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

Optimal. Leaf size=197 \[ -\frac{(a B+5 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{4/3}}+\frac{(a B+5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{4/3}}-\frac{(a B+5 A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3} b^{4/3}}+\frac{x (a B+5 A b)}{18 a^2 b \left (a+b x^3\right )}+\frac{x (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

[Out]

((A*b - a*B)*x)/(6*a*b*(a + b*x^3)^2) + ((5*A*b + a*B)*x)/(18*a^2*b*(a + b*x^3)) - ((5*A*b + a*B)*ArcTan[(a^(1
/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)*b^(4/3)) + ((5*A*b + a*B)*Log[a^(1/3) + b^(1/3)*x])/
(27*a^(8/3)*b^(4/3)) - ((5*A*b + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(8/3)*b^(4/3))

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Rubi [A]  time = 0.1026, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {385, 199, 200, 31, 634, 617, 204, 628} \[ -\frac{(a B+5 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{4/3}}+\frac{(a B+5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{4/3}}-\frac{(a B+5 A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3} b^{4/3}}+\frac{x (a B+5 A b)}{18 a^2 b \left (a+b x^3\right )}+\frac{x (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(a + b*x^3)^3,x]

[Out]

((A*b - a*B)*x)/(6*a*b*(a + b*x^3)^2) + ((5*A*b + a*B)*x)/(18*a^2*b*(a + b*x^3)) - ((5*A*b + a*B)*ArcTan[(a^(1
/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)*b^(4/3)) + ((5*A*b + a*B)*Log[a^(1/3) + b^(1/3)*x])/
(27*a^(8/3)*b^(4/3)) - ((5*A*b + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(8/3)*b^(4/3))

Rule 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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}{\left (a+b x^3\right )^3} \, dx &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) \int \frac{1}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) x}{18 a^2 b \left (a+b x^3\right )}+\frac{(5 A b+a B) \int \frac{1}{a+b x^3} \, dx}{9 a^2 b}\\ &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) x}{18 a^2 b \left (a+b x^3\right )}+\frac{(5 A b+a B) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{8/3} b}+\frac{(5 A b+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}{27 a^{8/3} b}\\ &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) x}{18 a^2 b \left (a+b x^3\right )}+\frac{(5 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{4/3}}-\frac{(5 A b+a B) \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}{54 a^{8/3} b^{4/3}}+\frac{(5 A b+a B) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{18 a^{7/3} b}\\ &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) x}{18 a^2 b \left (a+b x^3\right )}+\frac{(5 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{4/3}}-\frac{(5 A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{4/3}}+\frac{(5 A b+a B) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{9 a^{8/3} b^{4/3}}\\ &=\frac{(A b-a B) x}{6 a b \left (a+b x^3\right )^2}+\frac{(5 A b+a B) x}{18 a^2 b \left (a+b x^3\right )}-\frac{(5 A b+a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3} b^{4/3}}+\frac{(5 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{4/3}}-\frac{(5 A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.122757, size = 175, normalized size = 0.89 \[ \frac{-(a B+5 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{9 a^{5/3} \sqrt [3]{b} x (a B-A b)}{\left (a+b x^3\right )^2}+\frac{3 a^{2/3} \sqrt [3]{b} x (a B+5 A b)}{a+b x^3}+2 (a B+5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt{3} (a B+5 A b) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{54 a^{8/3} b^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^(5/3)*b^(1/3)*(-(A*b) + a*B)*x)/(a + b*x^3)^2 + (3*a^(2/3)*b^(1/3)*(5*A*b + a*B)*x)/(a + b*x^3) - 2*Sqr
t[3]*(5*A*b + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 2*(5*A*b + a*B)*Log[a^(1/3) + b^(1/3)*x] - (5
*A*b + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(8/3)*b^(4/3))

