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

Optimal. Leaf size=188 \[ \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x \sqrt [4]{x^4-x^3} \sqrt [4]{a-b}}{x^2 \sqrt {a-b}-\sqrt {b} \sqrt {x^4-x^3}}\right )}{b^{3/4} \sqrt [4]{a-b}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {x^2 \sqrt [4]{a-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} \sqrt {x^4-x^3}}{\sqrt {2} \sqrt [4]{a-b}}}{x \sqrt [4]{x^4-x^3}}\right )}{b^{3/4} \sqrt [4]{a-b}} \]

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Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 0.74, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2056, 93, 212, 208, 205} \begin {gather*} -\frac {2 \sqrt [4]{x-1} x^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{x} \sqrt [4]{b-a}}{\sqrt [4]{b} \sqrt [4]{x-1}}\right )}{b^{3/4} \sqrt [4]{x^4-x^3} \sqrt [4]{b-a}}-\frac {2 \sqrt [4]{x-1} x^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{x} \sqrt [4]{b-a}}{\sqrt [4]{b} \sqrt [4]{x-1}}\right )}{b^{3/4} \sqrt [4]{x^4-x^3} \sqrt [4]{b-a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

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

Rule 208

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

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

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.26 \begin {gather*} \frac {4 (x-1) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b (x-1)}{(a-b) x}\right )}{3 \sqrt [4]{(x-1) x^3} (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1 + x)*Hypergeometric2F1[3/4, 1, 7/4, -((b*(-1 + x))/((a - b)*x))])/(3*(a - b)*((-1 + x)*x^3)^(1/4))

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IntegrateAlgebraic [A]  time = 1.30, size = 188, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a-b} \sqrt [4]{b} x \sqrt [4]{-x^3+x^4}}{\sqrt {a-b} x^2-\sqrt {b} \sqrt {-x^3+x^4}}\right )}{\sqrt [4]{a-b} b^{3/4}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a-b} x^2}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} \sqrt {-x^3+x^4}}{\sqrt {2} \sqrt [4]{a-b}}}{x \sqrt [4]{-x^3+x^4}}\right )}{\sqrt [4]{a-b} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 262, normalized size = 1.39 \begin {gather*} -4 \, \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {1}{4}} \arctan \left (\frac {b x \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {1}{4}} \sqrt {-\frac {{\left (a b - b^{2}\right )} x^{2} \sqrt {-\frac {1}{a b^{3} - b^{4}}} - \sqrt {x^{4} - x^{3}}}{x^{2}}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} b \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {1}{4}}}{x}\right ) + \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a b^{2} - b^{3}\right )} x \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {3}{4}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (a b^{2} - b^{3}\right )} x \left (-\frac {1}{a b^{3} - b^{4}}\right )^{\frac {3}{4}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 317, normalized size = 1.69 \begin {gather*} -\frac {2 \, {\left (a b^{3} - b^{4}\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a - b}{b}\right )^{\frac {1}{4}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a - b}{b}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a b^{3} - \sqrt {2} b^{4}} - \frac {2 \, {\left (a b^{3} - b^{4}\right )}^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a - b}{b}\right )^{\frac {1}{4}} - 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a - b}{b}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a b^{3} - \sqrt {2} b^{4}} + \frac {{\left (a b^{3} - b^{4}\right )}^{\frac {3}{4}} \log \left (\sqrt {2} \left (\frac {a - b}{b}\right )^{\frac {1}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {\frac {a - b}{b}} + \sqrt {-\frac {1}{x} + 1}\right )}{\sqrt {2} a b^{3} - \sqrt {2} b^{4}} - \frac {{\left (a b^{3} - b^{4}\right )}^{\frac {3}{4}} \log \left (-\sqrt {2} \left (\frac {a - b}{b}\right )^{\frac {1}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {\frac {a - b}{b}} + \sqrt {-\frac {1}{x} + 1}\right )}{\sqrt {2} a b^{3} - \sqrt {2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a x -b \right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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