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

Optimal. Leaf size=117 \[ \frac {3}{4} a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {1}{2} \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {1}{2} a^{4/3} \log (x)+\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3} \]

[Out]

3/2*a*(b*x^2+a)^(1/3)+3/8*(b*x^2+a)^(4/3)-1/2*a^(4/3)*ln(x)+3/4*a^(4/3)*ln(a^(1/3)-(b*x^2+a)^(1/3))-1/2*a^(4/3
)*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 50, 57, 617, 204, 31} \[ \frac {3}{4} a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {1}{2} \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {1}{2} a^{4/3} \log (x)+\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x,x]

[Out]

(3*a*(a + b*x^2)^(1/3))/2 + (3*(a + b*x^2)^(4/3))/8 - (Sqrt[3]*a^(4/3)*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/
(Sqrt[3]*a^(1/3))])/2 - (a^(4/3)*Log[x])/2 + (3*a^(4/3)*Log[a^(1/3) - (a + b*x^2)^(1/3)])/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 {\left (a+b x^2\right )^{4/3}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{8} \left (a+b x^2\right )^{4/3}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3}+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^2\right )\\ &=\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3}-\frac {1}{2} a^{4/3} \log (x)-\frac {1}{4} \left (3 a^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )-\frac {1}{4} \left (3 a^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )\\ &=\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3}-\frac {1}{2} a^{4/3} \log (x)+\frac {3}{4} a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )+\frac {1}{2} \left (3 a^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )\\ &=\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3}-\frac {1}{2} \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\frac {1}{2} a^{4/3} \log (x)+\frac {3}{4} a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 144, normalized size = 1.23 \[ \frac {1}{8} \left (4 a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-2 a^{4/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )-4 \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )+3 b x^2 \sqrt [3]{a+b x^2}+15 a \sqrt [3]{a+b x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x,x]

[Out]

(15*a*(a + b*x^2)^(1/3) + 3*b*x^2*(a + b*x^2)^(1/3) - 4*Sqrt[3]*a^(4/3)*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1
/3))/Sqrt[3]] + 4*a^(4/3)*Log[a^(1/3) - (a + b*x^2)^(1/3)] - 2*a^(4/3)*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3)
+ (a + b*x^2)^(2/3)])/8

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fricas [A]  time = 0.98, size = 111, normalized size = 0.95 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (b x^{2} + 5 \, a\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*(2*sqrt(3)*(b*x^2 + a)^(1/3)*a^(2/3) + sqrt(3)*a)/a) - 1/4*a^(4/3)*log((b*x^2
+ a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(4/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/8*(b*x^2
+ 5*a)*(b*x^2 + a)^(1/3)

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giac [A]  time = 1.46, size = 110, normalized size = 0.94 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) + \frac {3}{8} \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) - 1/4*a^(4/3)*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(4/3)*log(abs((b*x^2 + a)^(1/3) - a^(1/3))) + 3/8*(b*x^2
+ a)^(4/3) + 3/2*(b*x^2 + a)^(1/3)*a

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.06, size = 109, normalized size = 0.93 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) - 1/4*a^(4/3)*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(4/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/8*(b*x^2 + a)
^(4/3) + 3/2*(b*x^2 + a)^(1/3)*a

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mupad [B]  time = 5.00, size = 133, normalized size = 1.14 \[ \frac {3\,a\,{\left (b\,x^2+a\right )}^{1/3}}{2}+\frac {3\,{\left (b\,x^2+a\right )}^{4/3}}{8}+\frac {a^{4/3}\,\ln \left (\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {9\,a^{7/3}}{2}\right )}{2}-\frac {a^{4/3}\,\ln \left (\frac {9\,a^{7/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+a^{4/3}\,\ln \left (9\,a^{7/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(4/3)/x,x)

[Out]

(3*a*(a + b*x^2)^(1/3))/2 + (3*(a + b*x^2)^(4/3))/8 + (a^(4/3)*log((9*a^2*(a + b*x^2)^(1/3))/2 - (9*a^(7/3))/2
))/2 - (a^(4/3)*log((9*a^(7/3)*((3^(1/2)*1i)/2 + 1/2))/2 + (9*a^2*(a + b*x^2)^(1/3))/2)*((3^(1/2)*1i)/2 + 1/2)
)/2 + a^(4/3)*log(9*a^(7/3)*((3^(1/2)*1i)/4 - 1/4) - (9*a^2*(a + b*x^2)^(1/3))/2)*((3^(1/2)*1i)/4 - 1/4)

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sympy [C]  time = 1.30, size = 49, normalized size = 0.42 \[ - \frac {b^{\frac {4}{3}} x^{\frac {8}{3}} \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {4}{3} \\ - \frac {1}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (- \frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(4/3)*x**(8/3)*gamma(-4/3)*hyper((-4/3, -4/3), (-1/3,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(-1/3))

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