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

Optimal. Leaf size=55 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{3 b^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{3 b^{3/4}} \]

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Rubi [A]  time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 63, 212, 206, 203} \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{3 b^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{3 b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(b + a*x^3)^(3/4)),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/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]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.84 \begin {gather*} -\frac {2 \left (\tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )\right )}{3 b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(b + a*x^3)^(3/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 55, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{3 b^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{3 b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(b + a*x^3)^(3/4)),x]

[Out]

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

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fricas [B]  time = 0.54, size = 110, normalized size = 2.00 \begin {gather*} \frac {4}{3} \, \frac {1}{b^{3}}^{\frac {1}{4}} \arctan \left (\sqrt {b^{2} \sqrt {\frac {1}{b^{3}}} + \sqrt {a x^{3} + b}} b^{2} \frac {1}{b^{3}}^{\frac {3}{4}} - {\left (a x^{3} + b\right )}^{\frac {1}{4}} b^{2} \frac {1}{b^{3}}^{\frac {3}{4}}\right ) - \frac {1}{3} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right ) + \frac {1}{3} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (-b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 186, normalized size = 3.38 \begin {gather*} -\frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{3 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{3 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{6 \, b} + \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(2)*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/b - 1/3*sqrt
(2)*(-b)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/b - 1/6*sqrt(2)*(-b)
^(1/4)*log(sqrt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/b + 1/6*sqrt(2)*(-b)^(1/4)*log(-
sqrt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/b

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (a \,x^{3}+b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 57, normalized size = 1.04 \begin {gather*} -\frac {2 \, \arctan \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{3 \, b^{\frac {3}{4}}} + \frac {\log \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{3} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{3 \, b^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.71, size = 39, normalized size = 0.71 \begin {gather*} -\frac {2\,\mathrm {atan}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{3\,b^{3/4}}-\frac {2\,\mathrm {atanh}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{3\,b^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b + a*x^3)^(3/4)),x)

[Out]

- (2*atan((b + a*x^3)^(1/4)/b^(1/4)))/(3*b^(3/4)) - (2*atanh((b + a*x^3)^(1/4)/b^(1/4)))/(3*b^(3/4))

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sympy [C]  time = 0.93, size = 41, normalized size = 0.75 \begin {gather*} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {9}{4}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(3/4)*hyper((3/4, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3/4)*x**(9/4)*gamma(7/4))

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