3.2.55 \(\int \frac {1}{x^2 \sqrt [3]{(a+b x)^2}} \, dx\) [155]

3.2.55.1 Optimal result
3.2.55.2 Mathematica [A] (verified)
3.2.55.3 Rubi [A] (verified)
3.2.55.4 Maple [F]
3.2.55.5 Fricas [B] (verification not implemented)
3.2.55.6 Sympy [F]
3.2.55.7 Maxima [F]
3.2.55.8 Giac [B] (verification not implemented)
3.2.55.9 Mupad [F(-1)]

3.2.55.1 Optimal result

Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^2 \sqrt [3]{(a+b x)^2}} \, dx=-\frac {\sqrt [3]{a+b x}}{a x}-\frac {2 b \left (-\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a+b x}}{2 \sqrt [3]{a}+\sqrt [3]{a+b x}}\right )+\frac {3}{2} \log \left (\frac {-\sqrt [3]{a}+\sqrt [3]{a+b x}}{\sqrt [3]{x}}\right )\right )}{3 a \sqrt [3]{a^2}} \]

output
-(b*x+a)^(1/3)/a/x-2/3*b/a/(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))))
 
3.2.55.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.51 \[ \int \frac {1}{x^2 \sqrt [3]{(a+b x)^2}} \, dx=\frac {-3 a^{5/3}-3 a^{2/3} b x+2 \sqrt {3} b x (a+b x)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 b x (a+b x)^{2/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )+b x (a+b x)^{2/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{3 a^{5/3} x \sqrt [3]{(a+b x)^2}} \]

input
Integrate[1/(x^2*((a + b*x)^2)^(1/3)),x]
 
output
(-3*a^(5/3) - 3*a^(2/3)*b*x + 2*Sqrt[3]*b*x*(a + b*x)^(2/3)*ArcTan[(1 + (2 
*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]] - 2*b*x*(a + b*x)^(2/3)*Log[a^(1/3) - 
(a + b*x)^(1/3)] + b*x*(a + b*x)^(2/3)*Log[a^(2/3) + a^(1/3)*(a + b*x)^(1/ 
3) + (a + b*x)^(2/3)])/(3*a^(5/3)*x*((a + b*x)^2)^(1/3))
 
3.2.55.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2008, 52, 69, 16, 1082, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2008

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {(a+b x)^{2/3} \left (-\frac {2 b \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{\sqrt [3]{(a+b x)^2}}\)

input
Int[1/(x^2*((a + b*x)^2)^(1/3)),x]
 
output
((a + b*x)^(2/3)*(-((a + b*x)^(1/3)/(a*x)) - (2*b*(-((Sqrt[3]*ArcTan[(1 + 
(2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x]/(2*a^(2/3)) + (3* 
Log[a^(1/3) - (a + b*x)^(1/3)])/(2*a^(2/3))))/(3*a)))/((a + b*x)^2)^(1/3)
 

3.2.55.3.1 Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 52
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 69
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[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 217
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 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 2008
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp 
on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p)   Int[u*(a + b*x)^(Expon[Px, x]*p), 
 x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /;  !IntegerQ[p] && PolyQ[Px, x 
] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 
3.2.55.4 Maple [F]

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

input
int(1/x^2/((b*x+a)^2)^(1/3),x)
 
output
int(1/x^2/((b*x+a)^2)^(1/3),x)
 
3.2.55.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (80) = 160\).

Time = 0.25 (sec) , antiderivative size = 325, normalized size of antiderivative = 3.12 \[ \int \frac {1}{x^2 \sqrt [3]{(a+b x)^2}} \, dx=-\frac {2 \, \sqrt {3} {\left (a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )} - 2 \, \sqrt {3} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} a\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, {\left (a b x + a^{2}\right )}}\right ) + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {2}{3}} a^{2} - {\left (b^{2} x^{2} + a b x\right )} \left (-a^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {2}{3}} a^{2} - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} {\left (a b x + a^{2}\right )} \left (-a^{2}\right )^{\frac {1}{3}} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a^{2}\right )^{\frac {2}{3}}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + 2 \, {\left (b^{2} x^{2} + a b x\right )} \left (-a^{2}\right )^{\frac {2}{3}} \log \left (\frac {\left (-a^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {1}{3}} a}{b x + a}\right )}{3 \, {\left (a^{3} b x^{2} + a^{4} x\right )}} \]

input
integrate(1/x^2/((b*x+a)^2)^(1/3),x, algorithm="fricas")
 
output
-1/3*(2*sqrt(3)*(a*b^2*x^2 + a^2*b*x)*sqrt(-(-a^2)^(1/3))*arctan(-1/3*(sqr 
t(3)*(-a^2)^(1/3)*(b*x + a) - 2*sqrt(3)*(b^2*x^2 + 2*a*b*x + a^2)^(1/3)*a) 
*sqrt(-(-a^2)^(1/3))/(a*b*x + a^2)) + 3*(b^2*x^2 + 2*a*b*x + a^2)^(2/3)*a^ 
2 - (b^2*x^2 + a*b*x)*(-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)) + 2*(b^2*x^2 + a 
*b*x)*(-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)))/(a^3*b*x^2 + a^4*x)
 
3.2.55.6 Sympy [F]

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

input
integrate(1/x**2/((b*x+a)**2)**(1/3),x)
 
output
Integral(1/(x**2*((a + b*x)**2)**(1/3)), x)
 
3.2.55.7 Maxima [F]

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

input
integrate(1/x^2/((b*x+a)^2)^(1/3),x, algorithm="maxima")
 
output
integrate(1/(((b*x + a)^2)^(1/3)*x^2), x)
 
3.2.55.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (80) = 160\).

Time = 3.13 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.37 \[ \int \frac {1}{x^2 \sqrt [3]{(a+b x)^2}} \, dx=\frac {\frac {2 \, \sqrt {3} \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} + \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}}\right )}}{3 \, \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}}}\right )}{a^{2}} + \frac {\left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{2} \log \left ({\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {2}{3}} + {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} + \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {2}{3}}\right )}{a^{2}} - \frac {2 \, \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} b^{2} \log \left ({\left | {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} - \left (a \mathrm {sgn}\left (b x + a\right )\right )^{\frac {1}{3}} \right |}\right )}{a^{2}} - \frac {3 \, {\left (b x \mathrm {sgn}\left (b x + a\right ) + a \mathrm {sgn}\left (b x + a\right )\right )}^{\frac {1}{3}} b}{a x}}{3 \, b \mathrm {sgn}\left (b x + a\right )} \]

input
integrate(1/x^2/((b*x+a)^2)^(1/3),x, algorithm="giac")
 
output
1/3*(2*sqrt(3)*(a*sgn(b*x + a))^(1/3)*b^2*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^2 + (a*sgn(b*x + a))^(1/3)*b^2*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^2 - 2*(a*sgn(b*x + a))^(1/3)*b^2*log(abs 
((b*x*sgn(b*x + a) + a*sgn(b*x + a))^(1/3) - (a*sgn(b*x + a))^(1/3)))/a^2 
- 3*(b*x*sgn(b*x + a) + a*sgn(b*x + a))^(1/3)*b/(a*x))/(b*sgn(b*x + a))
 
3.2.55.9 Mupad [F(-1)]

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

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