3.14.50 \(\int \frac {-b+a x^3}{x^3 \sqrt [4]{b x+a x^4}} \, dx\) [1350]

Optimal. Leaf size=97 \[ \frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {2}{3} a^{3/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )+\frac {2}{3} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right ) \]

[Out]

4/9*(a*x^4+b*x)^(3/4)/x^3+2/3*a^(3/4)*arctan(a^(1/4)*(a*x^4+b*x)^(3/4)/(a*x^3+b))+2/3*a^(3/4)*arctanh(a^(1/4)*
(a*x^4+b*x)^(3/4)/(a*x^3+b))

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Rubi [A]
time = 0.10, antiderivative size = 143, normalized size of antiderivative = 1.47, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2063, 2036, 335, 281, 246, 218, 212, 209} \begin {gather*} \frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3+b} \text {ArcTan}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a x^4+b x}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a x^4+b x}}+\frac {4 \left (a x^4+b x\right )^{3/4}}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b*x + a*x^4)^(3/4))/(9*x^3) + (2*a^(3/4)*x^(1/4)*(b + a*x^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1
/4)])/(3*(b*x + a*x^4)^(1/4)) + (2*a^(3/4)*x^(1/4)*(b + a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/
4)])/(3*(b*x + a*x^4)^(1/4))

Rule 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {-b+a x^3}{x^3 \sqrt [4]{b x+a x^4}} \, dx &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+a \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {\left (a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{b+a x^3}} \, dx}{\sqrt [4]{b x+a x^4}}\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{b+a x^{12}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}}\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^4}} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{b x+a x^4}}\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}\\ &=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 x^3}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{b+a x^3} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{b+a x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}\\ \end {align*}

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Mathematica [A]
time = 9.03, size = 122, normalized size = 1.26 \begin {gather*} \frac {2 \left (2 \left (b+a x^3\right )+3 a^{3/4} x^{9/4} \sqrt [4]{b+a x^3} \text {ArcTan}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+3 a^{3/4} x^{9/4} \sqrt [4]{b+a x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{9 x^2 \sqrt [4]{x \left (b+a x^3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*(b + a*x^3) + 3*a^(3/4)*x^(9/4)*(b + a*x^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)] + 3*a^(3/4
)*x^(9/4)*(b + a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)]))/(9*x^2*(x*(b + a*x^3))^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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((a*x^3-b)/x^3/(a*x^4+b*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{3} - b}{x^{3} \sqrt [4]{x \left (a x^{3} + b\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (77) = 154\).
time = 0.42, size = 185, normalized size = 1.91 \begin {gather*} \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right ) + \frac {4}{9} \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^3)^(1/4))/(-a)^(1/4)) + 1/3*sqrt(2)
*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1/4))/(-a)^(1/4)) - 1/6*sqrt(2)*(-a)^(3/4
)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) + 1/6*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)
*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) + 4/9*(a + b/x^3)^(3/4)

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Mupad [B]
time = 1.14, size = 58, normalized size = 0.60 \begin {gather*} \frac {4\,{\left (a\,x^4+b\,x\right )}^{3/4}}{9\,x^3}+\frac {4\,a\,x\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^3}{b}\right )}{3\,{\left (a\,x^4+b\,x\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(b*x + a*x^4)^(3/4))/(9*x^3) + (4*a*x*((a*x^3)/b + 1)^(1/4)*hypergeom([1/4, 1/4], 5/4, -(a*x^3)/b))/(3*(b*x
+ a*x^4)^(1/4))

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