\(\int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx\) [156]

Optimal result

Integrand size = 15, antiderivative size = 118 \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\left (-\frac {1}{2 a x^2}+\frac {5 b}{6 a^2 x}\right ) \sqrt [3]{a+b x}+\frac {5 b^2 \left (-\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a+b x}}{2 \sqrt [3]{a}+\sqrt [3]{a+b x}}\right )+\frac {3}{2} \log \left (\frac {-\sqrt [3]{a}+\sqrt [3]{a+b x}}{\sqrt [3]{x}}\right )\right )}{9 \left (a^2\right )^{4/3}} \]

[Out]

(-1/2/a/x^2+5/6*b/a^2/x)*(b*x+a)^(1/3)+5/9*b^2/a^2/(a^2)^(1/3)*(3/2*ln(((b*x+a)^(1/3)-a^(1/3))/x^(1/3))-3^(1/2
)*arctan(3^(1/2)*(b*x+a)^(1/3)/((b*x+a)^(1/3)+2*a^(1/3))))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.69, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1973, 44, 59, 632, 210, 31} \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=-\frac {5 b^2 \left (\frac {b x}{a}+1\right )^{2/3} \arctan \left (\frac {2 \sqrt [3]{\frac {b x}{a}+1}+1}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 \sqrt [3]{(a+b x)^2}}-\frac {5 b^2 \log (x) \left (\frac {b x}{a}+1\right )^{2/3}}{18 a^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b^2 \left (\frac {b x}{a}+1\right )^{2/3} \log \left (1-\sqrt [3]{\frac {b x}{a}+1}\right )}{6 a^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b (a+b x)}{6 a^2 x \sqrt [3]{(a+b x)^2}}-\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}} \]

[In]

Int[1/(x^3*((a + b*x)^2)^(1/3)),x]

[Out]

-1/2*(a + b*x)/(a*x^2*((a + b*x)^2)^(1/3)) + (5*b*(a + b*x))/(6*a^2*x*((a + b*x)^2)^(1/3)) - (5*b^2*(1 + (b*x)
/a)^(2/3)*ArcTan[(1 + 2*(1 + (b*x)/a)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*a^2*((a + b*x)^2)^(1/3)) - (5*b^2*(1 + (b*x)
/a)^(2/3)*Log[x])/(18*a^2*((a + b*x)^2)^(1/3)) + (5*b^2*(1 + (b*x)/a)^(2/3)*Log[1 - (1 + (b*x)/a)^(1/3)])/(6*a
^2*((a + b*x)^2)^(1/3))

Rule 31

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

Rule 44

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+\frac {b x}{a}\right )^{2/3} \int \frac {1}{x^3 \left (1+\frac {b x}{a}\right )^{2/3}} \, dx}{\sqrt [3]{(a+b x)^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}}-\frac {\left (5 b \left (1+\frac {b x}{a}\right )^{2/3}\right ) \int \frac {1}{x^2 \left (1+\frac {b x}{a}\right )^{2/3}} \, dx}{6 a \sqrt [3]{(a+b x)^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b (a+b x)}{6 a^2 x \sqrt [3]{(a+b x)^2}}+\frac {\left (5 b^2 \left (1+\frac {b x}{a}\right )^{2/3}\right ) \int \frac {1}{x \left (1+\frac {b x}{a}\right )^{2/3}} \, dx}{9 a^2 \sqrt [3]{(a+b x)^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b (a+b x)}{6 a^2 x \sqrt [3]{(a+b x)^2}}-\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \log (x)}{18 a^2 \sqrt [3]{(a+b x)^2}}-\frac {\left (5 b^2 \left (1+\frac {b x}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+\frac {b x}{a}}\right )}{6 a^2 \sqrt [3]{(a+b x)^2}}-\frac {\left (5 b^2 \left (1+\frac {b x}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+\frac {b x}{a}}\right )}{6 a^2 \sqrt [3]{(a+b x)^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b (a+b x)}{6 a^2 x \sqrt [3]{(a+b x)^2}}-\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \log (x)}{18 a^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \log \left (1-\sqrt [3]{1+\frac {b x}{a}}\right )}{6 a^2 \sqrt [3]{(a+b x)^2}}+\frac {\left (5 b^2 \left (1+\frac {b x}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+\frac {b x}{a}}\right )}{3 a^2 \sqrt [3]{(a+b x)^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b (a+b x)}{6 a^2 x \sqrt [3]{(a+b x)^2}}-\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \arctan \left (\frac {1+2 \sqrt [3]{1+\frac {b x}{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 \sqrt [3]{(a+b x)^2}}-\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \log (x)}{18 a^2 \sqrt [3]{(a+b x)^2}}+\frac {5 b^2 \left (1+\frac {b x}{a}\right )^{2/3} \log \left (1-\sqrt [3]{1+\frac {b x}{a}}\right )}{6 a^2 \sqrt [3]{(a+b x)^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.55 \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\frac {-9 a^{8/3}+6 a^{5/3} b x+15 a^{2/3} b^2 x^2-10 \sqrt {3} b^2 x^2 (a+b x)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+10 b^2 x^2 (a+b x)^{2/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-5 b^2 x^2 (a+b x)^{2/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{18 a^{8/3} x^2 \sqrt [3]{(a+b x)^2}} \]

