3.1066 \(\int \frac{(a+b x^4)^{5/4}}{x^6} \, dx\)

Optimal. Leaf size=92 \[ -\frac{1}{2} b^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac{1}{2} b^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac{b \sqrt [4]{a+b x^4}}{x}-\frac{\left (a+b x^4\right )^{5/4}}{5 x^5} \]

[Out]

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

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Rubi [A]  time = 0.0345465, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {277, 331, 298, 203, 206} \[ -\frac{1}{2} b^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac{1}{2} b^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac{b \sqrt [4]{a+b x^4}}{x}-\frac{\left (a+b x^4\right )^{5/4}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^6,x]

[Out]

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

Rule 277

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

Rubi steps

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

Mathematica [C]  time = 0.009268, size = 52, normalized size = 0.57 \[ -\frac{a \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{5}{4},-\frac{5}{4};-\frac{1}{4};-\frac{b x^4}{a}\right )}{5 x^5 \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^6,x]

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{6}} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(5/4)/x^6,x)

[Out]

int((b*x^4+a)^(5/4)/x^6,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="fricas")

[Out]

Timed out

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Sympy [C]  time = 2.84639, size = 46, normalized size = 0.5 \begin{align*} \frac{a^{\frac{5}{4}} \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{5}{4} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(5/4)/x**6,x)

[Out]

a**(5/4)*gamma(-5/4)*hyper((-5/4, -5/4), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4))

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Giac [B]  time = 1.17696, size = 316, normalized size = 3.43 \begin{align*} \frac{1}{4} \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} b \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right ) + \frac{1}{4} \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} b \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right ) + \frac{1}{8} \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} b \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right ) - \frac{1}{8} \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} b \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right ) - \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}}{\left (b + \frac{a}{x^{4}}\right )}}{5 \, x} - \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}} b}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="giac")

[Out]

1/4*sqrt(2)*(-b)^(1/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4)) + 1/4*sqr
t(2)*(-b)^(1/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4)) + 1/8*sqrt(2)*(
-b)^(1/4)*b*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2) - 1/8*sqrt(2)*(-b)^(1
/4)*b*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2) - 1/5*(b*x^4 + a)^(1/4)*(b
+ a/x^4)/x - (b*x^4 + a)^(1/4)*b/x