3.13.58 \(\int \sqrt [4]{b x^5+a x^8} \, dx\)

Optimal. Leaf size=91 \[ \frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a x^8+b x^5}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a x^8+b x^5}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5} \]

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Rubi [A]  time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.64, 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, 2032, 329, 275, 331, 298, 203, 206} \begin {gather*} -\frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 a^{3/4} \left (a x^8+b x^5\right )^{3/4}}+\frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 a^{3/4} \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^5 + a*x^8)^(1/4),x]

[Out]

(x*(b*x^5 + a*x^8)^(1/4))/3 - (b*x^(15/4)*(b + a*x^3)^(3/4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(6*a^
(3/4)*(b*x^5 + a*x^8)^(3/4)) + (b*x^(15/4)*(b + a*x^3)^(3/4)*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(6*
a^(3/4)*(b*x^5 + a*x^8)^(3/4))

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 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 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 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 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 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^5+a x^8} \, dx &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {1}{4} b \int \frac {x^5}{\left (b x^5+a x^8\right )^{3/4}} \, dx\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \int \frac {x^{5/4}}{\left (b+a x^3\right )^{3/4}} \, dx}{4 \left (b x^5+a x^8\right )^{3/4}}\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\left (b+a x^{12}\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{\left (b x^5+a x^8\right )^{3/4}}\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,x^{3/4}\right )}{3 \left (b x^5+a x^8\right )^{3/4}}\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \left (b x^5+a x^8\right )^{3/4}}\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 \sqrt {a} \left (b x^5+a x^8\right )^{3/4}}-\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 \sqrt {a} \left (b x^5+a x^8\right )^{3/4}}\\ &=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}-\frac {b x^{15/4} \left (b+a x^3\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 a^{3/4} \left (b x^5+a x^8\right )^{3/4}}+\frac {b x^{15/4} \left (b+a x^3\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 a^{3/4} \left (b x^5+a x^8\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 53, normalized size = 0.58 \begin {gather*} \frac {4 x \sqrt [4]{x^5 \left (a x^3+b\right )} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x^3}{b}\right )}{9 \sqrt [4]{\frac {a x^3}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^5 + a*x^8)^(1/4),x]

[Out]

(4*x*(x^5*(b + a*x^3))^(1/4)*Hypergeometric2F1[-1/4, 3/4, 7/4, -((a*x^3)/b)])/(9*(1 + (a*x^3)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.46, size = 91, normalized size = 1.00 \begin {gather*} \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x^5 + a*x^8)^(1/4),x]

[Out]

(x*(b*x^5 + a*x^8)^(1/4))/3 + (b*ArcTan[(b*x^5 + a*x^8)^(1/4)/(a^(1/4)*x^2)])/(6*a^(3/4)) + (b*ArcTanh[(b*x^5
+ a*x^8)^(1/4)/(a^(1/4)*x^2)])/(6*a^(3/4))

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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^8+b*x^5)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.29, size = 213, normalized size = 2.34 \begin {gather*} \frac {8 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} b x^{3} + \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^{3}}\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^{3}}\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^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{a} + \frac {\sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b*x^5)^(1/4),x, algorithm="giac")

[Out]

1/24*(8*(a + b/x^3)^(1/4)*b*x^3 + 2*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x
^3)^(1/4))/(-a)^(1/4))/a + 2*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1
/4))/(-a)^(1/4))/a + sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x
^3))/a + sqrt(2)*b^2*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3))/(-a)^(3/4))/b

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \left (a \,x^{8}+b \,x^{5}\right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8+b*x^5)^(1/4),x)

[Out]

int((a*x^8+b*x^5)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (a x^{8} + b x^{5}\right )}^{\frac {1}{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b*x^5)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^8 + b*x^5)^(1/4), x)

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mupad [B]  time = 0.93, size = 42, normalized size = 0.46 \begin {gather*} \frac {4\,x\,{\left (a\,x^8+b\,x^5\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^3}{b}\right )}{9\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8 + b*x^5)^(1/4),x)

[Out]

(4*x*(a*x^8 + b*x^5)^(1/4)*hypergeom([-1/4, 3/4], 7/4, -(a*x^3)/b))/(9*((a*x^3)/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^{8} + b x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**8+b*x**5)**(1/4),x)

[Out]

Integral((a*x**8 + b*x**5)**(1/4), x)

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