4}}\right ) \log{\left (x + \frac{- a e - 2 c^{2} \left (- \frac{a e}{2 c^{2}} - \frac{d \sqrt{- a c^{5}}}{2 c^{4}}\right )}{c d} \right )} + \left (- \frac{a e}{2 c^{2}} + \frac{d \sqrt{- a c^{5}}}{2 c^{4}}\right ) \log{\left (x + \frac{- a e - 2 c^{2} \left (- \frac{a e}{2 c^{2}} + \frac{d \sqrt{- a c^{5}}}{2 c^{4}}\right )}{c d} \right )} + \frac{d x}{c} + \frac{e x^{2}}{2 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a*e/(2*c**2) - d*sqrt(-a*c**5)/(2*c**4))*log(x + (-a*e - 2*c**2*(-a*e/(2*c**2) - d*sqrt(-a*c**5)/(2*c**4)))/
(c*d)) + (-a*e/(2*c**2) + d*sqrt(-a*c**5)/(2*c**4))*log(x + (-a*e - 2*c**2*(-a*e/(2*c**2) + d*sqrt(-a*c**5)/(2
*c**4)))/(c*d)) + d*x/c + e*x**2/(2*c)

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Giac [A]  time = 1.13193, size = 76, normalized size = 1.25 \begin{align*} -\frac{a d \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{\sqrt{a c} c} - \frac{a e \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{c x^{2} e + 2 \, c d x}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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