3.17.17 \(\int \frac {b+a x^6}{x^6 \sqrt [4]{b x+a x^4}} \, dx\)

Optimal. Leaf size=110 \[ \frac {2}{3} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4+b x\right )^{3/4}}{a x^3+b}\right )+\frac {2}{3} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4+b x\right )^{3/4}}{a x^3+b}\right )+\frac {4 \left (a x^4+b x\right )^{3/4} \left (4 a x^3-3 b\right )}{63 b x^6} \]

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Rubi [A]  time = 0.26, antiderivative size = 167, normalized size of antiderivative = 1.52, number of steps used = 11, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2052, 2011, 329, 275, 240, 212, 206, 203, 2016, 2014} \begin {gather*} \frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3+b} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a x^4+b x}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a x^4+b x}}-\frac {4 \left (a x^4+b x\right )^{3/4}}{21 x^6}+\frac {16 a \left (a x^4+b x\right )^{3/4}}{63 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^6)/(x^6*(b*x + a*x^4)^(1/4)),x]

[Out]

(-4*(b*x + a*x^4)^(3/4))/(21*x^6) + (16*a*(b*x + a*x^4)^(3/4))/(63*b*x^3) + (2*a^(3/4)*x^(1/4)*(b + a*x^3)^(1/
4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(3*(b*x + a*x^4)^(1/4)) + (2*a^(3/4)*x^(1/4)*(b + a*x^3)^(1/4)
*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(3*(b*x + a*x^4)^(1/4))

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 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 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 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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

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 2016

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*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

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 {b+a x^6}{x^6 \sqrt [4]{b x+a x^4}} \, dx &=\int \left (\frac {a}{\sqrt [4]{b x+a x^4}}+\frac {b}{x^6 \sqrt [4]{b x+a x^4}}\right ) \, dx\\ &=a \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx+b \int \frac {1}{x^6 \sqrt [4]{b x+a x^4}} \, dx\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}-\frac {1}{7} (4 a) \int \frac {1}{x^3 \sqrt [4]{b x+a x^4}} \, dx+\frac {\left (a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{b+a x^3}} \, dx}{\sqrt [4]{b x+a x^4}}\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}+\frac {16 a \left (b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{b+a x^{12}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}}\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}+\frac {16 a \left (b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^4}} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{b x+a x^4}}\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}+\frac {16 a \left (b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}+\frac {16 a \left (b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\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 )}{3 \sqrt [4]{b x+a x^4}}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\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 )}{3 \sqrt [4]{b x+a x^4}}\\ &=-\frac {4 \left (b x+a x^4\right )^{3/4}}{21 x^6}+\frac {16 a \left (b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{b+a x^3} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{b+a x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 138, normalized size = 1.25 \begin {gather*} \frac {2 \left (21 a^{3/4} b x^{21/4} \sqrt [4]{a x^3+b} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )+21 a^{3/4} b x^{21/4} \sqrt [4]{a x^3+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )+8 a^2 x^6+2 a b x^3-6 b^2\right )}{63 b x^5 \sqrt [4]{x \left (a x^3+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^6)/(x^6*(b*x + a*x^4)^(1/4)),x]

[Out]

(2*(-6*b^2 + 2*a*b*x^3 + 8*a^2*x^6 + 21*a^(3/4)*b*x^(21/4)*(b + a*x^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x
^3)^(1/4)] + 21*a^(3/4)*b*x^(21/4)*(b + a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)]))/(63*b*x^5*
(x*(b + a*x^3))^(1/4))

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IntegrateAlgebraic [A]  time = 0.70, size = 110, normalized size = 1.00 \begin {gather*} \frac {4 \left (-3 b+4 a x^3\right ) \left (b x+a x^4\right )^{3/4}}{63 b x^6}+\frac {2}{3} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )+\frac {2}{3} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^6)/(x^6*(b*x + a*x^4)^(1/4)),x]

[Out]

(4*(-3*b + 4*a*x^3)*(b*x + a*x^4)^(3/4))/(63*b*x^6) + (2*a^(3/4)*ArcTan[(a^(1/4)*(b*x + a*x^4)^(3/4))/(b + a*x
^3)])/3 + (2*a^(3/4)*ArcTanh[(a^(1/4)*(b*x + a*x^4)^(3/4))/(b + a*x^3)])/3

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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^6+b)/x^6/(a*x^4+b*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.29, size = 209, normalized size = 1.90 \begin {gather*} \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 ) - \frac {4 \, {\left (3 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {7}{4}} b^{6} - 7 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{4}} a b^{6}\right )}}{63 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^6/(a*x^4+b*x)^(1/4),x, algorithm="giac")

[Out]

1/3*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^3)^(1/4))/(-a)^(1/4)) + 1/3*sqrt(2)
*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1/4))/(-a)^(1/4)) - 1/6*sqrt(2)*(-a)^(3/4
)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) + 1/6*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)
*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) - 4/63*(3*(a + b/x^3)^(7/4)*b^6 - 7*(a + b/x^3)^(3
/4)*a*b^6)/b^7

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{6}+b}{x^{6} \left (a \,x^{4}+b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^6+b)/x^6/(a*x^4+b*x)^(1/4),x)

[Out]

int((a*x^6+b)/x^6/(a*x^4+b*x)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} + b}{{\left (a x^{4} + b x\right )}^{\frac {1}{4}} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^6/(a*x^4+b*x)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^6 + b)/((a*x^4 + b*x)^(1/4)*x^6), x)

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mupad [B]  time = 1.24, size = 71, normalized size = 0.65 \begin {gather*} \frac {4\,a\,x\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^3}{b}\right )}{3\,{\left (a\,x^4+b\,x\right )}^{1/4}}-\frac {4\,{\left (a\,x^4+b\,x\right )}^{3/4}\,\left (3\,b-4\,a\,x^3\right )}{63\,b\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^6)/(x^6*(b*x + a*x^4)^(1/4)),x)

[Out]

(4*a*x*((a*x^3)/b + 1)^(1/4)*hypergeom([1/4, 1/4], 5/4, -(a*x^3)/b))/(3*(b*x + a*x^4)^(1/4)) - (4*(b*x + a*x^4
)^(3/4)*(3*b - 4*a*x^3))/(63*b*x^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} + b}{x^{6} \sqrt [4]{x \left (a x^{3} + b\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**6+b)/x**6/(a*x**4+b*x)**(1/4),x)

[Out]

Integral((a*x**6 + b)/(x**6*(x*(a*x**3 + b))**(1/4)), x)

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