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

Optimal. Leaf size=237 \[ -\frac {\sqrt {2} \sqrt [4]{a-b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x \sqrt [4]{x^4-x^3} \sqrt [4]{a-b}}{x^2 \sqrt {a-b}-\sqrt {b} \sqrt {x^4-x^3}}\right )}{a \sqrt [4]{b}}-\frac {\sqrt {2} \sqrt [4]{a-b} \tanh ^{-1}\left (\frac {\frac {x^2 \sqrt [4]{a-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} \sqrt {x^4-x^3}}{\sqrt {2} \sqrt [4]{a-b}}}{x \sqrt [4]{x^4-x^3}}\right )}{a \sqrt [4]{b}}-\frac {2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )}{a}+\frac {2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )}{a} \]

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Rubi [A]  time = 0.25, antiderivative size = 233, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {2042, 105, 63, 240, 212, 206, 203, 93, 298, 205, 208} \begin {gather*} \frac {2 \sqrt [4]{x^4-x^3} \sqrt [4]{b-a} \tan ^{-1}\left (\frac {\sqrt [4]{x} \sqrt [4]{b-a}}{\sqrt [4]{b} \sqrt [4]{x-1}}\right )}{a \sqrt [4]{b} \sqrt [4]{x-1} x^{3/4}}-\frac {2 \sqrt [4]{x^4-x^3} \sqrt [4]{b-a} \tanh ^{-1}\left (\frac {\sqrt [4]{x} \sqrt [4]{b-a}}{\sqrt [4]{b} \sqrt [4]{x-1}}\right )}{a \sqrt [4]{b} \sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{x^4-x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{a \sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{x^4-x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{a \sqrt [4]{x-1} x^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-x^3 + x^4)^(1/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/(a*(-1 + x)^(1/4)*x^(3/4)) + (2*(-a + b)^(1/4)*(-x^3 + x
^4)^(1/4)*ArcTan[((-a + b)^(1/4)*x^(1/4))/(b^(1/4)*(-1 + x)^(1/4))])/(a*b^(1/4)*(-1 + x)^(1/4)*x^(3/4)) + (2*(
-x^3 + x^4)^(1/4)*ArcTanh[(-1 + x)^(1/4)/x^(1/4)])/(a*(-1 + x)^(1/4)*x^(3/4)) - (2*(-a + b)^(1/4)*(-x^3 + x^4)
^(1/4)*ArcTanh[((-a + b)^(1/4)*x^(1/4))/(b^(1/4)*(-1 + x)^(1/4))])/(a*b^(1/4)*(-1 + x)^(1/4)*x^(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 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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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 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 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 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 240

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 2042

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[(e^IntPart[m]*(e*x)^FracPart[m]*(a*x^j + b*x^(j + n))^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p]), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 68, normalized size = 0.29 \begin {gather*} \frac {4 \sqrt [4]{(x-1) x^3} \left (\sqrt [4]{x} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};1-x\right )-\, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {b-b x}{a x-b x}\right )\right )}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.08, size = 237, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )}{a}-\frac {\sqrt {2} \sqrt [4]{a-b} \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 )}{a \sqrt [4]{b}}+\frac {2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )}{a}-\frac {\sqrt {2} \sqrt [4]{a-b} \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 )}{a \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 297, normalized size = 1.25 \begin {gather*} \frac {4 \, a \left (-\frac {a - b}{a^{4} b}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{3} b x \sqrt {\frac {a^{2} x^{2} \sqrt {-\frac {a - b}{a^{4} b}} + \sqrt {x^{4} - x^{3}}}{x^{2}}} \left (-\frac {a - b}{a^{4} b}\right )^{\frac {3}{4}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} a^{3} b \left (-\frac {a - b}{a^{4} b}\right )^{\frac {3}{4}}}{{\left (a - b\right )} x}\right ) - a \left (-\frac {a - b}{a^{4} b}\right )^{\frac {1}{4}} \log \left (\frac {a x \left (-\frac {a - b}{a^{4} b}\right )^{\frac {1}{4}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + a \left (-\frac {a - b}{a^{4} b}\right )^{\frac {1}{4}} \log \left (-\frac {a x \left (-\frac {a - b}{a^{4} b}\right )^{\frac {1}{4}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 3.00, size = 325, normalized size = 1.37 \begin {gather*} -\frac {2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )}{a} - \frac {\log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right )}{a} + \frac {\log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right )}{a} + \frac {\sqrt {2} {\left (a b^{3} - b^{4}\right )}^{\frac {1}{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 )}{a b} + \frac {\sqrt {2} {\left (a b^{3} - b^{4}\right )}^{\frac {1}{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 )}{a b} + \frac {\sqrt {2} {\left (a b^{3} - b^{4}\right )}^{\frac {1}{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 )}{2 \, a b} - \frac {\sqrt {2} {\left (a b^{3} - b^{4}\right )}^{\frac {1}{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 )}{2 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x \left (a x -b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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