3.5.19 \(\int \frac {1}{x^3 (a+b x)^{4/3}} \, dx\) [419]

3.5.19.1 Optimal result
3.5.19.2 Mathematica [A] (verified)
3.5.19.3 Rubi [A] (verified)
3.5.19.4 Maple [A] (verified)
3.5.19.5 Fricas [A] (verification not implemented)
3.5.19.6 Sympy [C] (verification not implemented)
3.5.19.7 Maxima [A] (verification not implemented)
3.5.19.8 Giac [A] (verification not implemented)
3.5.19.9 Mupad [B] (verification not implemented)
3.5.19.10 Reduce [B] (verification not implemented)

3.5.19.1 Optimal result

Integrand size = 13, antiderivative size = 149 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {14 b^2}{3 a^3 \sqrt [3]{a+b x}}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}+\frac {7 b}{6 a^2 x \sqrt [3]{a+b x}}+\frac {14 b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{10/3}}-\frac {7 b^2 \log (x)}{9 a^{10/3}}+\frac {7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{10/3}} \]

output
14/3*b^2/a^3/(b*x+a)^(1/3)-1/2/a/x^2/(b*x+a)^(1/3)+7/6*b/a^2/x/(b*x+a)^(1/ 
3)-7/9*b^2*ln(x)/a^(10/3)+7/3*b^2*ln(a^(1/3)-(b*x+a)^(1/3))/a^(10/3)+14/9* 
b^2*arctan(1/3*(a^(1/3)+2*(b*x+a)^(1/3))/a^(1/3)*3^(1/2))/a^(10/3)*3^(1/2)
 
3.5.19.2 Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {\frac {3 \sqrt [3]{a} \left (-3 a^2+7 a b x+28 b^2 x^2\right )}{x^2 \sqrt [3]{a+b x}}+28 \sqrt {3} b^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+28 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-14 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{18 a^{10/3}} \]

input
Integrate[1/(x^3*(a + b*x)^(4/3)),x]
 
output
((3*a^(1/3)*(-3*a^2 + 7*a*b*x + 28*b^2*x^2))/(x^2*(a + b*x)^(1/3)) + 28*Sq 
rt[3]*b^2*ArcTan[(1 + (2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]] + 28*b^2*Log[a 
^(1/3) - (a + b*x)^(1/3)] - 14*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x)^(1/3) + 
 (a + b*x)^(2/3)])/(18*a^(10/3))
 
3.5.19.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {52, 52, 61, 67, 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.

\(\displaystyle \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 b \int \frac {1}{x^2 (a+b x)^{4/3}}dx}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 b \left (-\frac {4 b \int \frac {1}{x (a+b x)^{4/3}}dx}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 b \left (-\frac {4 b \left (\frac {\int \frac {1}{x \sqrt [3]{a+b x}}dx}{a}+\frac {3}{a \sqrt [3]{a+b x}}\right )}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 67

\(\displaystyle -\frac {7 b \left (-\frac {4 b \left (\frac {\frac {3}{2} \int \frac {1}{a^{2/3}+\sqrt [3]{a+b x} \sqrt [3]{a}+(a+b x)^{2/3}}d\sqrt [3]{a+b x}-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{a+b x}}d\sqrt [3]{a+b x}}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}}{a}+\frac {3}{a \sqrt [3]{a+b x}}\right )}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {7 b \left (-\frac {4 b \left (\frac {\frac {3}{2} \int \frac {1}{a^{2/3}+\sqrt [3]{a+b x} \sqrt [3]{a}+(a+b x)^{2/3}}d\sqrt [3]{a+b x}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}}{a}+\frac {3}{a \sqrt [3]{a+b x}}\right )}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {7 b \left (-\frac {4 b \left (\frac {-\frac {3 \int \frac {1}{-(a+b x)^{2/3}-3}d\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}}{a}+\frac {3}{a \sqrt [3]{a+b x}}\right )}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {7 b \left (-\frac {4 b \left (\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}}{a}+\frac {3}{a \sqrt [3]{a+b x}}\right )}{3 a}-\frac {1}{a x \sqrt [3]{a+b x}}\right )}{6 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}}\)

input
Int[1/(x^3*(a + b*x)^(4/3)),x]
 
output
-1/2*1/(a*x^2*(a + b*x)^(1/3)) - (7*b*(-(1/(a*x*(a + b*x)^(1/3))) - (4*b*( 
3/(a*(a + b*x)^(1/3)) + ((Sqrt[3]*ArcTan[(1 + (2*(a + b*x)^(1/3))/a^(1/3)) 
/Sqrt[3]])/a^(1/3) - Log[x]/(2*a^(1/3)) + (3*Log[a^(1/3) - (a + b*x)^(1/3) 
])/(2*a^(1/3)))/a))/(3*a)))/(6*a)
 