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Maple [A]  time = 0.01, size = 249, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( b{x}^{3}+a \right ) ^{2}} \left ({\frac{ \left ( 5\,Ab+Ba \right ){x}^{4}}{18\,{a}^{2}}}+{\frac{ \left ( 4\,Ab-Ba \right ) x}{9\,ab}} \right ) }+{\frac{5\,A}{27\,{a}^{2}b}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{B}{27\,a{b}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{5\,A}{54\,{a}^{2}b}\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{B}{54\,a{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{5\,\sqrt{3}A}{27\,{a}^{2}b}\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{\sqrt{3}B}{27\,a{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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/18*(5*A*b+B*a)/a^2*x^4+1/9*(4*A*b-B*a)/a/b*x)/(b*x^3+a)^2+5/27/a^2/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*A+1/27/a
/b^2/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*B-5/54/a^2/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A-1/54/a/b^2/(a/
b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B+5/27/a^2/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*
x-1))*A+1/27/a/b^2/(a/b)^(2/3)*3^(1/2)*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((B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.57998, size = 1648, normalized size = 8.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(3*(B*a^3*b^2 + 5*A*a^2*b^3)*x^4 + 3*sqrt(1/3)*((B*a^2*b^3 + 5*A*a*b^4)*x^6 + B*a^4*b + 5*A*a^3*b^2 + 2*
(B*a^3*b^2 + 5*A*a^2*b^3)*x^3)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)
*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - ((B*a*b^2 + 5*A*b^3)*x
^6 + B*a^3 + 5*A*a^2*b + 2*(B*a^2*b + 5*A*a*b^2)*x^3)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1
/3)*a) + 2*((B*a*b^2 + 5*A*b^3)*x^6 + B*a^3 + 5*A*a^2*b + 2*(B*a^2*b + 5*A*a*b^2)*x^3)*(a^2*b)^(2/3)*log(a*b*x
+ (a^2*b)^(2/3)) - 6*(B*a^4*b - 4*A*a^3*b^2)*x)/(a^4*b^4*x^6 + 2*a^5*b^3*x^3 + a^6*b^2), 1/54*(3*(B*a^3*b^2 +
5*A*a^2*b^3)*x^4 + 6*sqrt(1/3)*((B*a^2*b^3 + 5*A*a*b^4)*x^6 + B*a^4*b + 5*A*a^3*b^2 + 2*(B*a^3*b^2 + 5*A*a^2*
b^3)*x^3)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a
^2) - ((B*a*b^2 + 5*A*b^3)*x^6 + B*a^3 + 5*A*a^2*b + 2*(B*a^2*b + 5*A*a*b^2)*x^3)*(a^2*b)^(2/3)*log(a*b*x^2 -
(a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2*((B*a*b^2 + 5*A*b^3)*x^6 + B*a^3 + 5*A*a^2*b + 2*(B*a^2*b + 5*A*a*b^2)*
x^3)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)) - 6*(B*a^4*b - 4*A*a^3*b^2)*x)/(a^4*b^4*x^6 + 2*a^5*b^3*x^3 + a^
6*b^2)]

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Sympy [A]  time = 0.986885, size = 133, normalized size = 0.68 \begin{align*} \frac{x^{4} \left (5 A b^{2} + B a b\right ) + x \left (8 A a b - 2 B a^{2}\right )}{18 a^{4} b + 36 a^{3} b^{2} x^{3} + 18 a^{2} b^{3} x^{6}} + \operatorname{RootSum}{\left (19683 t^{3} a^{8} b^{4} - 125 A^{3} b^{3} - 75 A^{2} B a b^{2} - 15 A B^{2} a^{2} b - B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{27 t a^{3} b}{5 A b + B a} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x**4*(5*A*b**2 + B*a*b) + x*(8*A*a*b - 2*B*a**2))/(18*a**4*b + 36*a**3*b**2*x**3 + 18*a**2*b**3*x**6) + RootS
um(19683*_t**3*a**8*b**4 - 125*A**3*b**3 - 75*A**2*B*a*b**2 - 15*A*B**2*a**2*b - B**3*a**3, Lambda(_t, _t*log(
27*_t*a**3*b/(5*A*b + B*a) + x)))

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Giac [A]  time = 1.1472, size = 273, normalized size = 1.39 \begin{align*} -\frac{{\left (B a + 5 \, A b\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{27 \, a^{3} b} + \frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a + 5 \, \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 )}{27 \, a^{3} b^{2}} + \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a + 5 \, \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 )}{54 \, a^{3} b^{2}} + \frac{B a b x^{4} + 5 \, A b^{2} x^{4} - 2 \, B a^{2} x + 8 \, A a b x}{18 \,{\left (b x^{3} + a\right )}^{2} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(B*a + 5*A*b)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b) + 1/27*sqrt(3)*((-a*b^2)^(1/3)*B*a + 5*(-a
*b^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b^2) + 1/54*((-a*b^2)^(1/3)*B*a +
5*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b^2) + 1/18*(B*a*b*x^4 + 5*A*b^2*x^4 - 2*B
*a^2*x + 8*A*a*b*x)/((b*x^3 + a)^2*a^2*b)