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

Optimal. Leaf size=124 \[ -\frac{b^2 \log (x)}{9 a^{2/3}}+\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}}-\frac{2 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3}}-\frac{(a+b x)^{4/3}}{2 x^2}-\frac{2 b \sqrt [3]{a+b x}}{3 x} \]

[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))

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Rubi [A]  time = 0.0443007, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {47, 57, 617, 204, 31} \[ -\frac{b^2 \log (x)}{9 a^{2/3}}+\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{2/3}}-\frac{2 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3}}-\frac{(a+b x)^{4/3}}{2 x^2}-\frac{2 b \sqrt [3]{a+b x}}{3 x} \]

Antiderivative was successfully verified.

[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 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 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{(a+b x)^{4/3}}{x^3} \, dx &=-\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 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{3 a^{2/3}}-\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 )}{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 ) \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^{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 [C]  time = 0.0217566, size = 35, normalized size = 0.28 \[ -\frac{3 b^2 (a+b x)^{7/3} \, _2F_1\left (\frac{7}{3},3;\frac{10}{3};\frac{b x}{a}+1\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(4/3)/x^3,x]

[Out]

(-3*b^2*(a + b*x)^(7/3)*Hypergeometric2F1[7/3, 3, 10/3, 1 + (b*x)/a])/(7*a^3)

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Maple [A]  time = 0.009, size = 111, normalized size = 0.9 \begin{align*} -{\frac{7}{6\,{x}^{2}} \left ( bx+a \right ) ^{{\frac{4}{3}}}}+{\frac{2\,a}{3\,{x}^{2}}\sqrt [3]{bx+a}}+{\frac{2\,{b}^{2}}{9}\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){a}^{-{\frac{2}{3}}}}-{\frac{{b}^{2}}{9}\ln \left ( \left ( bx+a \right ) ^{{\frac{2}{3}}}+\sqrt [3]{a}\sqrt [3]{bx+a}+{a}^{{\frac{2}{3}}} \right ){a}^{-{\frac{2}{3}}}}-{\frac{2\,{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{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(4/3)/x^3,x)

[Out]

-7/6*(b*x+a)^(4/3)/x^2+2/3/x^2*a*(b*x+a)^(1/3)+2/9*b^2/a^(2/3)*ln((b*x+a)^(1/3)-a^(1/3))-1/9*b^2/a^(2/3)*ln((b
*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))-2/9*b^2/a^(2/3)*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)^(4/3)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6479, size = 455, normalized size = 3.67 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 4.11215, size = 2266, normalized size = 18.27 \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)**(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))

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Giac [A]  time = 1.88783, size = 171, normalized size = 1.38 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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