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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0448865, antiderivative size = 78, normalized size of antiderivative = 1., 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 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])

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 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^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*}

Mathematica [A]  time = 0.0509549, size = 75, normalized size = 0.96 \[ \frac{a d+a e x}{2 c^2 \left (a+c x^2\right )}+\frac{d \log \left (a+c x^2\right )}{2 c^2}-\frac{3 \sqrt{a} e \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 c^{5/2}}+\frac{e x}{c^2} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 76, normalized size = 1. \begin{align*}{\frac{ex}{{c}^{2}}}+{\frac{aex}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}+{\frac{ad}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}+{\frac{d\ln \left ( c{x}^{2}+a \right ) }{2\,{c}^{2}}}-{\frac{3\,ae}{2\,{c}^{2}}\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^3*(e*x+d)/(c*x^2+a)^2,x)

[Out]

e*x/c^2+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(x*c
/(a*c)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.55213, size = 417, normalized size = 5.35 \begin{align*} \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 ] \end{align*}

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)]

________________________________________________________________________________________

Sympy [B]  time = 0.89868, size = 162, normalized size = 2.08 \begin{align*} \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}} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.1099, size = 90, normalized size = 1.15 \begin{align*} -\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}} \end{align*}

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)