3.86 \(\int \frac{A+B x^3}{x^6 (a+b x^3)^2} \, dx\)

Optimal. Leaf size=215 \[ -\frac{b^{2/3} (8 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{11/3}}+\frac{b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3}}-\frac{b^{2/3} (8 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{11/3}}+\frac{8 A b-5 a B}{6 a^3 x^2}-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )} \]

[Out]

-(8*A*b - 5*a*B)/(15*a^2*b*x^5) + (8*A*b - 5*a*B)/(6*a^3*x^2) + (A*b - a*B)/(3*a*b*x^5*(a + b*x^3)) - (b^(2/3)
*(8*A*b - 5*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(11/3)) + (b^(2/3)*(8*A*b - 5
*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(11/3)) - (b^(2/3)*(8*A*b - 5*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2
/3)*x^2])/(18*a^(11/3))

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Rubi [A]  time = 0.127009, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {457, 325, 200, 31, 634, 617, 204, 628} \[ -\frac{b^{2/3} (8 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{11/3}}+\frac{b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3}}-\frac{b^{2/3} (8 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{11/3}}+\frac{8 A b-5 a B}{6 a^3 x^2}-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^6*(a + b*x^3)^2),x]

[Out]

-(8*A*b - 5*a*B)/(15*a^2*b*x^5) + (8*A*b - 5*a*B)/(6*a^3*x^2) + (A*b - a*B)/(3*a*b*x^5*(a + b*x^3)) - (b^(2/3)
*(8*A*b - 5*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(11/3)) + (b^(2/3)*(8*A*b - 5
*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(11/3)) - (b^(2/3)*(8*A*b - 5*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2
/3)*x^2])/(18*a^(11/3))

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{A+B x^3}{x^6 \left (a+b x^3\right )^2} \, dx &=\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}+\frac{(8 A b-5 a B) \int \frac{1}{x^6 \left (a+b x^3\right )} \, dx}{3 a b}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}-\frac{(8 A b-5 a B) \int \frac{1}{x^3 \left (a+b x^3\right )} \, dx}{3 a^2}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{8 A b-5 a B}{6 a^3 x^2}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}+\frac{(b (8 A b-5 a B)) \int \frac{1}{a+b x^3} \, dx}{3 a^3}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{8 A b-5 a B}{6 a^3 x^2}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}+\frac{(b (8 A b-5 a B)) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 a^{11/3}}+\frac{(b (8 A b-5 a B)) \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}{9 a^{11/3}}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{8 A b-5 a B}{6 a^3 x^2}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}+\frac{b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3}}-\frac{\left (b^{2/3} (8 A b-5 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}{18 a^{11/3}}+\frac{(b (8 A b-5 a B)) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 a^{10/3}}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{8 A b-5 a B}{6 a^3 x^2}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}+\frac{b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3}}-\frac{b^{2/3} (8 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{11/3}}+\frac{\left (b^{2/3} (8 A b-5 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 )}{3 a^{11/3}}\\ &=-\frac{8 A b-5 a B}{15 a^2 b x^5}+\frac{8 A b-5 a B}{6 a^3 x^2}+\frac{A b-a B}{3 a b x^5 \left (a+b x^3\right )}-\frac{b^{2/3} (8 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{11/3}}+\frac{b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3}}-\frac{b^{2/3} (8 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{11/3}}\\ \end{align*}

Mathematica [A]  time = 0.142564, size = 183, normalized size = 0.85 \[ \frac{5 b^{2/3} (5 a B-8 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{30 a^{2/3} b x (a B-A b)}{a+b x^3}-\frac{45 a^{2/3} (a B-2 A b)}{x^2}-\frac{18 a^{5/3} A}{x^5}+10 b^{2/3} (8 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-10 \sqrt{3} b^{2/3} (8 A b-5 a B) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{90 a^{11/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^6*(a + b*x^3)^2),x]

[Out]

((-18*a^(5/3)*A)/x^5 - (45*a^(2/3)*(-2*A*b + a*B))/x^2 - (30*a^(2/3)*b*(-(A*b) + a*B)*x)/(a + b*x^3) - 10*Sqrt
[3]*b^(2/3)*(8*A*b - 5*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 10*b^(2/3)*(8*A*b - 5*a*B)*Log[a^(1/
3) + b^(1/3)*x] + 5*b^(2/3)*(-8*A*b + 5*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(90*a^(11/3))

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Maple [A]  time = 0.013, size = 252, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}Ax}{3\,{a}^{3} \left ( b{x}^{3}+a \right ) }}-{\frac{bBx}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{8\,Ab}{9\,{a}^{3}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{4\,Ab}{9\,{a}^{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{8\,Ab\sqrt{3}}{9\,{a}^{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}}}}-{\frac{5\,B}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{5\,B}{18\,{a}^{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{5\,B\sqrt{3}}{9\,{a}^{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}{5\,{a}^{2}{x}^{5}}}+{\frac{Ab}{{a}^{3}{x}^{2}}}-{\frac{B}{2\,{a}^{2}{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^6/(b*x^3+a)^2,x)

[Out]

