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

Optimal. Leaf size=177 \[ \frac {1}{12} \sqrt [3]{a} b \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{a x^3+b x}+\left (a x^3+b x\right )^{2/3}\right )-\frac {1}{6} \sqrt [3]{a} b \log \left (\sqrt [3]{a x^3+b x}-\sqrt [3]{a} x\right )-\frac {\sqrt [3]{a} b \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{a} x}{2 \sqrt [3]{a x^3+b x}+\sqrt [3]{a} x}\right )}{2 \sqrt {3}}+\frac {\sqrt [3]{a x^3+b x} \left (4 a x^4-3 a x^2-3 b\right )}{8 x^3} \]

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Rubi [A]  time = 0.41, antiderivative size = 267, normalized size of antiderivative = 1.51, number of steps used = 14, number of rules used = 13, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {2052, 2004, 2032, 329, 275, 331, 292, 31, 634, 617, 204, 628, 2014} \begin {gather*} \frac {\sqrt [3]{a} b x^{2/3} \left (a x^2+b\right )^{2/3} \log \left (\frac {a^{2/3} x^{4/3}}{\left (a x^2+b\right )^{2/3}}+\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{a x^2+b}}+1\right )}{12 \left (a x^3+b x\right )^{2/3}}+\frac {1}{2} a x \sqrt [3]{a x^3+b x}-\frac {3 \left (a x^3+b x\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (a x^2+b\right )^{2/3} \log \left (1-\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{a x^2+b}}\right )}{6 \left (a x^3+b x\right )^{2/3}}-\frac {\sqrt [3]{a} b x^{2/3} \left (a x^2+b\right )^{2/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a} x^{2/3}}{\sqrt [3]{a x^2+b}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a x^3+b x\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(b*x + a*x^3)^(1/3))/2 - (3*(b*x + a*x^3)^(4/3))/(8*x^4) - (a^(1/3)*b*x^(2/3)*(b + a*x^2)^(2/3)*ArcTan[(1
+ (2*a^(1/3)*x^(2/3))/(b + a*x^2)^(1/3))/Sqrt[3]])/(2*Sqrt[3]*(b*x + a*x^3)^(2/3)) - (a^(1/3)*b*x^(2/3)*(b +
a*x^2)^(2/3)*Log[1 - (a^(1/3)*x^(2/3))/(b + a*x^2)^(1/3)])/(6*(b*x + a*x^3)^(2/3)) + (a^(1/3)*b*x^(2/3)*(b + a
*x^2)^(2/3)*Log[1 + (a^(2/3)*x^(4/3))/(b + a*x^2)^(2/3) + (a^(1/3)*x^(2/3))/(b + a*x^2)^(1/3)])/(12*(b*x + a*x
^3)^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

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

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 2014

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

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx &=\int \left (a \sqrt [3]{b x+a x^3}+\frac {b \sqrt [3]{b x+a x^3}}{x^4}\right ) \, dx\\ &=a \int \sqrt [3]{b x+a x^3} \, dx+b \int \frac {\sqrt [3]{b x+a x^3}}{x^4} \, dx\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {1}{3} (a b) \int \frac {x}{\left (b x+a x^3\right )^{2/3}} \, dx\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (b+a x^2\right )^{2/3}} \, dx}{3 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (b+a x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (b+a x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{2 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x}{1-a x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{2 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a^{2/3} b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt [3]{a} x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{6 \left (b x+a x^3\right )^{2/3}}-\frac {\left (a^{2/3} b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1-\sqrt [3]{a} x}{1+\sqrt [3]{a} x+a^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{6 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (1-\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{6 \left (b x+a x^3\right )^{2/3}}+\frac {\left (\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a}+2 a^{2/3} x}{1+\sqrt [3]{a} x+a^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{12 \left (b x+a x^3\right )^{2/3}}-\frac {\left (a^{2/3} b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [3]{a} x+a^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{4 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (1-\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{6 \left (b x+a x^3\right )^{2/3}}+\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (1+\frac {a^{2/3} x^{4/3}}{\left (b+a x^2\right )^{2/3}}+\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{12 \left (b x+a x^3\right )^{2/3}}+\frac {\left (\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{2 \left (b x+a x^3\right )^{2/3}}\\ &=\frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}}{\sqrt {3}}\right )}{2 \sqrt {3} \left (b x+a x^3\right )^{2/3}}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (1-\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{6 \left (b x+a x^3\right )^{2/3}}+\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (1+\frac {a^{2/3} x^{4/3}}{\left (b+a x^2\right )^{2/3}}+\frac {\sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}\right )}{12 \left (b x+a x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 69, normalized size = 0.39 \begin {gather*} \frac {3 \sqrt [3]{x \left (a x^2+b\right )} \left (\frac {2 a x^4 \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {a x^2}{b}\right )}{\sqrt [3]{\frac {a x^2}{b}+1}}-a x^2-b\right )}{8 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x*(b + a*x^2))^(1/3)*(-b - a*x^2 + (2*a*x^4*Hypergeometric2F1[-1/3, 2/3, 5/3, -((a*x^2)/b)])/(1 + (a*x^2)/
b)^(1/3)))/(8*x^3)

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IntegrateAlgebraic [A]  time = 0.57, size = 177, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{b x+a x^3} \left (-3 b-3 a x^2+4 a x^4\right )}{8 x^3}-\frac {\sqrt [3]{a} b \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{a} x}{\sqrt [3]{a} x+2 \sqrt [3]{b x+a x^3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \sqrt [3]{a} b \log \left (-\sqrt [3]{a} x+\sqrt [3]{b x+a x^3}\right )+\frac {1}{12} \sqrt [3]{a} b \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{b x+a x^3}+\left (b x+a x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*x + a*x^3)^(1/3)*(-3*b - 3*a*x^2 + 4*a*x^4))/(8*x^3) - (a^(1/3)*b*ArcTan[(Sqrt[3]*a^(1/3)*x)/(a^(1/3)*x +
2*(b*x + a*x^3)^(1/3))])/(2*Sqrt[3]) - (a^(1/3)*b*Log[-(a^(1/3)*x) + (b*x + a*x^3)^(1/3)])/6 + (a^(1/3)*b*Log[
a^(2/3)*x^2 + a^(1/3)*x*(b*x + a*x^3)^(1/3) + (b*x + a*x^3)^(2/3)])/12

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{3}+b x \right )^{\frac {1}{3}} \left (a \,x^{4}+b \right )}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.85, size = 65, normalized size = 0.37 \begin {gather*} \frac {3\,a\,x\,{\left (a\,x^3+b\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -\frac {a\,x^2}{b}\right )}{4\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/3}}-\frac {3\,{\left (a\,x^3+b\,x\right )}^{1/3}\,\left (a\,x^2+b\right )}{8\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*x*(b*x + a*x^3)^(1/3)*hypergeom([-1/3, 2/3], 5/3, -(a*x^2)/b))/(4*((a*x^2)/b + 1)^(1/3)) - (3*(b*x + a*x^
3)^(1/3)*(b + a*x^2))/(8*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{x \left (a x^{2} + b\right )} \left (a x^{4} + b\right )}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a*x**2 + b))**(1/3)*(a*x**4 + b)/x**4, x)

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