3.5.19.3.1 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 61
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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 67
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   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]
 
3.5.19.4 Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79

method result size
risch \(-\frac {\left (b x +a \right )^{\frac {2}{3}} \left (-10 b x +3 a \right )}{6 a^{3} x^{2}}+\frac {b^{2} \left (\frac {27}{\left (b x +a \right )^{\frac {1}{3}}}+\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}}-\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}}+\frac {14 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{a^{\frac {1}{3}}}\right )}{9 a^{3}}\) \(118\)
derivativedivides \(3 b^{2} \left (-\frac {\frac {-\frac {5 \left (b x +a \right )^{\frac {5}{3}}}{9}+\frac {13 a \left (b x +a \right )^{\frac {2}{3}}}{18}}{b^{2} x^{2}}-\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {1}{3}}}+\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {1}{3}}}-\frac {14 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {1}{3}}}}{a^{3}}+\frac {1}{a^{3} \left (b x +a \right )^{\frac {1}{3}}}\right )\) \(126\)
default \(3 b^{2} \left (-\frac {\frac {-\frac {5 \left (b x +a \right )^{\frac {5}{3}}}{9}+\frac {13 a \left (b x +a \right )^{\frac {2}{3}}}{18}}{b^{2} x^{2}}-\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {1}{3}}}+\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {1}{3}}}-\frac {14 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {1}{3}}}}{a^{3}}+\frac {1}{a^{3} \left (b x +a \right )^{\frac {1}{3}}}\right )\) \(126\)
pseudoelliptic \(\frac {\frac {14 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b x +a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, b^{2} x^{2} \left (b x +a \right )^{\frac {1}{3}}}{9}+\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) b^{2} x^{2} \left (b x +a \right )^{\frac {1}{3}}}{9}-\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) b^{2} x^{2} \left (b x +a \right )^{\frac {1}{3}}}{9}+\frac {14 b^{2} x^{2} a^{\frac {1}{3}}}{3}+\frac {7 a^{\frac {4}{3}} b x}{6}-\frac {a^{\frac {7}{3}}}{2}}{a^{\frac {10}{3}} x^{2} \left (b x +a \right )^{\frac {1}{3}}}\) \(145\)

input
int(1/x^3/(b*x+a)^(4/3),x,method=_RETURNVERBOSE)
 
output
-1/6*(b*x+a)^(2/3)*(-10*b*x+3*a)/a^3/x^2+1/9*b^2/a^3*(27/(b*x+a)^(1/3)+14/ 
a^(1/3)*ln((b*x+a)^(1/3)-a^(1/3))-7/a^(1/3)*ln((b*x+a)^(2/3)+a^(1/3)*(b*x+ 
a)^(1/3)+a^(2/3))+14*3^(1/2)/a^(1/3)*arctan(1/3*3^(1/2)*(2/a^(1/3)*(b*x+a) 
^(1/3)+1)))
 
3.5.19.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.73 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\left [\frac {42 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x}\right ) - 14 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} a^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 28 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} a^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (28 \, a b^{2} x^{2} + 7 \, a^{2} b x - 3 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{18 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, -\frac {14 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} a^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 28 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} a^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - \frac {84 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - 3 \, {\left (28 \, a b^{2} x^{2} + 7 \, a^{2} b x - 3 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{18 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]

input
integrate(1/x^3/(b*x+a)^(4/3),x, algorithm="fricas")
 
output
[1/18*(42*sqrt(1/3)*(a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-1/a^(2/3))*log((2*b*x 
+ 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*a^(2/3) - (b*x + a)^(1/3)*a - a^(4/3))*sq 
rt(-1/a^(2/3)) - 3*(b*x + a)^(1/3)*a^(2/3) + 3*a)/x) - 14*(b^3*x^3 + a*b^2 
*x^2)*a^(2/3)*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3)) + 2 
8*(b^3*x^3 + a*b^2*x^2)*a^(2/3)*log((b*x + a)^(1/3) - a^(1/3)) + 3*(28*a*b 
^2*x^2 + 7*a^2*b*x - 3*a^3)*(b*x + a)^(2/3))/(a^4*b*x^3 + a^5*x^2), -1/18* 
(14*(b^3*x^3 + a*b^2*x^2)*a^(2/3)*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^ 
(1/3) + a^(2/3)) - 28*(b^3*x^3 + a*b^2*x^2)*a^(2/3)*log((b*x + a)^(1/3) - 
a^(1/3)) - 84*sqrt(1/3)*(a*b^3*x^3 + a^2*b^2*x^2)*arctan(sqrt(1/3)*(2*(b*x 
 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(1/3) - 3*(28*a*b^2*x^2 + 7*a^2*b*x - 3* 
a^3)*(b*x + a)^(2/3))/(a^4*b*x^3 + a^5*x^2)]
 
