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

Optimal. Leaf size=78 \[ -\frac {3 \sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 c^{5/2}}+\frac {d \log \left (a+c x^2\right )}{2 c^2}-\frac {x^2 (d+e x)}{2 c \left (a+c x^2\right )}+\frac {3 e x}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {819, 774, 635, 205, 260} \[ \frac {d \log \left (a+c x^2\right )}{2 c^2}-\frac {3 \sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 c^{5/2}}-\frac {x^2 (d+e x)}{2 c \left (a+c x^2\right )}+\frac {3 e x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x))/(a + c*x^2)^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.96 \[ -\frac {3 \sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 c^{5/2}}+\frac {a d+a e x}{2 c^2 \left (a+c x^2\right )}+\frac {d \log \left (a+c x^2\right )}{2 c^2}+\frac {e x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 192, normalized size = 2.46 \[ \left [\frac {4 \, c e x^{3} + 6 \, a e x + 3 \, {\left (c e x^{2} + a e\right )} \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {a}{c}} - a}{c x^{2} + a}\right ) + 2 \, a d + 2 \, {\left (c d x^{2} + a d\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (c^{3} x^{2} + a c^{2}\right )}}, \frac {2 \, c e x^{3} + 3 \, a e x - 3 \, {\left (c e x^{2} + a e\right )} \sqrt {\frac {a}{c}} \arctan \left (\frac {c x \sqrt {\frac {a}{c}}}{a}\right ) + a d + {\left (c d x^{2} + a d\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*c*e*x^3 + 6*a*e*x + 3*(c*e*x^2 + a*e)*sqrt(-a/c)*log((c*x^2 - 2*c*x*sqrt(-a/c) - a)/(c*x^2 + a)) + 2*a
*d + 2*(c*d*x^2 + a*d)*log(c*x^2 + a))/(c^3*x^2 + a*c^2), 1/2*(2*c*e*x^3 + 3*a*e*x - 3*(c*e*x^2 + a*e)*sqrt(a/
c)*arctan(c*x*sqrt(a/c)/a) + a*d + (c*d*x^2 + a*d)*log(c*x^2 + a))/(c^3*x^2 + a*c^2)]

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giac [A]  time = 0.15, size = 67, normalized size = 0.86 \[ -\frac {3 \, a \arctan \left (\frac {c x}{\sqrt {a c}}\right ) e}{2 \, \sqrt {a c} c^{2}} + \frac {x e}{c^{2}} + \frac {d \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {a x e + a d}{2 \, {\left (c x^{2} + a\right )} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 76, normalized size = 0.97 \[ \frac {a e x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 a e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}+\frac {a d}{2 \left (c \,x^{2}+a \right ) c^{2}}+\frac {d \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {e x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.16, size = 67, normalized size = 0.86 \[ -\frac {3 \, a e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} c^{2}} + \frac {a e x + a d}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}} + \frac {e x}{c^{2}} + \frac {d \log \left (c x^{2} + a\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.08, size = 65, normalized size = 0.83 \[ \frac {\frac {a\,d}{2}+\frac {a\,e\,x}{2}}{c^3\,x^2+a\,c^2}+\frac {d\,\ln \left (c\,x^2+a\right )}{2\,c^2}+\frac {e\,x}{c^2}-\frac {3\,\sqrt {a}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x))/(a + c*x^2)^2,x)

[Out]

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

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sympy [B]  time = 0.67, size = 162, normalized size = 2.08 \[ \left (\frac {d}{2 c^{2}} - \frac {3 e \sqrt {- a c^{5}}}{4 c^{5}}\right ) \log {\left (x + \frac {- 4 c^{2} \left (\frac {d}{2 c^{2}} - \frac {3 e \sqrt {- a c^{5}}}{4 c^{5}}\right ) + 2 d}{3 e} \right )} + \left (\frac {d}{2 c^{2}} + \frac {3 e \sqrt {- a c^{5}}}{4 c^{5}}\right ) \log {\left (x + \frac {- 4 c^{2} \left (\frac {d}{2 c^{2}} + \frac {3 e \sqrt {- a c^{5}}}{4 c^{5}}\right ) + 2 d}{3 e} \right )} + \frac {a d + a e x}{2 a c^{2} + 2 c^{3} x^{2}} + \frac {e x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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