\(\int \frac {(a+b x)^{4/3}}{x^3} \, dx\) [391]
Optimal result
Integrand size = 13, antiderivative size = 124 \[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=-\frac {2 b \sqrt [3]{a+b x}}{3 x}-\frac {(a+b x)^{4/3}}{2 x^2}-\frac {2 b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b^2 \log (x)}{9 a^{2/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}}
\]
[Out]
-2/3*b*(b*x+a)^(1/3)/x-1/2*(b*x+a)^(4/3)/x^2-1/9*b^2*ln(x)/a^(2/3)+1/3*b^2*ln(a^(1/3)-(b*x+a)^(1/3))/a^(2/3)-2
/9*b^2*arctan(1/3*(a^(1/3)+2*(b*x+a)^(1/3))/a^(1/3)*3^(1/2))/a^(2/3)*3^(1/2)
Rubi [A] (verified)
Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {43, 59, 631, 210, 31}
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=-\frac {2 b^2 \arctan \left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b^2 \log (x)}{9 a^{2/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}}-\frac {(a+b x)^{4/3}}{2 x^2}-\frac {2 b \sqrt [3]{a+b x}}{3 x}
\]
[In]
Int[(a + b*x)^(4/3)/x^3,x]
[Out]
(-2*b*(a + b*x)^(1/3))/(3*x) - (a + b*x)^(4/3)/(2*x^2) - (2*b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*
a^(1/3))])/(3*Sqrt[3]*a^(2/3)) - (b^2*Log[x])/(9*a^(2/3)) + (b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(3*a^(2/3))
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && GtQ[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 631
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]
Rubi steps \begin{align*}
\text {integral}& = -\frac {(a+b x)^{4/3}}{2 x^2}+\frac {1}{3} (2 b) \int \frac {\sqrt [3]{a+b x}}{x^2} \, dx \\ & = -\frac {2 b \sqrt [3]{a+b x}}{3 x}-\frac {(a+b x)^{4/3}}{2 x^2}+\frac {1}{9} \left (2 b^2\right ) \int \frac {1}{x (a+b x)^{2/3}} \, dx \\ & = -\frac {2 b \sqrt [3]{a+b x}}{3 x}-\frac {(a+b x)^{4/3}}{2 x^2}-\frac {b^2 \log (x)}{9 a^{2/3}}-\frac {b^2 \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{3 a^{2/3}}-\frac {b^2 \text {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{3 \sqrt [3]{a}} \\ & = -\frac {2 b \sqrt [3]{a+b x}}{3 x}-\frac {(a+b x)^{4/3}}{2 x^2}-\frac {b^2 \log (x)}{9 a^{2/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}\right )}{3 a^{2/3}} \\ & = -\frac {2 b \sqrt [3]{a+b x}}{3 x}-\frac {(a+b x)^{4/3}}{2 x^2}-\frac {2 b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b^2 \log (x)}{9 a^{2/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=\frac {1}{18} \left (-\frac {3 \sqrt [3]{a+b x} (3 a+7 b x)}{x^2}-\frac {4 \sqrt {3} b^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {4 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{2/3}}-\frac {2 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{a^{2/3}}\right )
\]
[In]
Integrate[(a + b*x)^(4/3)/x^3,x]
[Out]
((-3*(a + b*x)^(1/3)*(3*a + 7*b*x))/x^2 - (4*Sqrt[3]*b^2*ArcTan[(1 + (2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]])/a^
(2/3) + (4*b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/a^(2/3) - (2*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x)^(1/3) + (a + b
*x)^(2/3)])/a^(2/3))/18
Maple [A] (verified)
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89
| | |
| method | result | size |
| | |
| derivativedivides |
\(3 b^{2} \left (-\frac {\frac {7 \left (b x +a \right )^{\frac {4}{3}}}{18}-\frac {2 a \left (b x +a \right )^{\frac {1}{3}}}{9}}{b^{2} x^{2}}+\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {2}{3}}}-\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {2}{3}}}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {2}{3}}}\right )\) |
\(110\) |
| default |
\(3 b^{2} \left (-\frac {\frac {7 \left (b x +a \right )^{\frac {4}{3}}}{18}-\frac {2 a \left (b x +a \right )^{\frac {1}{3}}}{9}}{b^{2} x^{2}}+\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {2}{3}}}-\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {2}{3}}}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {2}{3}}}\right )\) |
\(110\) |
| pseudoelliptic |
