3.289 \(\int \frac{x^4 (d+e x)}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{3 \sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 c^{5/2}}-\frac{a e \log \left (a+c x^2\right )}{c^3}-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}+\frac{3 d x}{2 c^2}+\frac{e x^2}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.0701616, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {819, 801, 635, 205, 260} \[ -\frac{3 \sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 c^{5/2}}-\frac{a e \log \left (a+c x^2\right )}{c^3}-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}+\frac{3 d x}{2 c^2}+\frac{e x^2}{c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x))/(a + c*x^2)^2,x]

[Out]

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

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 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]

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^4 (d+e x)}{\left (a+c x^2\right )^2} \, dx &=-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}+\frac{\int \frac{x^2 (3 a d+4 a e x)}{a+c x^2} \, dx}{2 a c}\\ &=-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}+\frac{\int \left (\frac{3 a d}{c}+\frac{4 a e x}{c}-\frac{3 a^2 d+4 a^2 e x}{c \left (a+c x^2\right )}\right ) \, dx}{2 a c}\\ &=\frac{3 d x}{2 c^2}+\frac{e x^2}{c^2}-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}-\frac{\int \frac{3 a^2 d+4 a^2 e x}{a+c x^2} \, dx}{2 a c^2}\\ &=\frac{3 d x}{2 c^2}+\frac{e x^2}{c^2}-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}-\frac{(3 a d) \int \frac{1}{a+c x^2} \, dx}{2 c^2}-\frac{(2 a e) \int \frac{x}{a+c x^2} \, dx}{c^2}\\ &=\frac{3 d x}{2 c^2}+\frac{e x^2}{c^2}-\frac{x^3 (d+e x)}{2 c \left (a+c x^2\right )}-\frac{3 \sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 c^{5/2}}-\frac{a e \log \left (a+c x^2\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.062003, size = 77, normalized size = 0.91 \[ \frac{\frac{a (c d x-a e)}{a+c x^2}-3 \sqrt{a} \sqrt{c} d \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )-2 a e \log \left (a+c x^2\right )+2 c d x+c e x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x))/(a + c*x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.008, size = 88, normalized size = 1. \begin{align*}{\frac{e{x}^{2}}{2\,{c}^{2}}}+{\frac{dx}{{c}^{2}}}+{\frac{adx}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}-{\frac{{a}^{2}e}{2\,{c}^{3} \left ( c{x}^{2}+a \right ) }}-{\frac{ae\ln \left ( c{x}^{2}+a \right ) }{{c}^{3}}}-{\frac{3\,ad}{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^4*(e*x+d)/(c*x^2+a)^2,x)

[Out]

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

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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^4*(e*x+d)/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52858, size = 528, normalized size = 6.21 \begin{align*} \left [\frac{2 \, c^{2} e x^{4} + 4 \, c^{2} d x^{3} + 2 \, a c e x^{2} + 6 \, a c d x - 2 \, a^{2} e + 3 \,{\left (c^{2} d x^{2} + a c d\right )} \sqrt{-\frac{a}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{a}{c}} - a}{c x^{2} + a}\right ) - 4 \,{\left (a c e x^{2} + a^{2} e\right )} \log \left (c x^{2} + a\right )}{4 \,{\left (c^{4} x^{2} + a c^{3}\right )}}, \frac{c^{2} e x^{4} + 2 \, c^{2} d x^{3} + a c e x^{2} + 3 \, a c d x - a^{2} e - 3 \,{\left (c^{2} d x^{2} + a c d\right )} \sqrt{\frac{a}{c}} \arctan \left (\frac{c x \sqrt{\frac{a}{c}}}{a}\right ) - 2 \,{\left (a c e x^{2} + a^{2} e\right )} \log \left (c x^{2} + a\right )}{2 \,{\left (c^{4} x^{2} + a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.906701, size = 189, normalized size = 2.22 \begin{align*} \left (- \frac{a e}{c^{3}} - \frac{3 d \sqrt{- a c^{7}}}{4 c^{6}}\right ) \log{\left (x + \frac{- 4 a e - 4 c^{3} \left (- \frac{a e}{c^{3}} - \frac{3 d \sqrt{- a c^{7}}}{4 c^{6}}\right )}{3 c d} \right )} + \left (- \frac{a e}{c^{3}} + \frac{3 d \sqrt{- a c^{7}}}{4 c^{6}}\right ) \log{\left (x + \frac{- 4 a e - 4 c^{3} \left (- \frac{a e}{c^{3}} + \frac{3 d \sqrt{- a c^{7}}}{4 c^{6}}\right )}{3 c d} \right )} + \frac{- a^{2} e + a c d x}{2 a c^{3} + 2 c^{4} x^{2}} + \frac{d x}{c^{2}} + \frac{e x^{2}}{2 c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)/(c*x**2+a)**2,x)

[Out]

(-a*e/c**3 - 3*d*sqrt(-a*c**7)/(4*c**6))*log(x + (-4*a*e - 4*c**3*(-a*e/c**3 - 3*d*sqrt(-a*c**7)/(4*c**6)))/(3
*c*d)) + (-a*e/c**3 + 3*d*sqrt(-a*c**7)/(4*c**6))*log(x + (-4*a*e - 4*c**3*(-a*e/c**3 + 3*d*sqrt(-a*c**7)/(4*c
**6)))/(3*c*d)) + (-a**2*e + a*c*d*x)/(2*a*c**3 + 2*c**4*x**2) + d*x/c**2 + e*x**2/(2*c**2)

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Giac [A]  time = 1.15664, size = 117, normalized size = 1.38 \begin{align*} -\frac{3 \, a d \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{2 \, \sqrt{a c} c^{2}} - \frac{a e \log \left (c x^{2} + a\right )}{c^{3}} + \frac{c^{2} x^{2} e + 2 \, c^{2} d x}{2 \, c^{4}} + \frac{a c d x - a^{2} e}{2 \,{\left (c x^{2} + a\right )} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(c*x^2+a)^2,x, algorithm="giac")

[Out]

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