3.13 \(\int \frac{d+e x^3}{x^4 (a+b x^3+c x^6)} \, dx\)

Optimal. Leaf size=112 \[ -\frac{\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 a^2 \sqrt{b^2-4 a c}}+\frac{(b d-a e) \log \left (a+b x^3+c x^6\right )}{6 a^2}-\frac{\log (x) (b d-a e)}{a^2}-\frac{d}{3 a x^3} \]

[Out]

-d/(3*a*x^3) - ((b^2*d - 2*a*c*d - a*b*e)*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*a^2*Sqrt[b^2 - 4*a*c])
- ((b*d - a*e)*Log[x])/a^2 + ((b*d - a*e)*Log[a + b*x^3 + c*x^6])/(6*a^2)

________________________________________________________________________________________

Rubi [A]  time = 0.197486, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1474, 800, 634, 618, 206, 628} \[ -\frac{\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 a^2 \sqrt{b^2-4 a c}}+\frac{(b d-a e) \log \left (a+b x^3+c x^6\right )}{6 a^2}-\frac{\log (x) (b d-a e)}{a^2}-\frac{d}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^3)/(x^4*(a + b*x^3 + c*x^6)),x]

[Out]

-d/(3*a*x^3) - ((b^2*d - 2*a*c*d - a*b*e)*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*a^2*Sqrt[b^2 - 4*a*c])
- ((b*d - a*e)*Log[x])/a^2 + ((b*d - a*e)*Log[a + b*x^3 + c*x^6])/(6*a^2)

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{d+e x^3}{x^4 \left (a+b x^3+c x^6\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{d+e x}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{d}{a x^2}+\frac{-b d+a e}{a^2 x}+\frac{b^2 d-a c d-a b e+c (b d-a e) x}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^3\right )\\ &=-\frac{d}{3 a x^3}-\frac{(b d-a e) \log (x)}{a^2}+\frac{\operatorname{Subst}\left (\int \frac{b^2 d-a c d-a b e+c (b d-a e) x}{a+b x+c x^2} \, dx,x,x^3\right )}{3 a^2}\\ &=-\frac{d}{3 a x^3}-\frac{(b d-a e) \log (x)}{a^2}+\frac{(b d-a e) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^3\right )}{6 a^2}+\frac{\left (b^2 d-2 a c d-a b e\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^3\right )}{6 a^2}\\ &=-\frac{d}{3 a x^3}-\frac{(b d-a e) \log (x)}{a^2}+\frac{(b d-a e) \log \left (a+b x^3+c x^6\right )}{6 a^2}-\frac{\left (b^2 d-2 a c d-a b e\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^3\right )}{3 a^2}\\ &=-\frac{d}{3 a x^3}-\frac{\left (b^2 d-2 a c d-a b e\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 a^2 \sqrt{b^2-4 a c}}-\frac{(b d-a e) \log (x)}{a^2}+\frac{(b d-a e) \log \left (a+b x^3+c x^6\right )}{6 a^2}\\ \end{align*}

Mathematica [C]  time = 0.0521102, size = 130, normalized size = 1.16 \[ \frac{\text{RootSum}\left [\text{$\#$1}^3 b+\text{$\#$1}^6 c+a\& ,\frac{-\text{$\#$1}^3 a c e \log (x-\text{$\#$1})+\text{$\#$1}^3 b c d \log (x-\text{$\#$1})-a b e \log (x-\text{$\#$1})-a c d \log (x-\text{$\#$1})+b^2 d \log (x-\text{$\#$1})}{2 \text{$\#$1}^3 c+b}\& \right ]}{3 a^2}+\frac{\log (x) (a e-b d)}{a^2}-\frac{d}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^3)/(x^4*(a + b*x^3 + c*x^6)),x]

[Out]

-d/(3*a*x^3) + ((-(b*d) + a*e)*Log[x])/a^2 + RootSum[a + b*#1^3 + c*#1^6 & , (b^2*d*Log[x - #1] - a*c*d*Log[x
- #1] - a*b*e*Log[x - #1] + b*c*d*Log[x - #1]*#1^3 - a*c*e*Log[x - #1]*#1^3)/(b + 2*c*#1^3) & ]/(3*a^2)

