3.58 \(\int \frac{x^4 (A+B x^3)}{a+b x^3} \, dx\)

Optimal. Leaf size=167 \[ -\frac{a^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}}+\frac{a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}+\frac{a^{2/3} (A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{8/3}}+\frac{x^2 (A b-a B)}{2 b^2}+\frac{B x^5}{5 b} \]

[Out]

((A*b - a*B)*x^2)/(2*b^2) + (B*x^5)/(5*b) + (a^(2/3)*(A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/
3))])/(Sqrt[3]*b^(8/3)) + (a^(2/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(8/3)) - (a^(2/3)*(A*b - a*B)*Lo
g[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(8/3))

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Rubi [A]  time = 0.120236, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {459, 321, 292, 31, 634, 617, 204, 628} \[ -\frac{a^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}}+\frac{a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}+\frac{a^{2/3} (A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{8/3}}+\frac{x^2 (A b-a B)}{2 b^2}+\frac{B x^5}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^3))/(a + b*x^3),x]

[Out]

((A*b - a*B)*x^2)/(2*b^2) + (B*x^5)/(5*b) + (a^(2/3)*(A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/
3))])/(Sqrt[3]*b^(8/3)) + (a^(2/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(8/3)) - (a^(2/3)*(A*b - a*B)*Lo
g[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(8/3))

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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^4 \left (A+B x^3\right )}{a+b x^3} \, dx &=\frac{B x^5}{5 b}-\frac{(-5 A b+5 a B) \int \frac{x^4}{a+b x^3} \, dx}{5 b}\\ &=\frac{(A b-a B) x^2}{2 b^2}+\frac{B x^5}{5 b}-\frac{(a (A b-a B)) \int \frac{x}{a+b x^3} \, dx}{b^2}\\ &=\frac{(A b-a B) x^2}{2 b^2}+\frac{B x^5}{5 b}+\frac{\left (a^{2/3} (A b-a B)\right ) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{7/3}}-\frac{\left (a^{2/3} (A b-a B)\right ) \int \frac{\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 b^{7/3}}\\ &=\frac{(A b-a B) x^2}{2 b^2}+\frac{B x^5}{5 b}+\frac{a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}-\frac{\left (a^{2/3} (A b-a B)\right ) \int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 b^{8/3}}-\frac{(a (A b-a B)) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b^{7/3}}\\ &=\frac{(A b-a B) x^2}{2 b^2}+\frac{B x^5}{5 b}+\frac{a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}-\frac{a^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}}-\frac{\left (a^{2/3} (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b^{8/3}}\\ &=\frac{(A b-a B) x^2}{2 b^2}+\frac{B x^5}{5 b}+\frac{a^{2/3} (A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{8/3}}+\frac{a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}-\frac{a^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}}\\ \end{align*}

Mathematica [A]  time = 0.0816788, size = 154, normalized size = 0.92 \[ \frac{5 a^{2/3} (a B-A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-10 a^{2/3} (a B-A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-10 \sqrt{3} a^{2/3} (a B-A b) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )+15 b^{2/3} x^2 (A b-a B)+6 b^{5/3} B x^5}{30 b^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^3))/(a + b*x^3),x]

[Out]

(15*b^(2/3)*(A*b - a*B)*x^2 + 6*b^(5/3)*B*x^5 - 10*Sqrt[3]*a^(2/3)*(-(A*b) + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^
(1/3))/Sqrt[3]] - 10*a^(2/3)*(-(A*b) + a*B)*Log[a^(1/3) + b^(1/3)*x] + 5*a^(2/3)*(-(A*b) + a*B)*Log[a^(2/3) -
a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(30*b^(8/3))

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Maple [A]  time = 0.004, size = 226, normalized size = 1.4 \begin{align*}{\frac{B{x}^{5}}{5\,b}}+{\frac{A{x}^{2}}{2\,b}}-{\frac{B{x}^{2}a}{2\,{b}^{2}}}+{\frac{aA}{3\,{b}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{{a}^{2}B}{3\,{b}^{3}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{aA}{6\,{b}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}B}{6\,{b}^{3}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{a\sqrt{3}A}{3\,{b}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}\sqrt{3}B}{3\,{b}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^3+A)/(b*x^3+a),x)