3.5.19.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.53 (sec) , antiderivative size = 2793, normalized size of antiderivative = 18.74 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\text {Too large to display} \]

input
integrate(1/x**3/(b*x+a)**(4/3),x)
 
output
54*a**(13/3)*b**(5/3)*exp(2*I*pi/3)*gamma(-1/3)/(-54*a**(22/3)*(a/b + x)** 
(1/3)*exp(2*I*pi/3)*gamma(2/3) + 162*a**(19/3)*b*(a/b + x)**(4/3)*exp(2*I* 
pi/3)*gamma(2/3) - 162*a**(16/3)*b**2*(a/b + x)**(7/3)*exp(2*I*pi/3)*gamma 
(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) - 20 
1*a**(10/3)*b**(8/3)*(a/b + x)*exp(2*I*pi/3)*gamma(-1/3)/(-54*a**(22/3)*(a 
/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 162*a**(19/3)*b*(a/b + x)**(4/3) 
*exp(2*I*pi/3)*gamma(2/3) - 162*a**(16/3)*b**2*(a/b + x)**(7/3)*exp(2*I*pi 
/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2 
/3)) + 231*a**(7/3)*b**(11/3)*(a/b + x)**2*exp(2*I*pi/3)*gamma(-1/3)/(-54* 
a**(22/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 162*a**(19/3)*b*(a/b 
 + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3) - 162*a**(16/3)*b**2*(a/b + x)**(7/3 
)*exp(2*I*pi/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*p 
i/3)*gamma(2/3)) - 84*a**(4/3)*b**(14/3)*(a/b + x)**3*exp(2*I*pi/3)*gamma( 
-1/3)/(-54*a**(22/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 162*a**(1 
9/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3) - 162*a**(16/3)*b**2*(a/b 
 + x)**(7/3)*exp(2*I*pi/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3 
)*exp(2*I*pi/3)*gamma(2/3)) + 28*a**4*b**2*(a/b + x)**(1/3)*exp(2*I*pi/3)* 
log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(-1/3)/(-54*a**(22/3)*(a/ 
b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 162*a**(19/3)*b*(a/b + x)**(4/3)* 
exp(2*I*pi/3)*gamma(2/3) - 162*a**(16/3)*b**2*(a/b + x)**(7/3)*exp(2*I*...
 
3.5.19.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {14 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {10}{3}}} + \frac {28 \, {\left (b x + a\right )}^{2} b^{2} - 49 \, {\left (b x + a\right )} a b^{2} + 18 \, a^{2} b^{2}}{6 \, {\left ({\left (b x + a\right )}^{\frac {7}{3}} a^{3} - 2 \, {\left (b x + a\right )}^{\frac {4}{3}} a^{4} + {\left (b x + a\right )}^{\frac {1}{3}} a^{5}\right )}} - \frac {7 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {10}{3}}} + \frac {14 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {10}{3}}} \]

input
integrate(1/x^3/(b*x+a)^(4/3),x, algorithm="maxima")
 
output
14/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3)) 
/a^(10/3) + 1/6*(28*(b*x + a)^2*b^2 - 49*(b*x + a)*a*b^2 + 18*a^2*b^2)/((b 
*x + a)^(7/3)*a^3 - 2*(b*x + a)^(4/3)*a^4 + (b*x + a)^(1/3)*a^5) - 7/9*b^2 
*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(10/3) + 14/9* 
b^2*log((b*x + a)^(1/3) - a^(1/3))/a^(10/3)
 
3.5.19.8 Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {14 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {10}{3}}} - \frac {7 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {10}{3}}} + \frac {14 \, b^{2} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{9 \, a^{\frac {10}{3}}} + \frac {3 \, b^{2}}{{\left (b x + a\right )}^{\frac {1}{3}} a^{3}} + \frac {10 \, {\left (b x + a\right )}^{\frac {5}{3}} b^{2} - 13 \, {\left (b x + a\right )}^{\frac {2}{3}} a b^{2}}{6 \, a^{3} b^{2} x^{2}} \]

input
integrate(1/x^3/(b*x+a)^(4/3),x, algorithm="giac")
 
output
14/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3)) 
/a^(10/3) - 7/9*b^2*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3 
))/a^(10/3) + 14/9*b^2*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(10/3) + 3*b^ 
2/((b*x + a)^(1/3)*a^3) + 1/6*(10*(b*x + a)^(5/3)*b^2 - 13*(b*x + a)^(2/3) 
*a*b^2)/(a^3*b^2*x^2)
 
