### 3.10 $$\int \frac{x^5 (d+e x^3)}{a+b x^3+c x^6} \, dx$$

Optimal. Leaf size=97 $\frac{\left (2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 c^2 \sqrt{b^2-4 a c}}+\frac{(c d-b e) \log \left (a+b x^3+c x^6\right )}{6 c^2}+\frac{e x^3}{3 c}$

[Out]

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

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Rubi [A]  time = 0.119684, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.24, Rules used = {1474, 773, 634, 618, 206, 628} $\frac{\left (2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 c^2 \sqrt{b^2-4 a c}}+\frac{(c d-b e) \log \left (a+b x^3+c x^6\right )}{6 c^2}+\frac{e x^3}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x^5*(d + e*x^3))/(a + b*x^3 + c*x^6),x]

[Out]

(e*x^3)/(3*c) + ((b*c*d - b^2*e + 2*a*c*e)*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*c^2*Sqrt[b^2 - 4*a*c])
+ ((c*d - b*e)*Log[a + b*x^3 + c*x^6])/(6*c^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 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

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{x^5 \left (d+e x^3\right )}{a+b x^3+c x^6} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x (d+e x)}{a+b x+c x^2} \, dx,x,x^3\right )\\ &=\frac{e x^3}{3 c}+\frac{\operatorname{Subst}\left (\int \frac{-a e+(c d-b e) x}{a+b x+c x^2} \, dx,x,x^3\right )}{3 c}\\ &=\frac{e x^3}{3 c}+\frac{(c d-b e) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^3\right )}{6 c^2}-\frac{\left (b c d-b^2 e+2 a c e\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^3\right )}{6 c^2}\\ &=\frac{e x^3}{3 c}+\frac{(c d-b e) \log \left (a+b x^3+c x^6\right )}{6 c^2}+\frac{\left (b c d-b^2 e+2 a c e\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^3\right )}{3 c^2}\\ &=\frac{e x^3}{3 c}+\frac{\left (b c d-b^2 e+2 a c e\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{\sqrt{b^2-4 a c}}\right )}{3 c^2 \sqrt{b^2-4 a c}}+\frac{(c d-b e) \log \left (a+b x^3+c x^6\right )}{6 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0702844, size = 93, normalized size = 0.96 $\frac{\frac{2 \left (-2 a c e+b^2 e-b c d\right ) \tan ^{-1}\left (\frac{b+2 c x^3}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+(c d-b e) \log \left (a+b x^3+c x^6\right )+2 c e x^3}{6 c^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x^5*(d + e*x^3))/(a + b*x^3 + c*x^6),x]

[Out]

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

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Maple [A]  time = 0.003, size = 175, normalized size = 1.8 \begin{align*}{\frac{e{x}^{3}}{3\,c}}-{\frac{\ln \left ( c{x}^{6}+b{x}^{3}+a \right ) be}{6\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{6}+b{x}^{3}+a \right ) d}{6\,c}}-{\frac{2\,ae}{3\,c}\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}e}{3\,{c}^{2}}\arctan \left ({(2\,c{x}^{3}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{bd}{3\,c}\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(x^5*(e*x^3+d)/(c*x^6+b*x^3+a),x)

[Out]

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

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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^5*(e*x^3+d)/(c*x^6+b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92803, size = 664, normalized size = 6.85 \begin{align*} \left [\frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e x^{3} +{\left (b c d -{\left (b^{2} - 2 \, a c\right )} e\right )} \sqrt{b^{2} - 4 \, a c} \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^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\right )} e\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e x^{3} + 2 \,{\left (b c d -{\left (b^{2} - 2 \, a c\right )} e\right )} \sqrt{-b^{2} + 4 \, a c} \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^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\right )} e\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^3+d)/(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 9.09558, size = 434, normalized size = 4.47 \begin{align*} \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right ) \log{\left (x^{3} + \frac{- a b e - 12 a c^{2} \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right ) + 2 a c d + 3 b^{2} c \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right )}{2 a c e - b^{2} e + b c d} \right )} + \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right ) \log{\left (x^{3} + \frac{- a b e - 12 a c^{2} \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right ) + 2 a c d + 3 b^{2} c \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{6 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{6 c^{2}}\right )}{2 a c e - b^{2} e + b c d} \right )} + \frac{e x^{3}}{3 c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*x**3+d)/(c*x**6+b*x**3+a),x)

[Out]

(-sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b*c*d)/(6*c**2*(4*a*c - b**2)) - (b*e - c*d)/(6*c**2))*log(x**3 + (-
a*b*e - 12*a*c**2*(-sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b*c*d)/(6*c**2*(4*a*c - b**2)) - (b*e - c*d)/(6*c*
*2)) + 2*a*c*d + 3*b**2*c*(-sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b*c*d)/(6*c**2*(4*a*c - b**2)) - (b*e - c*
d)/(6*c**2)))/(2*a*c*e - b**2*e + b*c*d)) + (sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b*c*d)/(6*c**2*(4*a*c - b
**2)) - (b*e - c*d)/(6*c**2))*log(x**3 + (-a*b*e - 12*a*c**2*(sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b*c*d)/(
6*c**2*(4*a*c - b**2)) - (b*e - c*d)/(6*c**2)) + 2*a*c*d + 3*b**2*c*(sqrt(-4*a*c + b**2)*(2*a*c*e - b**2*e + b
*c*d)/(6*c**2*(4*a*c - b**2)) - (b*e - c*d)/(6*c**2)))/(2*a*c*e - b**2*e + b*c*d)) + e*x**3/(3*c)

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Giac [A]  time = 1.35018, size = 128, normalized size = 1.32 \begin{align*} \frac{x^{3} e}{3 \, c} + \frac{{\left (c d - b e\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, c^{2}} - \frac{{\left (b c d - b^{2} e + 2 \, a c e\right )} \arctan \left (\frac{2 \, c x^{3} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt{-b^{2} + 4 \, a c} c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^3+d)/(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

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