1/3/a^3*b^2*x/(b*x^3+a)*A-1/3/a^2*b*x/(b*x^3+a)*B+8/9/a^3*b*A/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-4/9/a^3*b*A/(a/b)^
(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+8/9/a^3*b*A/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)
)-5/9/a^2*B/(a/b)^(2/3)*ln(x+(a/b)^(1/3))+5/18/a^2*B/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-5/9/a^2*B/(
a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))-1/5*A/a^2/x^5+1/a^3/x^2*A*b-1/2/a^2/x^2*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((B*x^3+A)/x^6/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65282, size = 633, normalized size = 2.94 \begin{align*} -\frac{15 \,{\left (5 \, B a b - 8 \, A b^{2}\right )} x^{6} + 9 \,{\left (5 \, B a^{2} - 8 \, A a b\right )} x^{3} + 18 \, A a^{2} + 10 \, \sqrt{3}{\left ({\left (5 \, B a b - 8 \, A b^{2}\right )} x^{8} +{\left (5 \, B a^{2} - 8 \, A a b\right )} x^{5}\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} a x \left (\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} - \sqrt{3} b}{3 \, b}\right ) - 5 \,{\left ({\left (5 \, B a b - 8 \, A b^{2}\right )} x^{8} +{\left (5 \, B a^{2} - 8 \, A a b\right )} x^{5}\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} + a^{2} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}}\right ) + 10 \,{\left ({\left (5 \, B a b - 8 \, A b^{2}\right )} x^{8} +{\left (5 \, B a^{2} - 8 \, A a b\right )} x^{5}\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b x + a \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right )}{90 \,{\left (a^{3} b x^{8} + a^{4} x^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

-1/90*(15*(5*B*a*b - 8*A*b^2)*x^6 + 9*(5*B*a^2 - 8*A*a*b)*x^3 + 18*A*a^2 + 10*sqrt(3)*((5*B*a*b - 8*A*b^2)*x^8
+ (5*B*a^2 - 8*A*a*b)*x^5)*(b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b^2/a^2)^(2/3) - sqrt(3)*b)/b) - 5*((5*
B*a*b - 8*A*b^2)*x^8 + (5*B*a^2 - 8*A*a*b)*x^5)*(b^2/a^2)^(1/3)*log(b^2*x^2 - a*b*x*(b^2/a^2)^(1/3) + a^2*(b^2
/a^2)^(2/3)) + 10*((5*B*a*b - 8*A*b^2)*x^8 + (5*B*a^2 - 8*A*a*b)*x^5)*(b^2/a^2)^(1/3)*log(b*x + a*(b^2/a^2)^(1
/3)))/(a^3*b*x^8 + a^4*x^5)

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Sympy [A]  time = 1.60649, size = 138, normalized size = 0.64 \begin{align*} \operatorname{RootSum}{\left (729 t^{3} a^{11} - 512 A^{3} b^{5} + 960 A^{2} B a b^{4} - 600 A B^{2} a^{2} b^{3} + 125 B^{3} a^{3} b^{2}, \left ( t \mapsto t \log{\left (- \frac{9 t a^{4}}{- 8 A b^{2} + 5 B a b} + x \right )} \right )\right )} - \frac{6 A a^{2} + x^{6} \left (- 40 A b^{2} + 25 B a b\right ) + x^{3} \left (- 24 A a b + 15 B a^{2}\right )}{30 a^{4} x^{5} + 30 a^{3} b x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**6/(b*x**3+a)**2,x)

[Out]

RootSum(729*_t**3*a**11 - 512*A**3*b**5 + 960*A**2*B*a*b**4 - 600*A*B**2*a**2*b**3 + 125*B**3*a**3*b**2, Lambd
a(_t, _t*log(-9*_t*a**4/(-8*A*b**2 + 5*B*a*b) + x))) - (6*A*a**2 + x**6*(-40*A*b**2 + 25*B*a*b) + x**3*(-24*A*
a*b + 15*B*a**2))/(30*a**4*x**5 + 30*a**3*b*x**8)

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Giac [A]  time = 1.14912, size = 278, normalized size = 1.29 \begin{align*} -\frac{\sqrt{3}{\left (5 \, \left (-a b^{2}\right )^{\frac{1}{3}} B a - 8 \, \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 )}{9 \, a^{4}} + \frac{{\left (5 \, B a b - 8 \, A b^{2}\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{9 \, a^{4}} - \frac{{\left (5 \, \left (-a b^{2}\right )^{\frac{1}{3}} B a - 8 \, \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 )}{18 \, a^{4}} - \frac{B a b x - A b^{2} x}{3 \,{\left (b x^{3} + a\right )} a^{3}} - \frac{5 \, B a x^{3} - 10 \, A b x^{3} + 2 \, A a}{10 \, a^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/9*sqrt(3)*(5*(-a*b^2)^(1/3)*B*a - 8*(-a*b^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3
))/a^4 + 1/9*(5*B*a*b - 8*A*b^2)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^4 - 1/18*(5*(-a*b^2)^(1/3)*B*a - 8*
(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^4 - 1/3*(B*a*b*x - A*b^2*x)/((b*x^3 + a)*a^3) -
1/10*(5*B*a*x^3 - 10*A*b*x^3 + 2*A*a)/(a^3*x^5)