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

Optimal. Leaf size=149 \[ \frac{14 b^2}{3 a^3 \sqrt [3]{a+b x}}-\frac{7 b^2 \log (x)}{9 a^{10/3}}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{10/3}}+\frac{14 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{10/3}}+\frac{7 b}{6 a^2 x \sqrt [3]{a+b x}}-\frac{1}{2 a x^2 \sqrt [3]{a+b x}} \]

[Out]

(14*b^2)/(3*a^3*(a + b*x)^(1/3)) - 1/(2*a*x^2*(a + b*x)^(1/3)) + (7*b)/(6*a^2*x*(a + b*x)^(1/3)) + (14*b^2*Arc
Tan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(10/3)) - (7*b^2*Log[x])/(9*a^(10/3)) + (7*
b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(3*a^(10/3))

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Rubi [A]  time = 0.0579336, antiderivative size = 147, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {51, 55, 617, 204, 31} \[ -\frac{7 b^2 \log (x)}{9 a^{10/3}}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{10/3}}+\frac{14 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{10/3}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}+\frac{14 b (a+b x)^{2/3}}{3 a^3 x}+\frac{3}{a x^2 \sqrt [3]{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(4/3)),x]

[Out]

3/(a*x^2*(a + b*x)^(1/3)) - (7*(a + b*x)^(2/3))/(2*a^2*x^2) + (14*b*(a + b*x)^(2/3))/(3*a^3*x) + (14*b^2*ArcTa
n[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(10/3)) - (7*b^2*Log[x])/(9*a^(10/3)) + (7*b^
2*Log[a^(1/3) - (a + b*x)^(1/3)])/(3*a^(10/3))

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 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[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 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{1}{x^3 (a+b x)^{4/3}} \, dx &=\frac{3}{a x^2 \sqrt [3]{a+b x}}+\frac{7 \int \frac{1}{x^3 \sqrt [3]{a+b x}} \, dx}{a}\\ &=\frac{3}{a x^2 \sqrt [3]{a+b x}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}-\frac{(14 b) \int \frac{1}{x^2 \sqrt [3]{a+b x}} \, dx}{3 a^2}\\ &=\frac{3}{a x^2 \sqrt [3]{a+b x}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}+\frac{14 b (a+b x)^{2/3}}{3 a^3 x}+\frac{\left (14 b^2\right ) \int \frac{1}{x \sqrt [3]{a+b x}} \, dx}{9 a^3}\\ &=\frac{3}{a x^2 \sqrt [3]{a+b x}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}+\frac{14 b (a+b x)^{2/3}}{3 a^3 x}-\frac{7 b^2 \log (x)}{9 a^{10/3}}-\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{3 a^{10/3}}+\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{3 a^3}\\ &=\frac{3}{a x^2 \sqrt [3]{a+b x}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}+\frac{14 b (a+b x)^{2/3}}{3 a^3 x}-\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}}-\frac{\left (14 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^{10/3}}\\ &=\frac{3}{a x^2 \sqrt [3]{a+b x}}-\frac{7 (a+b x)^{2/3}}{2 a^2 x^2}+\frac{14 b (a+b x)^{2/3}}{3 a^3 x}+\frac{14 b^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt{3}}\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}}\\ \end{align*}

Mathematica [C]  time = 0.0289249, size = 33, normalized size = 0.22 \[ \frac{3 b^2 \, _2F_1\left (-\frac{1}{3},3;\frac{2}{3};\frac{b x}{a}+1\right )}{a^3 \sqrt [3]{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(4/3)),x]

[Out]

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

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Maple [A]  time = 0.012, size = 131, normalized size = 0.9 \begin{align*} 3\,{\frac{{b}^{2}}{{a}^{3}\sqrt [3]{bx+a}}}+{\frac{5}{3\,{a}^{3}{x}^{2}} \left ( bx+a \right ) ^{{\frac{5}{3}}}}-{\frac{13}{6\,{a}^{2}{x}^{2}} \left ( bx+a \right ) ^{{\frac{2}{3}}}}+{\frac{14\,{b}^{2}}{9}\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){a}^{-{\frac{10}{3}}}}-{\frac{7\,{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{10}{3}}}}+{\frac{14\,{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{10}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x+a)^(4/3),x)

[Out]

3*b^2/a^3/(b*x+a)^(1/3)+5/3/a^3/x^2*(b*x+a)^(5/3)-13/6/a^2/x^2*(b*x+a)^(2/3)+14/9*b^2/a^(10/3)*ln((b*x+a)^(1/3
)-a^(1/3))-7/9*b^2/a^(10/3)*ln((b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))+14/9*b^2/a^(10/3)*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(1/x^3/(b*x+a)^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70796, size = 1064, normalized size = 7.14 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(4/3),x, algorithm="fricas")

[Out]

[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))*sqrt(-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)) + 28*(b^3*x^3 + a*b^2*x^2)*a^(2/3)*lo
g((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)]

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Sympy [C]  time = 4.41727, size = 2793, normalized size = 18.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x+a)**(4/3),x)

[Out]

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)*gam
ma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) - 201*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)*gam
ma(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)*e
xp(2*I*pi/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**(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)
) + 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*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)*exp_polar(2
*I*pi/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*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)*log(1 - b**(1/3)*(a/b +
x)**(1/3)*exp_polar(4*I*pi/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*pi/3
)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) - 84*a**3*b**3*(a/b + x)**(4/3)*e
xp(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*p
i/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)) - 84*a**3*b**3*(a
/b + x)**(4/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/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*pi/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*
I*pi/3)*gamma(2/3)) - 84*a**3*b**3*(a/b + x)**(4/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/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*pi/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a
/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) + 84*a**2*b**4*(a/b + x)**(7/3)*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*pi/3)*gamma(2/3) + 54*a
**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) + 84*a**2*b**4*(a/b + x)**(7/3)*exp(-2*I*pi/3)*log(1
- b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/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*pi/3)*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3)) + 84*a**2*b**4*
(a/b + x)**(7/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/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*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*b**5*(a/b + x)**(10/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(-1/3)/(-5
4*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)) - 28*a*b**5*(a/b + x)**(10/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(
2*I*pi/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*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*b**5*(a/b + x)**(10/3)*log(1 - b**(1/3)*(a/b +
x)**(1/3)*exp_polar(4*I*pi/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*pi/3)
*gamma(2/3) + 54*a**(13/3)*b**3*(a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(2/3))

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Giac [A]  time = 2.20288, size = 189, normalized size = 1.27 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(4/3),x, algorithm="giac")

[Out]

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)