[Out]

1/5*B*x^5/b+1/2/b*A*x^2-1/2/b^2*B*x^2*a+1/3*a/b^2/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*A-1/3*a^2/b^3/(a/b)^(1/3)*ln(x
+(a/b)^(1/3))*B-1/6*a/b^2/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A+1/6*a^2/b^3/(a/b)^(1/3)*ln(x^2-(a/b)
^(1/3)*x+(a/b)^(2/3))*B-1/3*a/b^2*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/3*a^2/b^3*3^
(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*B

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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*(B*x^3+A)/(b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47417, size = 390, normalized size = 2.34 \begin{align*} \frac{6 \, B b x^{5} - 15 \,{\left (B a - A b\right )} x^{2} + 10 \, \sqrt{3}{\left (B a - A b\right )} \left (\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} b x \left (\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} - \sqrt{3} a}{3 \, a}\right ) + 5 \,{\left (B a - A b\right )} \left (\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (a x^{2} - b x \left (\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} + a \left (\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}}\right ) - 10 \,{\left (B a - A b\right )} \left (\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (a x + b \left (\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}}\right )}{30 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^3+A)/(b*x^3+a),x, algorithm="fricas")

[Out]

1/30*(6*B*b*x^5 - 15*(B*a - A*b)*x^2 + 10*sqrt(3)*(B*a - A*b)*(a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a^2/b
^2)^(1/3) - sqrt(3)*a)/a) + 5*(B*a - A*b)*(a^2/b^2)^(1/3)*log(a*x^2 - b*x*(a^2/b^2)^(2/3) + a*(a^2/b^2)^(1/3))
- 10*(B*a - A*b)*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)))/b^2

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Sympy [A]  time = 0.781595, size = 112, normalized size = 0.67 \begin{align*} \frac{B x^{5}}{5 b} + \operatorname{RootSum}{\left (27 t^{3} b^{8} - A^{3} a^{2} b^{3} + 3 A^{2} B a^{3} b^{2} - 3 A B^{2} a^{4} b + B^{3} a^{5}, \left ( t \mapsto t \log{\left (\frac{9 t^{2} b^{5}}{A^{2} a b^{2} - 2 A B a^{2} b + B^{2} a^{3}} + x \right )} \right )\right )} - \frac{x^{2} \left (- A b + B a\right )}{2 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x**3+A)/(b*x**3+a),x)

[Out]

B*x**5/(5*b) + RootSum(27*_t**3*b**8 - A**3*a**2*b**3 + 3*A**2*B*a**3*b**2 - 3*A*B**2*a**4*b + B**3*a**5, Lamb
da(_t, _t*log(9*_t**2*b**5/(A**2*a*b**2 - 2*A*B*a**2*b + B**2*a**3) + x))) - x**2*(-A*b + B*a)/(2*b**2)

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Giac [A]  time = 1.16652, size = 279, normalized size = 1.67 \begin{align*} -\frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{2}{3}} B a - \left (-a b^{2}\right )^{\frac{2}{3}} A b\right )} \arctan \left (\frac{\sqrt{3}{\left (2 \, x + \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )}{3 \, b^{4}} + \frac{{\left (\left (-a b^{2}\right )^{\frac{2}{3}} B a - \left (-a b^{2}\right )^{\frac{2}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{6 \, b^{4}} - \frac{{\left (B a^{2} b^{3} \left (-\frac{a}{b}\right )^{\frac{1}{3}} - A a b^{4} \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{3 \, a b^{5}} + \frac{2 \, B b^{4} x^{5} - 5 \, B a b^{3} x^{2} + 5 \, A b^{4} x^{2}}{10 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^3+A)/(b*x^3+a),x, algorithm="giac")

[Out]

-1/3*sqrt(3)*((-a*b^2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b
^4 + 1/6*((-a*b^2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^4 - 1/3*(B*a^2*b
^3*(-a/b)^(1/3) - A*a*b^4*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^5) + 1/10*(2*B*b^4*x^5 -
5*B*a*b^3*x^2 + 5*A*b^4*x^2)/b^5