\(\frac {-9 \left (b x +a \right )^{\frac {1}{3}} a^{\frac {5}{3}}-4 b^{2} \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b x +a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, x^{2}-21 b x \left (b x +a \right )^{\frac {1}{3}} a^{\frac {2}{3}}+4 b^{2} \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{2}-2 b^{2} \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) x^{2}}{18 x^{2} a^{\frac {2}{3}}}\) |
\(122\) |
| | |
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[In]
int((b*x+a)^(4/3)/x^3,x,method=_RETURNVERBOSE)
[Out]
3*b^2*(-(7/18*(b*x+a)^(4/3)-2/9*a*(b*x+a)^(1/3))/b^2/x^2+2/27/a^(2/3)*ln((b*x+a)^(1/3)-a^(1/3))-1/27/a^(2/3)*l
n((b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))-2/27/a^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/a^(1/3)*(b*x+a)^(1/3
)+1)))
Fricas [A] (verification not implemented)
none
Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.31
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=-\frac {4 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{2} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right )}}{3 \, a^{2}}\right ) + 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right ) - 4 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) + 3 \, {\left (7 \, a^{2} b x + 3 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {1}{3}}}{18 \, a^{2} x^{2}}
\]
[In]
integrate((b*x+a)^(4/3)/x^3,x, algorithm="fricas")
[Out]
-1/18*(4*sqrt(3)*(a^2)^(1/6)*a*b^2*x^2*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(a^2)^(2/3)*(
b*x + a)^(1/3))/a^2) + 2*(a^2)^(2/3)*b^2*x^2*log((b*x + a)^(2/3)*a + (a^2)^(1/3)*a + (a^2)^(2/3)*(b*x + a)^(1/
3)) - 4*(a^2)^(2/3)*b^2*x^2*log((b*x + a)^(1/3)*a - (a^2)^(2/3)) + 3*(7*a^2*b*x + 3*a^3)*(b*x + a)^(1/3))/(a^2
*x^2)
Sympy [C] (verification not implemented)
Result contains complex when optimal does not.
Time = 1.91 (sec) , antiderivative size = 2266, normalized size of antiderivative = 18.27
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=\text {Too large to display}
\]
[In]
integrate((b*x+a)**(4/3)/x**3,x)
[Out]
28*a**(19/3)*b**2*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*
gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(
10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) + 28*a**(19/3)*b**2*log(1 - b**(1/3)*(a/b + x)**(
1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*
I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I
*pi/3)*gamma(10/3)) + 28*a**(19/3)*b**2*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a
**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*
a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) - 84*a
**(16/3)*b**3*(a/b + x)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*
pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*
gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) - 84*a**(16/3)*b**3*(a/b + x)*log(1 - b**(1
/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*
(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b
+ x)**3*exp(2*I*pi/3)*gamma(10/3)) - 84*a**(16/3)*b**3*(a/b + x)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(
1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*
I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I
*pi/3)*gamma(10/3)) + 84*a**(13/3)*b**4*(a/b + x)**2*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))
*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**
2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) + 84*a**(13/3)
*b**4*(a/b + x)**2*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I
*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)
*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) + 84*a**(13/3)*b**4*(a/b + x)**2*exp(-2*I*
pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(
10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3)
- 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) - 28*a**(10/3)*b**5*(a/b + x)**3*exp(2*I*pi/3)*log(1 -
b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2
