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

Optimal. Leaf size=49 \[ -\frac {\sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{3/2}}+\frac {d \log \left (a+c x^2\right )}{2 c}+\frac {e x}{c} \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {774, 635, 205, 260} \begin {gather*} -\frac {\sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{3/2}}+\frac {d \log \left (a+c x^2\right )}{2 c}+\frac {e x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x))/(a + c*x^2),x]

[Out]

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

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 774

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{3/2}}+\frac {d \log \left (a+c x^2\right )}{2 c}+\frac {e x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 108, normalized size = 2.20 \begin {gather*} \left [\frac {e \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {a}{c}} - a}{c x^{2} + a}\right ) + 2 \, e x + d \log \left (c x^{2} + a\right )}{2 \, c}, -\frac {2 \, e \sqrt {\frac {a}{c}} \arctan \left (\frac {c x \sqrt {\frac {a}{c}}}{a}\right ) - 2 \, e x - d \log \left (c x^{2} + a\right )}{2 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 44, normalized size = 0.90 \begin {gather*} -\frac {a \arctan \left (\frac {c x}{\sqrt {a c}}\right ) e}{\sqrt {a c} c} + \frac {x e}{c} + \frac {d \log \left (c x^{2} + a\right )}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 43, normalized size = 0.88 \begin {gather*} -\frac {a e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {d \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.15, size = 42, normalized size = 0.86 \begin {gather*} -\frac {a e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c} + \frac {e x}{c} + \frac {d \log \left (c x^{2} + a\right )}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.07, size = 39, normalized size = 0.80 \begin {gather*} \frac {d\,\ln \left (c\,x^2+a\right )}{2\,c}+\frac {e\,x}{c}-\frac {\sqrt {a}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(d + e*x))/(a + c*x^2),x)

[Out]

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

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sympy [B]  time = 0.30, size = 112, normalized size = 2.29 \begin {gather*} \left (\frac {d}{2 c} - \frac {e \sqrt {- a c^{3}}}{2 c^{3}}\right ) \log {\left (x + \frac {- 2 c \left (\frac {d}{2 c} - \frac {e \sqrt {- a c^{3}}}{2 c^{3}}\right ) + d}{e} \right )} + \left (\frac {d}{2 c} + \frac {e \sqrt {- a c^{3}}}{2 c^{3}}\right ) \log {\left (x + \frac {- 2 c \left (\frac {d}{2 c} + \frac {e \sqrt {- a c^{3}}}{2 c^{3}}\right ) + d}{e} \right )} + \frac {e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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