3.1144 \(\int \frac{1}{x (a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=70 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}}+\frac{1}{a \sqrt [4]{a+b x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0443256, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 51, 63, 298, 203, 206} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}}+\frac{1}{a \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^4)^(5/4)),x]

[Out]

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

Rule 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0064257, size = 33, normalized size = 0.47 \[ \frac{\, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\frac{b x^4}{a}+1\right )}{a \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^4)^(5/4)),x]

[Out]

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

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.64399, size = 455, normalized size = 6.5 \begin{align*} -\frac{4 \,{\left (a b x^{4} + a^{2}\right )} \frac{1}{a^{5}}^{\frac{1}{4}} \arctan \left (\sqrt{a^{3} \sqrt{\frac{1}{a^{5}}} + \sqrt{b x^{4} + a}} a \frac{1}{a^{5}}^{\frac{1}{4}} -{\left (b x^{4} + a\right )}^{\frac{1}{4}} a \frac{1}{a^{5}}^{\frac{1}{4}}\right ) +{\left (a b x^{4} + a^{2}\right )} \frac{1}{a^{5}}^{\frac{1}{4}} \log \left (a^{4} \frac{1}{a^{5}}^{\frac{3}{4}} +{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right ) -{\left (a b x^{4} + a^{2}\right )} \frac{1}{a^{5}}^{\frac{1}{4}} \log \left (-a^{4} \frac{1}{a^{5}}^{\frac{3}{4}} +{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right ) - 4 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{4 \,{\left (a b x^{4} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(a*b*x^4 + a^2)*(a^(-5))^(1/4)*arctan(sqrt(a^3*sqrt(a^(-5)) + sqrt(b*x^4 + a))*a*(a^(-5))^(1/4) - (b*x
^4 + a)^(1/4)*a*(a^(-5))^(1/4)) + (a*b*x^4 + a^2)*(a^(-5))^(1/4)*log(a^4*(a^(-5))^(3/4) + (b*x^4 + a)^(1/4)) -
(a*b*x^4 + a^2)*(a^(-5))^(1/4)*log(-a^4*(a^(-5))^(3/4) + (b*x^4 + a)^(1/4)) - 4*(b*x^4 + a)^(3/4))/(a*b*x^4 +
a^2)

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Sympy [C]  time = 1.20153, size = 39, normalized size = 0.56 \begin{align*} - \frac{\Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac{5}{4}} x^{5} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(5/4)*hyper((5/4, 5/4), (9/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(5/4)*x**5*gamma(9/4))

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Giac [B]  time = 1.12763, size = 269, normalized size = 3.84 \begin{align*} -\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 (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right )}{4 \, a^{2}} - \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 (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right )}{4 \, a^{2}} + \frac{\sqrt{2} \left (-a\right )^{\frac{3}{4}} \log \left (\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right )}{8 \, a^{2}} - \frac{\sqrt{2} \left (-a\right )^{\frac{3}{4}} \log \left (-\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right )}{8 \, a^{2}} + \frac{1}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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