*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*
I*pi/3)*gamma(10/3)) - 28*a**(10/3)*b**5*(a/b + x)**3*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a*
*(1/3))*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a
**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) - 28*a*
*(10/3)*b**5*(a/b + x)**3*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma
(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b
+ x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) + 84*a**6*b**(7/3)*(
a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/
3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3
)*gamma(10/3)) - 231*a**5*b**(10/3)*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/
3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(10/3) - 5
4*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3)) + 147*a**4*b**(13/3)*(a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(
7/3)/(54*a**7*exp(2*I*pi/3)*gamma(10/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3) + 162*a**5*b**2*(a/b
+ x)**2*exp(2*I*pi/3)*gamma(10/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(10/3))
Maxima [A] (verification not implemented)
none
Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=-\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {b^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {2}{3}}} + \frac {2 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {7 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{2} - 4 \, {\left (b x + a\right )}^{\frac {1}{3}} a b^{2}}{6 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + a^{2}\right )}}
\]
[In]
integrate((b*x+a)^(4/3)/x^3,x, algorithm="maxima")
[Out]
-2/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(2/3) - 1/9*b^2*log((b*x + a)^(2/
3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) + 2/9*b^2*log((b*x + a)^(1/3) - a^(1/3))/a^(2/3) - 1/6*(7*(b*x
+ a)^(4/3)*b^2 - 4*(b*x + a)^(1/3)*a*b^2)/((b*x + a)^2 - 2*(b*x + a)*a + a^2)
Giac [A] (verification not implemented)
none
Time = 0.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=-\frac {\frac {4 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {2 \, b^{3} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {4 \, b^{3} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {2}{3}}} + \frac {3 \, {\left (7 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{3} - 4 \, {\left (b x + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{b^{2} x^{2}}}{18 \, b}
\]
[In]
integrate((b*x+a)^(4/3)/x^3,x, algorithm="giac")
[Out]
-1/18*(4*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(2/3) + 2*b^3*log((b*x + a)^(
2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) - 4*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(2/3) + 3*(7*(
b*x + a)^(4/3)*b^3 - 4*(b*x + a)^(1/3)*a*b^3)/(b^2*x^2))/b
Mupad [B] (verification not implemented)
Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.40
\[
\int \frac {(a+b x)^{4/3}}{x^3} \, dx=\frac {2\,b^2\,\ln \left (2\,b^2\,{\left (a+b\,x\right )}^{1/3}-2\,a^{1/3}\,b^2\right )}{9\,a^{2/3}}-\frac {\frac {7\,b^2\,{\left (a+b\,x\right )}^{4/3}}{6}-\frac {2\,a\,b^2\,{\left (a+b\,x\right )}^{1/3}}{3}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}-\frac {\ln \left (2\,b^2\,{\left (a+b\,x\right )}^{1/3}+a^{1/3}\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{9\,a^{2/3}}+\frac {b^2\,\ln \left (2\,b^2\,{\left (a+b\,x\right )}^{1/3}-9\,a^{1/3}\,b^2\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}{a^{2/3}}
\]
[In]
int((a + b*x)^(4/3)/x^3,x)
[Out]
(2*b^2*log(2*b^2*(a + b*x)^(1/3) - 2*a^(1/3)*b^2))/(9*a^(2/3)) - ((7*b^2*(a + b*x)^(4/3))/6 - (2*a*b^2*(a + b*
x)^(1/3))/3)/((a + b*x)^2 - 2*a*(a + b*x) + a^2) - (log(2*b^2*(a + b*x)^(1/3) + a^(1/3)*(3^(1/2)*b^2*1i + b^2)
)*(3^(1/2)*b^2*1i + b^2))/(9*a^(2/3)) + (b^2*log(2*b^2*(a + b*x)^(1/3) - 9*a^(1/3)*b^2*((3^(1/2)*1i)/9 - 1/9))
*((3^(1/2)*1i)/9 - 1/9))/a^(2/3)