3.5.19.9 Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {\frac {3\,b^2}{a}+\frac {14\,b^2\,{\left (a+b\,x\right )}^2}{3\,a^3}-\frac {49\,b^2\,\left (a+b\,x\right )}{6\,a^2}}{{\left (a+b\,x\right )}^{7/3}-2\,a\,{\left (a+b\,x\right )}^{4/3}+a^2\,{\left (a+b\,x\right )}^{1/3}}+\frac {\ln \left (588\,a^3\,b^4\,{\left (a+b\,x\right )}^{1/3}-3\,a^{10/3}\,{\left (-7\,b^2+\sqrt {3}\,b^2\,7{}\mathrm {i}\right )}^2\right )\,\left (-7\,b^2+\sqrt {3}\,b^2\,7{}\mathrm {i}\right )}{9\,a^{10/3}}-\frac {\ln \left (588\,a^3\,b^4\,{\left (a+b\,x\right )}^{1/3}-3\,a^{10/3}\,{\left (7\,b^2+\sqrt {3}\,b^2\,7{}\mathrm {i}\right )}^2\right )\,\left (7\,b^2+\sqrt {3}\,b^2\,7{}\mathrm {i}\right )}{9\,a^{10/3}}+\frac {14\,b^2\,\ln \left (588\,a^3\,b^4\,{\left (a+b\,x\right )}^{1/3}-588\,a^{10/3}\,b^4\right )}{9\,a^{10/3}} \]

input
int(1/(x^3*(a + b*x)^(4/3)),x)
 
output
((3*b^2)/a + (14*b^2*(a + b*x)^2)/(3*a^3) - (49*b^2*(a + b*x))/(6*a^2))/(( 
a + b*x)^(7/3) - 2*a*(a + b*x)^(4/3) + a^2*(a + b*x)^(1/3)) + (log(588*a^3 
*b^4*(a + b*x)^(1/3) - 3*a^(10/3)*(3^(1/2)*b^2*7i - 7*b^2)^2)*(3^(1/2)*b^2 
*7i - 7*b^2))/(9*a^(10/3)) - (log(588*a^3*b^4*(a + b*x)^(1/3) - 3*a^(10/3) 
*(3^(1/2)*b^2*7i + 7*b^2)^2)*(3^(1/2)*b^2*7i + 7*b^2))/(9*a^(10/3)) + (14* 
b^2*log(588*a^3*b^4*(a + b*x)^(1/3) - 588*a^(10/3)*b^4))/(9*a^(10/3))
 
3.5.19.10 Reduce [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.69 \[ \int \frac {1}{x^3 (a+b x)^{4/3}} \, dx=\frac {-28 \left (b x +a \right )^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b^{2} x^{2}+28 \left (b x +a \right )^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b^{2} x^{2}+28 \left (b x +a \right )^{\frac {1}{3}} \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}\right ) b^{2} x^{2}+28 \left (b x +a \right )^{\frac {1}{3}} \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}\right ) b^{2} x^{2}-14 \left (b x +a \right )^{\frac {1}{3}} \mathrm {log}\left (-a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b^{2} x^{2}-14 \left (b x +a \right )^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b^{2} x^{2}-9 a^{\frac {7}{3}}+21 a^{\frac {4}{3}} b x +84 a^{\frac {1}{3}} b^{2} x^{2}}{18 a^{\frac {10}{3}} \left (b x +a \right )^{\frac {1}{3}} x^{2}} \]

input
int(1/((a + b*x)**(1/3)*x**3*(a + b*x)),x)
 
output
( - 28*(a + b*x)**(1/3)*sqrt(3)*atan((2*(a + b*x)**(1/6) + a**(1/6))/(a**( 
1/6)*sqrt(3)))*b**2*x**2 + 28*(a + b*x)**(1/3)*sqrt(3)*atan((2*(a + b*x)** 
(1/6) - a**(1/6))/(a**(1/6)*sqrt(3)))*b**2*x**2 + 28*(a + b*x)**(1/3)*log( 
(a + b*x)**(1/6) + a**(1/6))*b**2*x**2 + 28*(a + b*x)**(1/3)*log((a + b*x) 
**(1/6) - a**(1/6))*b**2*x**2 - 14*(a + b*x)**(1/3)*log( - a**(1/6)*(a + b 
*x)**(1/6) + (a + b*x)**(1/3) + a**(1/3))*b**2*x**2 - 14*(a + b*x)**(1/3)* 
log(a**(1/6)*(a + b*x)**(1/6) + (a + b*x)**(1/3) + a**(1/3))*b**2*x**2 - 9 
*a**(1/3)*a**2 + 21*a**(1/3)*a*b*x + 84*a**(1/3)*b**2*x**2)/(18*a**(1/3)*( 
a + b*x)**(1/3)*a**3*x**2)