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

Optimal. Leaf size=82 \[ -\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{2 a^{3/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{2 a^{3/4}}+\sqrt [4]{a x^4+b x^3} \]

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Rubi [A]  time = 0.15, antiderivative size = 136, normalized size of antiderivative = 1.66, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2021, 2032, 63, 331, 298, 203, 206} \begin {gather*} -\frac {b x^{9/4} (a x+b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4} \left (a x^4+b x^3\right )^{3/4}}+\frac {b x^{9/4} (a x+b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4} \left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^3 + a*x^4)^(1/4)/x,x]

[Out]

(b*x^3 + a*x^4)^(1/4) - (b*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(2*a^(3/4)*(b*x^
3 + a*x^4)^(3/4)) + (b*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(2*a^(3/4)*(b*x^3 +
a*x^4)^(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 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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 46, normalized size = 0.56 \begin {gather*} \frac {4 \sqrt [4]{x^3 (a x+b)} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x}{b}\right )}{3 \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^3 + a*x^4)^(1/4)/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.38, size = 82, normalized size = 1.00 \begin {gather*} \sqrt [4]{b x^3+a x^4}-\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 206, normalized size = 2.51 \begin {gather*} -\left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {a^{2} \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{2} \sqrt {\frac {b^{4}}{a^{3}}} x^{2} + \sqrt {a x^{4} + b x^{3}} b^{2}}{x^{2}}} - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{2} b \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}}}{b^{4} x}\right ) + \frac {1}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) - \frac {1}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 211, normalized size = 2.57 \begin {gather*} \frac {\frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a} + \frac {\sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}}} + 8 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} b x}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a + 2*sq
rt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a + sqrt(2)*(-a)
^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a + sqrt(2)*b^2*log(-sqrt(2)*(-a
)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/(-a)^(3/4) + 8*(a + b/x)^(1/4)*b*x)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b*x^3)^(1/4)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 + b*x^3)^(1/4)/x,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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