3.7.83 \(\int \frac {x^{4/3}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=125 \[ -\frac {2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{b^{7/3}}+\frac {2 \sqrt [3]{a} \log (a+b x)}{3 b^{7/3}}+\frac {4 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {x^{4/3}}{b (a+b x)}+\frac {4 \sqrt [3]{x}}{b^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 125, normalized size of antiderivative = 1.00, 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, 50, 58, 617, 204, 31} \begin {gather*} -\frac {2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{b^{7/3}}+\frac {2 \sqrt [3]{a} \log (a+b x)}{3 b^{7/3}}+\frac {4 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {x^{4/3}}{b (a+b x)}+\frac {4 \sqrt [3]{x}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(4/3)/(a + b*x)^2,x]

[Out]

(4*x^(1/3))/b^2 - x^(4/3)/(b*(a + b*x)) + (4*a^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/
(Sqrt[3]*b^(7/3)) - (2*a^(1/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/b^(7/3) + (2*a^(1/3)*Log[a + b*x])/(3*b^(7/3))

Rule 31

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[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
] && NegQ[(b*c - a*d)/b]

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 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]

Rubi steps

\begin {align*} \int \frac {x^{4/3}}{(a+b x)^2} \, dx &=-\frac {x^{4/3}}{b (a+b x)}+\frac {4 \int \frac {\sqrt [3]{x}}{a+b x} \, dx}{3 b}\\ &=\frac {4 \sqrt [3]{x}}{b^2}-\frac {x^{4/3}}{b (a+b x)}-\frac {(4 a) \int \frac {1}{x^{2/3} (a+b x)} \, dx}{3 b^2}\\ &=\frac {4 \sqrt [3]{x}}{b^2}-\frac {x^{4/3}}{b (a+b x)}+\frac {2 \sqrt [3]{a} \log (a+b x)}{3 b^{7/3}}-\frac {\left (2 a^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{x}\right )}{b^{8/3}}-\frac {\left (2 \sqrt [3]{a}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{b^{7/3}}\\ &=\frac {4 \sqrt [3]{x}}{b^2}-\frac {x^{4/3}}{b (a+b x)}-\frac {2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{b^{7/3}}+\frac {2 \sqrt [3]{a} \log (a+b x)}{3 b^{7/3}}-\frac {\left (4 \sqrt [3]{a}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}\right )}{b^{7/3}}\\ &=\frac {4 \sqrt [3]{x}}{b^2}-\frac {x^{4/3}}{b (a+b x)}+\frac {4 \sqrt [3]{a} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} b^{7/3}}-\frac {2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{b^{7/3}}+\frac {2 \sqrt [3]{a} \log (a+b x)}{3 b^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.22 \begin {gather*} \frac {3 x^{7/3} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};-\frac {b x}{a}\right )}{7 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(4/3)/(a + b*x)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 156, normalized size = 1.25 \begin {gather*} \frac {2 \sqrt [3]{a} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{3 b^{7/3}}-\frac {4 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 b^{7/3}}+\frac {4 \sqrt [3]{a} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}+\frac {4 a \sqrt [3]{x}+3 b x^{4/3}}{b^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(4/3)/(a + b*x)^2,x]

[Out]

(4*a*x^(1/3) + 3*b*x^(4/3))/(b^2*(a + b*x)) + (4*a^(1/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/
3))])/(Sqrt[3]*b^(7/3)) - (4*a^(1/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(3*b^(7/3)) + (2*a^(1/3)*Log[a^(2/3) - a^
(1/3)*b^(1/3)*x^(1/3) + b^(2/3)*x^(2/3)])/(3*b^(7/3))

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fricas [A]  time = 1.40, size = 147, normalized size = 1.18 \begin {gather*} \frac {4 \, \sqrt {3} {\left (b x + a\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - 2 \, {\left (b x + a\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) + 4 \, {\left (b x + a\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (3 \, b x + 4 \, a\right )} x^{\frac {1}{3}}}{3 \, {\left (b^{3} x + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)/(b*x+a)^2,x, algorithm="fricas")

[Out]