[In]

Integrate[1/(x^3*((a + b*x)^2)^(1/3)),x]

[Out]

(-9*a^(8/3) + 6*a^(5/3)*b*x + 15*a^(2/3)*b^2*x^2 - 10*Sqrt[3]*b^2*x^2*(a + b*x)^(2/3)*ArcTan[(1 + (2*(a + b*x)
^(1/3))/a^(1/3))/Sqrt[3]] + 10*b^2*x^2*(a + b*x)^(2/3)*Log[a^(1/3) - (a + b*x)^(1/3)] - 5*b^2*x^2*(a + b*x)^(2
/3)*Log[a^(2/3) + a^(1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)])/(18*a^(8/3)*x^2*((a + b*x)^2)^(1/3))

Maple [F]

\[\int \frac {1}{x^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {1}{3}}}d x\]

[In]

int(1/x^3/((b*x+a)^2)^(1/3),x)

[Out]

int(1/x^3/((b*x+a)^2)^(1/3),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (94) = 188\).

Time = 0.27 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.76 \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\frac {10 \, \sqrt {3} {\left (a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} {\left (a^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} {\left (b x + a\right )} + 2 \, \sqrt {3} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} a\right )}}{3 \, {\left (a b x + a^{2}\right )}}\right ) - 5 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} {\left (a^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {2}{3}} a^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} {\left (a b x + a^{2}\right )} {\left (a^{2}\right )}^{\frac {1}{3}} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} {\left (a^{2}\right )}^{\frac {2}{3}}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + 10 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} {\left (a^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (a^{2}\right )}^{\frac {1}{3}} {\left (b x + a\right )} - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} a}{b x + a}\right ) + 3 \, {\left (5 \, a^{2} b x - 3 \, a^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {2}{3}}}{18 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}} \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/3),x, algorithm="fricas")

[Out]

1/18*(10*sqrt(3)*(a*b^3*x^3 + a^2*b^2*x^2)*(a^2)^(1/6)*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*(b*x + a) +
2*sqrt(3)*(b^2*x^2 + 2*a*b*x + a^2)^(1/3)*a)/(a*b*x + a^2)) - 5*(b^3*x^3 + a*b^2*x^2)*(a^2)^(2/3)*log(((b^2*x
^2 + 2*a*b*x + a^2)^(2/3)*a^2 + (b^2*x^2 + 2*a*b*x + a^2)^(1/3)*(a*b*x + a^2)*(a^2)^(1/3) + (b^2*x^2 + 2*a*b*x
+ a^2)*(a^2)^(2/3))/(b^2*x^2 + 2*a*b*x + a^2)) + 10*(b^3*x^3 + a*b^2*x^2)*(a^2)^(2/3)*log(-((a^2)^(1/3)*(b*x
+ a) - (b^2*x^2 + 2*a*b*x + a^2)^(1/3)*a)/(b*x + a)) + 3*(5*a^2*b*x - 3*a^3)*(b^2*x^2 + 2*a*b*x + a^2)^(2/3))/
(a^4*b*x^3 + a^5*x^2)

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\int \frac {1}{x^{3} \sqrt [3]{\left (a + b x\right )^{2}}}\, dx \]

[In]

integrate(1/x**3/((b*x+a)**2)**(1/3),x)

[Out]

Integral(1/(x**3*((a + b*x)**2)**(1/3)), x)

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\int { \frac {1}{{\left ({\left (b x + a\right )}^{2}\right )}^{\frac {1}{3}} x^{3}} \,d x } \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/(((b*x + a)^2)^(1/3)*x^3), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (94) = 188\).

Time = 3.15 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.50 \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=-\frac {\frac {10 \, \sqrt {3} \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} + \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}}\right )}}{3 \, \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}}}\right )}{a^{3}} + \frac {5 \, \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{3} \log \left ({\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {2}{3}} + {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} + \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {2}{3}}\right )}{a^{3}} - \frac {10 \, \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{3} \log \left ({\left | {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} - \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} \right |}\right )}{a^{3}} - \frac {3 \, {\left (5 \, {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {4}{3}} b^{3} \mathrm {sgn}\left (b x + a\right ) - 8 \, {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right )^{2}}}{18 \, b \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt [3]{(a+b x)^2}} \, dx=\int \frac {1}{x^3\,{\left ({\left (a+b\,x\right )}^2\right )}^{1/3}} \,d x \]

[In]

int(1/(x^3*((a + b*x)^2)^(1/3)),x)

[Out]

int(1/(x^3*((a + b*x)^2)^(1/3)), x)