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

Optimal. Leaf size=99 \[ \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{16 a^{7/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{16 a^{7/4}}+\frac {(4 a x+b) \sqrt [4]{a x^4+b x^3}}{8 a} \]

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Rubi [A]  time = 0.20, antiderivative size = 168, normalized size of antiderivative = 1.70, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2004, 2024, 2032, 63, 331, 298, 203, 206} \begin {gather*} \frac {3 b^2 x^{9/4} (a x+b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{16 a^{7/4} \left (a x^4+b x^3\right )^{3/4}}-\frac {3 b^2 x^{9/4} (a x+b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{16 a^{7/4} \left (a x^4+b x^3\right )^{3/4}}+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}+\frac {b \sqrt [4]{a x^4+b x^3}}{8 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^3 + a*x^4)^(1/4),x]

[Out]

(b*(b*x^3 + a*x^4)^(1/4))/(8*a) + (x*(b*x^3 + a*x^4)^(1/4))/2 + (3*b^2*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)
*x^(1/4))/(b + a*x)^(1/4)])/(16*a^(7/4)*(b*x^3 + a*x^4)^(3/4)) - (3*b^2*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/
4)*x^(1/4))/(b + a*x)^(1/4)])/(16*a^(7/4)*(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 2004

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

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 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 \sqrt [4]{b x^3+a x^4} \, dx &=\frac {1}{2} x \sqrt [4]{b x^3+a x^4}+\frac {1}{8} b \int \frac {x^3}{\left (b x^3+a x^4\right )^{3/4}} \, dx\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}-\frac {\left (3 b^2\right ) \int \frac {x^2}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{32 a}\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}-\frac {\left (3 b^2 x^{9/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{32 a \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}-\frac {\left (3 b^2 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 )}{8 a \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}-\frac {\left (3 b^2 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 )}{8 a \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}-\frac {\left (3 b^2 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 )}{16 a^{3/2} \left (b x^3+a x^4\right )^{3/4}}+\frac {\left (3 b^2 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 )}{16 a^{3/2} \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {1}{2} x \sqrt [4]{b x^3+a x^4}+\frac {3 b^2 x^{9/4} (b+a x)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{16 a^{7/4} \left (b x^3+a x^4\right )^{3/4}}-\frac {3 b^2 x^{9/4} (b+a x)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{16 a^{7/4} \left (b x^3+a x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.47 \begin {gather*} \frac {4 x \sqrt [4]{x^3 (a x+b)} \, _2F_1\left (-\frac {1}{4},\frac {7}{4};\frac {11}{4};-\frac {a x}{b}\right )}{7 \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^3 + a*x^4)^(1/4),x]

[Out]

(4*x*(x^3*(b + a*x))^(1/4)*Hypergeometric2F1[-1/4, 7/4, 11/4, -((a*x)/b)])/(7*(1 + (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.42, size = 99, normalized size = 1.00 \begin {gather*} \frac {(b+4 a x) \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{16 a^{7/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{16 a^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + 4*a*x)*(b*x^3 + a*x^4)^(1/4))/(8*a) + (3*b^2*ArcTan[(a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/(16*a^(7/4)) - (
3*b^2*ArcTanh[(a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/(16*a^(7/4))

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fricas [B]  time = 0.95, size = 234, normalized size = 2.36 \begin {gather*} \frac {12 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{5} b^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {3}{4}} - a^{5} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{4} \sqrt {\frac {b^{8}}{a^{7}}} x^{2} + \sqrt {a x^{4} + b x^{3}} b^{4}}{x^{2}}}}{b^{8} x}\right ) - 3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (\frac {3 \, {\left (a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) + 3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) + 4 \, {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} {\left (4 \, a x + b\right )}}{32 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*a*(b^8/a^7)^(1/4)*arctan(-((a*x^4 + b*x^3)^(1/4)*a^5*b^2*(b^8/a^7)^(3/4) - a^5*(b^8/a^7)^(3/4)*x*sqrt
((a^4*sqrt(b^8/a^7)*x^2 + sqrt(a*x^4 + b*x^3)*b^4)/x^2))/(b^8*x)) - 3*a*(b^8/a^7)^(1/4)*log(3*(a^2*(b^8/a^7)^(
1/4)*x + (a*x^4 + b*x^3)^(1/4)*b^2)/x) + 3*a*(b^8/a^7)^(1/4)*log(-3*(a^2*(b^8/a^7)^(1/4)*x - (a*x^4 + b*x^3)^(
1/4)*b^2)/x) + 4*(a*x^4 + b*x^3)^(1/4)*(4*a*x + b))/a

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giac [B]  time = 0.19, size = 243, normalized size = 2.45 \begin {gather*} \frac {\frac {6 \, \sqrt {2} b^{3} \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 )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {6 \, \sqrt {2} b^{3} \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 )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {3 \, \sqrt {2} b^{3} \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}} a} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \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^{2}} + \frac {8 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} b^{3} + 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a b^{3}\right )} x^{2}}{a b^{2}}}{64 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(6*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 6
*sqrt(2)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 3*sqrt(
2)*b^3*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a) + 3*sqrt(2)*(-a)^(1/4
)*b^3*log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^2 + 8*((a + b/x)^(5/4)*b^3 + 3*(a
+ b/x)^(1/4)*a*b^3)*x^2/(a*b^2))/b

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \left (a \,x^{4}+b \,x^{3}\right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((a*x^4 + b*x^3)^(1/4), x)

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mupad [B]  time = 0.86, size = 38, normalized size = 0.38 \begin {gather*} \frac {4\,x\,{\left (a\,x^4+b\,x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ -\frac {a\,x}{b}\right )}{7\,{\left (\frac {a\,x}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 + b*x^3)^(1/4),x)

[Out]

(4*x*(a*x^4 + b*x^3)^(1/4)*hypergeom([-1/4, 7/4], 11/4, -(a*x)/b))/(7*((a*x)/b + 1)^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**4 + b*x**3)**(1/4), x)

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