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

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

[Out]

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)

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Rubi [A]
time = 0.04, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {44, 53, 57, 631, 210, 31} \begin {gather*} \frac {14 b^2 \text {ArcTan}\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^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}{3 a^3 \sqrt [3]{a+b x}}+\frac {7 b}{6 a^2 x \sqrt [3]{a+b x}}-\frac {1}{2 a x^2 \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(4/3)),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))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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_))^(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 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*} \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 ) \text {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 ) \text {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 ) \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^{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*}

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Mathematica [A]
time = 0.28, size = 142, normalized size = 0.95 \begin {gather*} \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 \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(4/3)),x]

[Out]

((3*a^(1/3)*(-3*a^2 + 7*a*b*x + 28*b^2*x^2))/(x^2*(a + b*x)^(1/3)) + 28*Sqrt[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))

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Maple [A]
time = 0.13, size = 126, normalized size = 0.85

method	result	size
risch	\(-\frac {\left (b x +a \right )^{\frac {2}{3}} \left (-10 b x +3 a \right )}{6 a^{3} x^{2}}+\frac {3 b^{2}}{a^{3} \left (b x +a \right )^{\frac {1}{3}}}+\frac {14 b^{2} \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{9 a^{\frac {10}{3}}}-\frac {7 b^{2} \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{9 a^{\frac {10}{3}}}+\frac {14 b^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{9 a^{\frac {10}{3}}}\)	\(124\)
derivativedivides	\(3 b^{2} \left (\frac {1}{a^{3} \left (b x +a \right )^{\frac {1}{3}}}-\frac {\frac {-\frac {5 \left (b x +a \right )^{\frac {5}{3}}}{9}+\frac {13 a \left (b x +a \right )^{\frac {2}{3}}}{18}}{b^{2} x^{2}}-\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {1}{3}}}+\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {1}{3}}}-\frac {14 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {1}{3}}}}{a^{3}}\right )\)	\(126\)
default	\(3 b^{2} \left (\frac {1}{a^{3} \left (b x +a \right )^{\frac {1}{3}}}-\frac {\frac {-\frac {5 \left (b x +a \right )^{\frac {5}{3}}}{9}+\frac {13 a \left (b x +a \right )^{\frac {2}{3}}}{18}}{b^{2} x^{2}}-\frac {14 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{27 a^{\frac {1}{3}}}+\frac {7 \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{27 a^{\frac {1}{3}}}-\frac {14 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{27 a^{\frac {1}{3}}}}{a^{3}}\right )\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b^2*(1/a^3/(b*x+a)^(1/3)-1/a^3*((-5/9*(b*x+a)^(5/3)+13/18*a*(b*x+a)^(2/3))/b^2/x^2-14/27/a^(1/3)*ln((b*x+a)^
(1/3)-a^(1/3))+7/27/a^(1/3)*ln((b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))-14/27*3^(1/2)/a^(1/3)*arctan(1/3*3
^(1/2)*(2/a^(1/3)*(b*x+a)^(1/3)+1))))

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Maxima [A]
time = 0.49, size = 158, normalized size = 1.06 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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) + 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)

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Fricas [A]
time = 1.25, size = 407, normalized size = 2.73 \begin {gather*} \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 {gather*}

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] Result contains complex when optimal does not.
time = 2.99, size = 2793, normalized size = 18.74 \begin {gather*} \text {Too large to display} \end {gather*}

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*...

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Giac [A]
time = 0.81, size = 140, normalized size = 0.94 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.13, size = 221, normalized size = 1.48 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(4/3)),x)

[Out]

((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))

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