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

Optimal. Leaf size=162 \[ \frac{\sqrt [3]{a} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}+\frac{x (A b-a B)}{b^2}-\frac{\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{\sqrt [3]{a} (A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{7/3}}+\frac{B x^4}{4 b} \]

[Out]

((A*b - a*B)*x)/b^2 + (B*x^4)/(4*b) + (a^(1/3)*(A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/
(Sqrt[3]*b^(7/3)) - (a^(1/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(7/3)) + (a^(1/3)*(A*b - a*B)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(7/3))

________________________________________________________________________________________

Rubi [A]  time = 0.115521, antiderivative size = 162, 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, 200, 31, 634, 617, 204, 628} \[ \frac{\sqrt [3]{a} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}+\frac{x (A b-a B)}{b^2}-\frac{\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{\sqrt [3]{a} (A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{7/3}}+\frac{B x^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x^3))/(a + b*x^3),x]

[Out]

((A*b - a*B)*x)/b^2 + (B*x^4)/(4*b) + (a^(1/3)*(A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/
(Sqrt[3]*b^(7/3)) - (a^(1/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(7/3)) + (a^(1/3)*(A*b - a*B)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(7/3))

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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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^3 \left (A+B x^3\right )}{a+b x^3} \, dx &=\frac{B x^4}{4 b}-\frac{(-4 A b+4 a B) \int \frac{x^3}{a+b x^3} \, dx}{4 b}\\ &=\frac{(A b-a B) x}{b^2}+\frac{B x^4}{4 b}-\frac{(a (A b-a B)) \int \frac{1}{a+b x^3} \, dx}{b^2}\\ &=\frac{(A b-a B) x}{b^2}+\frac{B x^4}{4 b}-\frac{\left (\sqrt [3]{a} (A b-a B)\right ) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^2}-\frac{\left (\sqrt [3]{a} (A b-a B)\right ) \int \frac{2 \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^2}\\ &=\frac{(A b-a B) x}{b^2}+\frac{B x^4}{4 b}-\frac{\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{\left (\sqrt [3]{a} (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^{7/3}}-\frac{\left (a^{2/3} (A b-a B)\right ) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b^2}\\ &=\frac{(A b-a B) x}{b^2}+\frac{B x^4}{4 b}-\frac{\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{\sqrt [3]{a} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}-\frac{\left (\sqrt [3]{a} (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^{7/3}}\\ &=\frac{(A b-a B) x}{b^2}+\frac{B x^4}{4 b}+\frac{\sqrt [3]{a} (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^{7/3}}-\frac{\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{\sqrt [3]{a} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}\\ \end{align*}

Mathematica [A]  time = 0.0784405, size = 152, normalized size = 0.94 \[ \frac{-2 \sqrt [3]{a} (a B-A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+12 \sqrt [3]{b} x (A b-a B)+4 \sqrt [3]{a} (a B-A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-4 \sqrt{3} \sqrt [3]{a} (a B-A b) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )+3 b^{4/3} B x^4}{12 b^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*b^(1/3)*(A*b - a*B)*x + 3*b^(4/3)*B*x^4 - 4*Sqrt[3]*a^(1/3)*(-(A*b) + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/
3))/Sqrt[3]] + 4*a^(1/3)*(-(A*b) + a*B)*Log[a^(1/3) + b^(1/3)*x] - 2*a^(1/3)*(-(A*b) + a*B)*Log[a^(2/3) - a^(1
/3)*b^(1/3)*x + b^(2/3)*x^2])/(12*b^(7/3))

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 221, normalized size = 1.4 \begin{align*}{\frac{B{x}^{4}}{4\,b}}+{\frac{Ax}{b}}-{\frac{Bax}{{b}^{2}}}-{\frac{aA}{3\,{b}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{{a}^{2}B}{3\,{b}^{3}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{aA}{6\,{b}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\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 ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\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 ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\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 ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*x^4/b+1/b*A*x-1/b^2*B*a*x-1/3*a/b^2/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*A+1/3*a^2/b^3/(a/b)^(2/3)*ln(x+(a/b)^(
1/3))*B+1/6*a/b^2/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A-1/6*a^2/b^3/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x
+(a/b)^(2/3))*B-1/3*a/b^2/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/3*a^2/b^3/(a/b)^(2/3
)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*B

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.52449, size = 343, normalized size = 2.12 \begin{align*} \frac{3 \, B b x^{4} - 4 \, \sqrt{3}{\left (B a - A b\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} b x \left (-\frac{a}{b}\right )^{\frac{2}{3}} - \sqrt{3} a}{3 \, a}\right ) + 2 \,{\left (B a - A b\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left (x^{2} + x \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right ) - 4 \,{\left (B a - A b\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left (x - \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right ) - 12 \,{\left (B a - A b\right )} x}{12 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*b*x^4 - 4*sqrt(3)*(B*a - A*b)*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(-a/b)^(2/3) - sqrt(3)*a)/a) +
2*(B*a - A*b)*(-a/b)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3)) - 4*(B*a - A*b)*(-a/b)^(1/3)*log(x - (-a/b
)^(1/3)) - 12*(B*a - A*b)*x)/b^2

________________________________________________________________________________________

Sympy [A]  time = 0.825358, size = 87, normalized size = 0.54 \begin{align*} \frac{B x^{4}}{4 b} + \operatorname{RootSum}{\left (27 t^{3} b^{7} + A^{3} a b^{3} - 3 A^{2} B a^{2} b^{2} + 3 A B^{2} a^{3} b - B^{3} a^{4}, \left ( t \mapsto t \log{\left (\frac{3 t b^{2}}{- A b + B a} + x \right )} \right )\right )} - \frac{x \left (- A b + B a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x**4/(4*b) + RootSum(27*_t**3*b**7 + A**3*a*b**3 - 3*A**2*B*a**2*b**2 + 3*A*B**2*a**3*b - B**3*a**4, Lambda(
_t, _t*log(3*_t*b**2/(-A*b + B*a) + x))) - x*(-A*b + B*a)/b**2

________________________________________________________________________________________

Giac [A]  time = 1.13934, size = 251, normalized size = 1.55 \begin{align*} \frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a - \left (-a b^{2}\right )^{\frac{1}{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^{3}} + \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a - \left (-a b^{2}\right )^{\frac{1}{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^{3}} - \frac{{\left (B a^{2} b^{2} - A a b^{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^{4}} + \frac{B b^{3} x^{4} - 4 \, B a b^{2} x + 4 \, A b^{3} x}{4 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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