3.3.80 \(\int \frac {x^3 (d+e x)}{a+c x^2} \, dx\)

Optimal. Leaf size=73 \[ \frac {a^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}-\frac {a d \log \left (a+c x^2\right )}{2 c^2}-\frac {a e x}{c^2}+\frac {d x^2}{2 c}+\frac {e x^3}{3 c} \]

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {a^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}-\frac {a d \log \left (a+c x^2\right )}{2 c^2}-\frac {a e x}{c^2}+\frac {d x^2}{2 c}+\frac {e x^3}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x))/(a + c*x^2),x]

[Out]

-((a*e*x)/c^2) + (d*x^2)/(2*c) + (e*x^3)/(3*c) + (a^(3/2)*e*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/c^(5/2) - (a*d*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 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]

Rubi steps

\begin {align*} \int \frac {x^3 (d+e x)}{a+c x^2} \, dx &=\int \left (-\frac {a e}{c^2}+\frac {d x}{c}+\frac {e x^2}{c}+\frac {a^2 e-a c d x}{c^2 \left (a+c x^2\right )}\right ) \, dx\\ &=-\frac {a e x}{c^2}+\frac {d x^2}{2 c}+\frac {e x^3}{3 c}+\frac {\int \frac {a^2 e-a c d x}{a+c x^2} \, dx}{c^2}\\ &=-\frac {a e x}{c^2}+\frac {d x^2}{2 c}+\frac {e x^3}{3 c}-\frac {(a d) \int \frac {x}{a+c x^2} \, dx}{c}+\frac {\left (a^2 e\right ) \int \frac {1}{a+c x^2} \, dx}{c^2}\\ &=-\frac {a e x}{c^2}+\frac {d x^2}{2 c}+\frac {e x^3}{3 c}+\frac {a^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}-\frac {a d \log \left (a+c x^2\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.88 \begin {gather*} \frac {a^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}+\frac {x (c x (3 d+2 e x)-6 a e)-3 a d \log \left (a+c x^2\right )}{6 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x))/(a + c*x^2),x]

[Out]

(a^(3/2)*e*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/c^(5/2) + (x*(-6*a*e + c*x*(3*d + 2*e*x)) - 3*a*d*Log[a + c*x^2])/(6*c
^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 (d+e x)}{a+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^3*(d + e*x))/(a + c*x^2),x]

[Out]

IntegrateAlgebraic[(x^3*(d + e*x))/(a + c*x^2), x]

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fricas [A]  time = 0.39, size = 144, normalized size = 1.97 \begin {gather*} \left [\frac {2 \, c e x^{3} + 3 \, c d x^{2} + 3 \, a e \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {a}{c}} - a}{c x^{2} + a}\right ) - 6 \, a e x - 3 \, a d \log \left (c x^{2} + a\right )}{6 \, c^{2}}, \frac {2 \, c e x^{3} + 3 \, c d x^{2} + 6 \, a e \sqrt {\frac {a}{c}} \arctan \left (\frac {c x \sqrt {\frac {a}{c}}}{a}\right ) - 6 \, a e x - 3 \, a d \log \left (c x^{2} + a\right )}{6 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+a),x, algorithm="fricas")

[Out]

[1/6*(2*c*e*x^3 + 3*c*d*x^2 + 3*a*e*sqrt(-a/c)*log((c*x^2 + 2*c*x*sqrt(-a/c) - a)/(c*x^2 + a)) - 6*a*e*x - 3*a
*d*log(c*x^2 + a))/c^2, 1/6*(2*c*e*x^3 + 3*c*d*x^2 + 6*a*e*sqrt(a/c)*arctan(c*x*sqrt(a/c)/a) - 6*a*e*x - 3*a*d
*log(c*x^2 + a))/c^2]

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giac [A]  time = 0.16, size = 71, normalized size = 0.97 \begin {gather*} \frac {a^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right ) e}{\sqrt {a c} c^{2}} - \frac {a d \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {2 \, c^{2} x^{3} e + 3 \, c^{2} d x^{2} - 6 \, a c x e}{6 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 65, normalized size = 0.89 \begin {gather*} \frac {e \,x^{3}}{3 c}+\frac {a^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}+\frac {d \,x^{2}}{2 c}-\frac {a d \ln \left (c \,x^{2}+a \right )}{2 c^{2}}-\frac {a e x}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)/(c*x^2+a),x)

[Out]

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

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maxima [A]  time = 1.26, size = 63, normalized size = 0.86 \begin {gather*} \frac {a^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} - \frac {a d \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {2 \, c e x^{3} + 3 \, c d x^{2} - 6 \, a e x}{6 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 59, normalized size = 0.81 \begin {gather*} \frac {d\,x^2}{2\,c}+\frac {e\,x^3}{3\,c}+\frac {a^{3/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{c^{5/2}}-\frac {a\,e\,x}{c^2}-\frac {a\,d\,\ln \left (c\,x^2+a\right )}{2\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x))/(a + c*x^2),x)

[Out]

(d*x^2)/(2*c) + (e*x^3)/(3*c) + (a^(3/2)*e*atan((c^(1/2)*x)/a^(1/2)))/c^(5/2) - (a*e*x)/c^2 - (a*d*log(a + c*x
^2))/(2*c^2)

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sympy [B]  time = 0.36, size = 167, normalized size = 2.29 \begin {gather*} - \frac {a e x}{c^{2}} + \left (- \frac {a d}{2 c^{2}} - \frac {e \sqrt {- a^{3} c^{5}}}{2 c^{5}}\right ) \log {\left (x + \frac {a d + 2 c^{2} \left (- \frac {a d}{2 c^{2}} - \frac {e \sqrt {- a^{3} c^{5}}}{2 c^{5}}\right )}{a e} \right )} + \left (- \frac {a d}{2 c^{2}} + \frac {e \sqrt {- a^{3} c^{5}}}{2 c^{5}}\right ) \log {\left (x + \frac {a d + 2 c^{2} \left (- \frac {a d}{2 c^{2}} + \frac {e \sqrt {- a^{3} c^{5}}}{2 c^{5}}\right )}{a e} \right )} + \frac {d x^{2}}{2 c} + \frac {e x^{3}}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)/(c*x**2+a),x)

[Out]

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

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