1/3*(4*sqrt(3)*(b*x + a)*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x^(1/3)*(-a/b)^(2/3) - sqrt(3)*a)/a) - 2*(b*x +
a)*(-a/b)^(1/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3)) + 4*(b*x + a)*(-a/b)^(1/3)*log(x^(1/3) - (-
a/b)^(1/3)) + 3*(3*b*x + 4*a)*x^(1/3))/(b^3*x + a*b^2)

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giac [A]  time = 1.07, size = 135, normalized size = 1.08 \begin {gather*} \frac {4 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, b^{2}} - \frac {4 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3}} + \frac {a x^{\frac {1}{3}}}{{\left (b x + a\right )} b^{2}} + \frac {3 \, x^{\frac {1}{3}}}{b^{2}} - \frac {2 \, \left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)/(b*x+a)^2,x, algorithm="giac")

[Out]

4/3*(-a/b)^(1/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/b^2 - 4/3*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x^(1/
3) + (-a/b)^(1/3))/(-a/b)^(1/3))/b^3 + a*x^(1/3)/((b*x + a)*b^2) + 3*x^(1/3)/b^2 - 2/3*(-a*b^2)^(1/3)*log(x^(2
/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/b^3

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maple [A]  time = 0.01, size = 123, normalized size = 0.98 \begin {gather*} \frac {a \,x^{\frac {1}{3}}}{\left (b x +a \right ) b^{2}}-\frac {4 \sqrt {3}\, a \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}-\frac {4 a \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {2 a \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {3 x^{\frac {1}{3}}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(4/3)/(b*x+a)^2,x)

[Out]

3*x^(1/3)/b^2+a/b^2*x^(1/3)/(b*x+a)-4/3*a/b^3/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))+2/3*a/b^3/(a/b)^(2/3)*ln(x^(
2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))-4/3*a/b^3/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x^(1/3)-
1))

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maxima [A]  time = 2.93, size = 133, normalized size = 1.06 \begin {gather*} \frac {a x^{\frac {1}{3}}}{b^{3} x + a b^{2}} - \frac {4 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {3 \, x^{\frac {1}{3}}}{b^{2}} + \frac {2 \, a \log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {4 \, a \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)/(b*x+a)^2,x, algorithm="maxima")

[Out]

a*x^(1/3)/(b^3*x + a*b^2) - 4/3*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x^(1/3) - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)
^(2/3)) + 3*x^(1/3)/b^2 + 2/3*a*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) - 4/3*a*log
(x^(1/3) + (a/b)^(1/3))/(b^3*(a/b)^(2/3))

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mupad [B]  time = 0.15, size = 142, normalized size = 1.14 \begin {gather*} \frac {3\,x^{1/3}}{b^2}+\frac {a\,x^{1/3}}{x\,b^3+a\,b^2}+\frac {4\,{\left (-a\right )}^{1/3}\,\ln \left (\frac {12\,{\left (-a\right )}^{4/3}}{b^{1/3}}+12\,a\,x^{1/3}\right )}{3\,b^{7/3}}-\frac {4\,{\left (-a\right )}^{1/3}\,\ln \left (12\,a\,x^{1/3}-\frac {12\,{\left (-a\right )}^{4/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^{1/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{7/3}}+\frac {{\left (-a\right )}^{1/3}\,\ln \left (12\,a\,x^{1/3}+\frac {9\,{\left (-a\right )}^{4/3}\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right )}{b^{1/3}}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right )}{b^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(4/3)/(a + b*x)^2,x)

[Out]

(3*x^(1/3))/b^2 + (a*x^(1/3))/(a*b^2 + b^3*x) + (4*(-a)^(1/3)*log((12*(-a)^(4/3))/b^(1/3) + 12*a*x^(1/3)))/(3*
b^(7/3)) - (4*(-a)^(1/3)*log(12*a*x^(1/3) - (12*(-a)^(4/3)*((3^(1/2)*1i)/2 + 1/2))/b^(1/3))*((3^(1/2)*1i)/2 +
1/2))/(3*b^(7/3)) + ((-a)^(1/3)*log(12*a*x^(1/3) + (9*(-a)^(4/3)*((3^(1/2)*2i)/3 - 2/3))/b^(1/3))*((3^(1/2)*2i
)/3 - 2/3))/b^(7/3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(4/3)/(b*x+a)**2,x)

[Out]

Timed out

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