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 191, normalized size = 1.7 \begin{align*} -{\frac{d}{3\,a{x}^{3}}}+{\frac{\ln \left ( x \right ) e}{a}}-{\frac{b\ln \left ( x \right ) d}{{a}^{2}}}-{\frac{\ln \left ( c{x}^{6}+b{x}^{3}+a \right ) e}{6\,a}}+{\frac{\ln \left ( c{x}^{6}+b{x}^{3}+a \right ) bd}{6\,{a}^{2}}}-{\frac{be}{3\,a}\arctan \left ({(2\,c{x}^{3}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{2\,cd}{3\,a}\arctan \left ({(2\,c{x}^{3}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}d}{3\,{a}^{2}}\arctan \left ({(2\,c{x}^{3}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^3+d)/x^4/(c*x^6+b*x^3+a),x)

[Out]

-1/3*d/a/x^3+1/a*ln(x)*e-1/a^2*ln(x)*b*d-1/6/a*ln(c*x^6+b*x^3+a)*e+1/6/a^2*ln(c*x^6+b*x^3+a)*b*d-1/3/a/(4*a*c-
b^2)^(1/2)*arctan((2*c*x^3+b)/(4*a*c-b^2)^(1/2))*b*e-2/3/a/(4*a*c-b^2)^(1/2)*arctan((2*c*x^3+b)/(4*a*c-b^2)^(1
/2))*c*d+1/3/a^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x^3+b)/(4*a*c-b^2)^(1/2))*b^2*d

________________________________________________________________________________________

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((e*x^3+d)/x^4/(c*x^6+b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 4.11086, size = 845, normalized size = 7.54 \begin{align*} \left [\frac{{\left (a b e -{\left (b^{2} - 2 \, a c\right )} d\right )} \sqrt{b^{2} - 4 \, a c} x^{3} \log \left (\frac{2 \, c^{2} x^{6} + 2 \, b c x^{3} + b^{2} - 2 \, a c +{\left (2 \, c x^{3} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{6} + b x^{3} + a}\right ) +{\left ({\left (b^{3} - 4 \, a b c\right )} d -{\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{3} \log \left (c x^{6} + b x^{3} + a\right ) - 6 \,{\left ({\left (b^{3} - 4 \, a b c\right )} d -{\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{3} \log \left (x\right ) - 2 \,{\left (a b^{2} - 4 \, a^{2} c\right )} d}{6 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{3}}, \frac{2 \,{\left (a b e -{\left (b^{2} - 2 \, a c\right )} d\right )} \sqrt{-b^{2} + 4 \, a c} x^{3} \arctan \left (-\frac{{\left (2 \, c x^{3} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) +{\left ({\left (b^{3} - 4 \, a b c\right )} d -{\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{3} \log \left (c x^{6} + b x^{3} + a\right ) - 6 \,{\left ({\left (b^{3} - 4 \, a b c\right )} d -{\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{3} \log \left (x\right ) - 2 \,{\left (a b^{2} - 4 \, a^{2} c\right )} d}{6 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^3+d)/x^4/(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**3+d)/x**4/(c*x**6+b*x**3+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.36632, size = 173, normalized size = 1.54 \begin{align*} \frac{{\left (b d - a e\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, a^{2}} - \frac{{\left (b d - a e\right )} \log \left ({\left | x \right |}\right )}{a^{2}} + \frac{{\left (b^{2} d - 2 \, a c d - a b e\right )} \arctan \left (\frac{2 \, c x^{3} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt{-b^{2} + 4 \, a c} a^{2}} + \frac{b d x^{3} - a x^{3} e - a d}{3 \, a^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^3+d)/x^4/(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

1/6*(b*d - a*e)*log(c*x^6 + b*x^3 + a)/a^2 - (b*d - a*e)*log(abs(x))/a^2 + 1/3*(b^2*d - 2*a*c*d - a*b*e)*arcta
n((2*c*x^3 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^2) + 1/3*(b*d*x^3 - a*x^3*e - a*d)/(a^2*x^3)