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

Optimal. Leaf size=67 \[ \frac {d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 \sqrt {a} c^{3/2}}+\frac {e \log \left (a+c x^2\right )}{2 c^2}-\frac {x (d+e x)}{2 c \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {819, 635, 205, 260} \begin {gather*} \frac {d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 \sqrt {a} c^{3/2}}+\frac {e \log \left (a+c x^2\right )}{2 c^2}-\frac {x (d+e x)}{2 c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x))/(a + c*x^2)^2,x]

[Out]

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

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]

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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 62, normalized size = 0.93 \begin {gather*} \frac {\frac {a e-c d x}{a+c x^2}+\frac {\sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}+e \log \left (a+c x^2\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 (d+e x)}{\left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 186, normalized size = 2.78 \begin {gather*} \left [-\frac {2 \, a c d x - 2 \, a^{2} e + {\left (c d x^{2} + a d\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (a c e x^{2} + a^{2} e\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}, -\frac {a c d x - a^{2} e - {\left (c d x^{2} + a d\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (a c e x^{2} + a^{2} e\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*c*d*x - 2*a^2*e + (c*d*x^2 + a*d)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(a*c
*e*x^2 + a^2*e)*log(c*x^2 + a))/(a*c^3*x^2 + a^2*c^2), -1/2*(a*c*d*x - a^2*e - (c*d*x^2 + a*d)*sqrt(a*c)*arcta
n(sqrt(a*c)*x/a) - (a*c*e*x^2 + a^2*e)*log(c*x^2 + a))/(a*c^3*x^2 + a^2*c^2)]

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giac [A]  time = 0.19, size = 62, normalized size = 0.93 \begin {gather*} \frac {d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} c} + \frac {e \log \left (c x^{2} + a\right )}{2 \, c^{2}} - \frac {d x - \frac {a e}{c}}{2 \, {\left (c x^{2} + a\right )} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 61, normalized size = 0.91 \begin {gather*} \frac {d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}+\frac {e \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {-\frac {d x}{2 c}+\frac {a e}{2 c^{2}}}{c \,x^{2}+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.16, size = 61, normalized size = 0.91 \begin {gather*} \frac {d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} c} - \frac {c d x - a e}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}} + \frac {e \log \left (c x^{2} + a\right )}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.04, size = 72, normalized size = 1.07 \begin {gather*} \frac {e\,\ln \left (c\,x^2+a\right )}{2\,c^2}-\frac {d\,x}{2\,\left (c^2\,x^2+a\,c\right )}+\frac {a\,e}{2\,\left (c^3\,x^2+a\,c^2\right )}+\frac {d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d + e*x))/(a + c*x^2)^2,x)

[Out]

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

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sympy [B]  time = 0.58, size = 162, normalized size = 2.42 \begin {gather*} \left (\frac {e}{2 c^{2}} - \frac {d \sqrt {- a c^{5}}}{4 a c^{4}}\right ) \log {\left (x + \frac {4 a c^{2} \left (\frac {e}{2 c^{2}} - \frac {d \sqrt {- a c^{5}}}{4 a c^{4}}\right ) - 2 a e}{c d} \right )} + \left (\frac {e}{2 c^{2}} + \frac {d \sqrt {- a c^{5}}}{4 a c^{4}}\right ) \log {\left (x + \frac {4 a c^{2} \left (\frac {e}{2 c^{2}} + \frac {d \sqrt {- a c^{5}}}{4 a c^{4}}\right ) - 2 a e}{c d} \right )} + \frac {a e - c d x}{2 a c^{2} + 2 c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e/(2*c**2) - d*sqrt(-a*c**5)/(4*a*c**4))*log(x + (4*a*c**2*(e/(2*c**2) - d*sqrt(-a*c**5)/(4*a*c**4)) - 2*a*e)
/(c*d)) + (e/(2*c**2) + d*sqrt(-a*c**5)/(4*a*c**4))*log(x + (4*a*c**2*(e/(2*c**2) + d*sqrt(-a*c**5)/(4*a*c**4)
) - 2*a*e)/(c*d)) + (a*e - c*d*x)/(2*a*c**2 + 2*c**3*x**2)

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