3.16.90 \(\int \frac {1}{(b+a x^3) \sqrt [4]{-b x+a x^4}} \, dx\)

Optimal. Leaf size=109 \[ \frac {2^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )}{3 \sqrt [4]{a} b}+\frac {2^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )}{3 \sqrt [4]{a} b} \]

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Rubi [A]  time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.46, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2056, 466, 465, 377, 212, 206, 203} \begin {gather*} \frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3-b} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{a x^4-b x}}+\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3-b} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{a x^4-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 465

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

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 76, normalized size = 0.70 \begin {gather*} \frac {4 x \sqrt [4]{1-\frac {a x^3}{b}} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {2 a x^3}{a x^3+b}\right )}{3 b \sqrt [4]{\frac {a x^3}{b}+1} \sqrt [4]{a x^4-b x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*x*(1 - (a*x^3)/b)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (2*a*x^3)/(b + a*x^3)])/(3*b*(1 + (a*x^3)/b)^(1/4)
*(-(b*x) + a*x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.40, size = 109, normalized size = 1.00 \begin {gather*} \frac {2^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 \sqrt [4]{a} b}+\frac {2^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 \sqrt [4]{a} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 107.75, size = 430, normalized size = 3.94 \begin {gather*} -\frac {2}{3} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, {\left (2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a b^{3} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b x^{2} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + {\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {a x^{4} - b x} a b x \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (3 \, a^{2} b^{3} x^{3} - a b^{4}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}}\right )} \sqrt {\sqrt {\frac {1}{2}} b^{2} \sqrt {\frac {1}{a b^{4}}}}\right )}}{a x^{3} + b}\right ) + \frac {1}{6} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x} a b^{3} x \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {1}{a b^{4}}} + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a b x^{3} - b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}}}{a x^{3} + b}\right ) - \frac {1}{6} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x} a b^{3} x \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} - 4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {1}{a b^{4}}} + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a b x^{3} - b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}}}{a x^{3} + b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.24, size = 209, normalized size = 1.92 \begin {gather*} \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 )}{3 \, a b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 )}{3 \, a b} + \frac {2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{6 \, \left (-a\right )^{\frac {1}{4}} b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{6 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a \,x^{3}+b \right ) \left (a \,x^{4}-b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*x^3+b)/(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*} \int \frac {1}{{\left (a x^{4} - b x\right )}^{\frac {1}{4}} {\left (a x^{3} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a\,x^4-b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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