3.19.18 \(\int \frac {b+a x^6}{x^3 (b+a x^3) \sqrt [4]{b x+a x^4}} \, dx\)

Optimal. Leaf size=123 \[ -\frac {4 \left (a x^4+b x\right )^{3/4} \left (4 a x^3+3 b x^3+b\right )}{9 b x^3 \left (a x^3+b\right )}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4+b x\right )^{3/4}}{a x^3+b}\right )}{3 \sqrt [4]{a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4+b x\right )^{3/4}}{a x^3+b}\right )}{3 \sqrt [4]{a}} \]

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Rubi [A]  time = 0.37, antiderivative size = 183, normalized size of antiderivative = 1.49, number of steps used = 12, number of rules used = 11, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2056, 1489, 271, 264, 288, 329, 275, 240, 212, 206, 203} \begin {gather*} -\frac {16 a x}{9 b \sqrt [4]{a x^4+b x}}-\frac {4 x}{3 \sqrt [4]{a x^4+b x}}-\frac {4}{9 x^2 \sqrt [4]{a x^4+b x}}+\frac {2 \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} \sqrt [4]{a x^4+b x}}+\frac {2 \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} \sqrt [4]{a x^4+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^6)/(x^3*(b + a*x^3)*(b*x + a*x^4)^(1/4)),x]

[Out]

-4/(9*x^2*(b*x + a*x^4)^(1/4)) - (4*x)/(3*(b*x + a*x^4)^(1/4)) - (16*a*x)/(9*b*(b*x + a*x^4)^(1/4)) + (2*x^(1/
4)*(b + a*x^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(3*a^(1/4)*(b*x + a*x^4)^(1/4)) + (2*x^(1/4)
*(b + a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(3*a^(1/4)*(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 264

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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 1489

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[Expan
dIntegrand[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {b+a x^6}{x^{13/4} \left (b+a x^3\right )^{5/4}} \, dx}{\sqrt [4]{b x+a x^4}}\\ &=\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \left (\frac {b}{x^{13/4} \left (b+a x^3\right )^{5/4}}+\frac {a x^{11/4}}{\left (b+a x^3\right )^{5/4}}\right ) \, dx}{\sqrt [4]{b x+a x^4}}\\ &=\frac {\left (a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {x^{11/4}}{\left (b+a x^3\right )^{5/4}} \, dx}{\sqrt [4]{b x+a x^4}}+\frac {\left (b \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {1}{x^{13/4} \left (b+a x^3\right )^{5/4}} \, dx}{\sqrt [4]{b x+a x^4}}\\ &=-\frac {4}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}+\frac {\left (\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 {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {1}{\sqrt [4]{x} \left (b+a x^3\right )^{5/4}} \, dx}{3 \sqrt [4]{b x+a x^4}}\\ &=-\frac {4}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}-\frac {16 a x}{9 b \sqrt [4]{b x+a x^4}}+\frac {\left (4 \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}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}-\frac {16 a x}{9 b \sqrt [4]{b x+a x^4}}+\frac {\left (4 \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}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}-\frac {16 a x}{9 b \sqrt [4]{b x+a x^4}}+\frac {\left (4 \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}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}-\frac {16 a x}{9 b \sqrt [4]{b x+a x^4}}+\frac {\left (2 \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 \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}{9 x^2 \sqrt [4]{b x+a x^4}}-\frac {4 x}{3 \sqrt [4]{b x+a x^4}}-\frac {16 a x}{9 b \sqrt [4]{b x+a x^4}}+\frac {2 \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]{a} \sqrt [4]{b x+a x^4}}+\frac {2 \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]{a} \sqrt [4]{b x+a x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 73, normalized size = 0.59 \begin {gather*} \frac {4 \left (3 a x^6 \sqrt [4]{\frac {a x^3}{b}+1} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};-\frac {a x^3}{b}\right )-5 \left (4 a x^3+b\right )\right )}{45 b x^2 \sqrt [4]{x \left (a x^3+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^6)/(x^3*(b + a*x^3)*(b*x + a*x^4)^(1/4)),x]

[Out]

(4*(-5*(b + 4*a*x^3) + 3*a*x^6*(1 + (a*x^3)/b)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((a*x^3)/b)]))/(45*b*x^
2*(x*(b + a*x^3))^(1/4))

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IntegrateAlgebraic [A]  time = 0.49, size = 123, normalized size = 1.00 \begin {gather*} -\frac {4 \left (b+4 a x^3+3 b x^3\right ) \left (b x+a x^4\right )^{3/4}}{9 b x^3 \left (b+a x^3\right )}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^6)/(x^3*(b + a*x^3)*(b*x + a*x^4)^(1/4)),x]

[Out]

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

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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^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.39, size = 217, normalized size = 1.76 \begin {gather*} \frac {\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 )}{3 \, a} + \frac {\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 )}{3 \, a} - \frac {\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 )}{6 \, a} + \frac {\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 )}{6 \, a} - \frac {4 \, {\left (a + b\right )}}{3 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} b} - \frac {4 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{4}}}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^3/(a*x^3+b)/(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))/a + 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))/a - 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))/a + 1/6*sqrt(2)*(-a)^(3/4)*log(-s
qrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3))/a - 4/3*(a + b)/((a + b/x^3)^(1/4)*b) - 4/9*
(a + b/x^3)^(3/4)/b

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{6}+b}{x^{3} \left (a \,x^{3}+b \right ) \left (a \,x^{4}+b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x^6+b)/x^3/(a*x^3+b)/(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}} {\left (a x^{3} + b\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a\,x^6+b}{x^3\,{\left (a\,x^4+b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} + b}{x^{3} \sqrt [4]{x \left (a x^{3} + b\right )} \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**3/(a*x**3+b)/(a*x**4+b*x)**(1/4),x)

[Out]

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

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