3.377 \(\int \frac{\sqrt [3]{a+b x}}{x^3} \, dx\)
Optimal. Leaf size=127 \[ \frac{b^2 \log (x)}{18 a^{5/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{5/3}}+\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3}}-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x} \]
[Out]
-(a + b*x)^(1/3)/(2*x^2) - (b*(a + b*x)^(1/3))/(6*a*x) + (b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^
(1/3))])/(3*Sqrt[3]*a^(5/3)) + (b^2*Log[x])/(18*a^(5/3)) - (b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(6*a^(5/3))
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Rubi [A] time = 0.0494358, antiderivative size = 127, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used =
{47, 51, 57, 617, 204, 31} \[ \frac{b^2 \log (x)}{18 a^{5/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{5/3}}+\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3}}-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x)^(1/3)/x^3,x]
[Out]
-(a + b*x)^(1/3)/(2*x^2) - (b*(a + b*x)^(1/3))/(6*a*x) + (b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^
(1/3))])/(3*Sqrt[3]*a^(5/3)) + (b^2*Log[x])/(18*a^(5/3)) - (b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(6*a^(5/3))
Rule 47
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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]
Rule 51
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] && 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 57
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 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 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \frac{\sqrt [3]{a+b x}}{x^3} \, dx &=-\frac{\sqrt [3]{a+b x}}{2 x^2}+\frac{1}{6} b \int \frac{1}{x^2 (a+b x)^{2/3}} \, dx\\ &=-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x}-\frac{b^2 \int \frac{1}{x (a+b x)^{2/3}} \, dx}{9 a}\\ &=-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x}+\frac{b^2 \log (x)}{18 a^{5/3}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a^{5/3}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a^{4/3}}\\ &=-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x}+\frac{b^2 \log (x)}{18 a^{5/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{5/3}}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}\right )}{3 a^{5/3}}\\ &=-\frac{\sqrt [3]{a+b x}}{2 x^2}-\frac{b \sqrt [3]{a+b x}}{6 a x}+\frac{b^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{5/3}}+\frac{b^2 \log (x)}{18 a^{5/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{5/3}}\\ \end{align*}
Mathematica [C] time = 0.0191414, size = 35, normalized size = 0.28 \[ -\frac{3 b^2 (a+b x)^{4/3} \, _2F_1\left (\frac{4}{3},3;\frac{7}{3};\frac{b x}{a}+1\right )}{4 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)^(1/3)/x^3,x]
[Out]
(-3*b^2*(a + b*x)^(4/3)*Hypergeometric2F1[4/3, 3, 7/3, 1 + (b*x)/a])/(4*a^3)
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Maple [A] time = 0.01, size = 113, normalized size = 0.9 \begin{align*} -{\frac{1}{6\,a{x}^{2}} \left ( bx+a \right ) ^{{\frac{4}{3}}}}-{\frac{1}{3\,{x}^{2}}\sqrt [3]{bx+a}}-{\frac{{b}^{2}}{9}\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){a}^{-{\frac{5}{3}}}}+{\frac{{b}^{2}}{18}\ln \left ( \left ( bx+a \right ) ^{{\frac{2}{3}}}+\sqrt [3]{a}\sqrt [3]{bx+a}+{a}^{{\frac{2}{3}}} \right ){a}^{-{\frac{5}{3}}}}+{\frac{{b}^{2}\sqrt{3}}{9}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt [3]{bx+a}}{\sqrt [3]{a}}}+1 \right ) } \right ){a}^{-{\frac{5}{3}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(1/3)/x^3,x)
[Out]
-1/6/x^2/a*(b*x+a)^(4/3)-1/3*(b*x+a)^(1/3)/x^2-1/9*b^2/a^(5/3)*ln((b*x+a)^(1/3)-a^(1/3))+1/18*b^2/a^(5/3)*ln((
b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))+1/9*b^2/a^(5/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/a^(1/3)*(b*x+a)^(1/3
)+1))
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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+a)^(1/3)/x^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.89489, size = 481, normalized size = 3.79 \begin{align*} \frac{2 \, \sqrt{3} a b^{2} x^{2} \sqrt{-\left (-a^{2}\right )^{\frac{1}{3}}} \arctan \left (-\frac{{\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 )} \sqrt{-\left (-a^{2}\right )^{\frac{1}{3}}}}{3 \, a^{2}}\right ) + \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 ) - 2 \, \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 (a^{2} b x + 3 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{1}{3}}}{18 \, a^{3} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/3)/x^3,x, algorithm="fricas")
[Out]
1/18*(2*sqrt(3)*a*b^2*x^2*sqrt(-(-a^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-a^2)^(1/3)*a - 2*sqrt(3)*(-a^2)^(2/3)*(b*
x + a)^(1/3))*sqrt(-(-a^2)^(1/3))/a^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)) - 2*(-a^2)^(2/3)*b^2*x^2*log((b*x + a)^(1/3)*a - (-a^2)^(2/3)) - 3*(a^2*b*x + 3*a^3)*(b
*x + a)^(1/3))/(a^3*x^2)
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Sympy [C] time = 3.92463, size = 2266, normalized size = 17.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(1/3)/x**3,x)
[Out]
-4*a**(16/3)*b**2*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*
gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3)
- 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) - 4*a**(16/3)*b**2*log(1 - b**(1/3)*(a/b + x)**(1/3)*ex
p_polar(2*I*pi/3)/a**(1/3))*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*g
amma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma
(7/3)) - 4*a**(16/3)*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
(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x
)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) + 12*a**(13/3)*b**3*(a/b +
x)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 8
1*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**
3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) + 12*a**(13/3)*b**3*(a/b + x)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_p
olar(2*I*pi/3)/a**(1/3))*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamm
a(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/
3)) + 12*a**(13/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*
(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) - 12*a**(10/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gam
ma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) -
27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) - 12*a**(10/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(
2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*
pi/3)*gamma(7/3)) - 12*a**(10/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3
) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) +
4*a**(7/3)*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(4/3)/(27*a**7*ex
p(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3
)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) + 4*a**(7/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(
a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x
)**3*exp(2*I*pi/3)*gamma(7/3)) + 4*a**(7/3)*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(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3
)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*ga
mma(7/3)) - 12*a**5*b**(7/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*
a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*
(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) + 6*a**4*b**(10/3)*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(4/3)/(27*a**7*e
xp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*b**2*(a/b + x)**2*exp(2*I*pi/
3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3)) + 6*a**3*b**(13/3)*(a/b + x)**(7/3)*exp(2*
I*pi/3)*gamma(4/3)/(27*a**7*exp(2*I*pi/3)*gamma(7/3) - 81*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(7/3) + 81*a**5*
b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(7/3) - 27*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(7/3))
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Giac [A] time = 1.98182, size = 173, normalized size = 1.36 \begin{align*} \frac{\frac{2 \, \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{5}{3}}} + \frac{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{5}{3}}} - \frac{2 \, b^{3} \log \left ({\left |{\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{5}{3}}} - \frac{3 \,{\left ({\left (b x + a\right )}^{\frac{4}{3}} b^{3} + 2 \,{\left (b x + a\right )}^{\frac{1}{3}} a b^{3}\right )}}{a b^{2} x^{2}}}{18 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/3)/x^3,x, algorithm="giac")
[Out]
1/18*(2*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) + b^3*log((b*x + a)^(2/3
) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*((b*x +
a)^(4/3)*b^3 + 2*(b*x + a)^(1/3)*a*b^3)/(a*b^2*x^2))/b