3.18.84 \(\int \frac {b+a x^8}{x^6 (-b+a x^4) (b+a x^4)^{3/4}} \, dx\)

Optimal. Leaf size=120 \[ \frac {\left (a^{5/4}+\sqrt [4]{a} b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} b^2}-\frac {\left (a^{5/4}+\sqrt [4]{a} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} b^2}+\frac {\left (a x^4+b\right )^{5/4}}{5 b^2 x^5} \]

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Rubi [A]  time = 0.74, antiderivative size = 132, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {6725, 271, 264, 494, 298, 203, 206} \begin {gather*} \frac {a \sqrt [4]{a x^4+b}}{5 b^2 x}+\frac {\sqrt [4]{a} (a+b) \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} b^2}-\frac {\sqrt [4]{a} (a+b) \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} b^2}+\frac {\sqrt [4]{a x^4+b}}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^8)/(x^6*(-b + a*x^4)*(b + a*x^4)^(3/4)),x]

[Out]

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

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 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 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 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C]  time = 0.12, size = 65, normalized size = 0.54 \begin {gather*} \frac {3 \left (a x^4+b\right )^2-5 a x^8 (a+b) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {2 a x^4}{a x^4+b}\right )}{15 b^2 x^5 \left (a x^4+b\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^8)/(x^6*(-b + a*x^4)*(b + a*x^4)^(3/4)),x]

[Out]

(3*(b + a*x^4)^2 - 5*a*(a + b)*x^8*Hypergeometric2F1[3/4, 1, 7/4, (2*a*x^4)/(b + a*x^4)])/(15*b^2*x^5*(b + a*x
^4)^(3/4))

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IntegrateAlgebraic [A]  time = 1.17, size = 120, normalized size = 1.00 \begin {gather*} \frac {\left (b+a x^4\right )^{5/4}}{5 b^2 x^5}+\frac {\left (a^{5/4}+\sqrt [4]{a} b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} b^2}-\frac {\left (a^{5/4}+\sqrt [4]{a} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^8)/(x^6*(-b + a*x^4)*(b + a*x^4)^(3/4)),x]

[Out]

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

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{8} + b}{{\left (a x^{4} + b\right )}^{\frac {3}{4}} {\left (a x^{4} - b\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b)/x^6/(a*x^4-b)/(a*x^4+b)^(3/4),x, algorithm="giac")

[Out]

integrate((a*x^8 + b)/((a*x^4 + b)^(3/4)*(a*x^4 - b)*x^6), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{8}+b}{x^{6} \left (a \,x^{4}-b \right ) \left (a \,x^{4}+b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8+b)/x^6/(a*x^4-b)/(a*x^4+b)^(3/4),x)

[Out]

int((a*x^8+b)/x^6/(a*x^4-b)/(a*x^4+b)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{8} + b}{{\left (a x^{4} + b\right )}^{\frac {3}{4}} {\left (a x^{4} - b\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b)/x^6/(a*x^4-b)/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

integrate((a*x^8 + b)/((a*x^4 + b)^(3/4)*(a*x^4 - b)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {a\,x^8+b}{x^6\,{\left (a\,x^4+b\right )}^{3/4}\,\left (b-a\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b + a*x^8)/(x^6*(b + a*x^4)^(3/4)*(b - a*x^4)),x)

[Out]

-int((b + a*x^8)/(x^6*(b + a*x^4)^(3/4)*(b - a*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{8} + b}{x^{6} \left (a x^{4} - b\right ) \left (a x^{4} + b\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**8+b)/x**6/(a*x**4-b)/(a*x**4+b)**(3/4),x)

[Out]

Integral((a*x**8 + b)/(x**6*(a*x**4 - b)*(a*x**4 + b)**(3/4)), x)

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