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/(b*x^4+a)^(1/4)+1/2*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)-1/2*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)

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Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [B]  time = 0.89, size = 164, normalized size = 2.34 \[ -\frac {4 \, {\left (a b x^{4} + a^{2}\right )} \frac {1}{a^{5}}^{\frac {1}{4}} \arctan \left (\sqrt {a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {b x^{4} + a}} a \frac {1}{a^{5}}^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a \frac {1}{a^{5}}^{\frac {1}{4}}\right ) + {\left (a b x^{4} + a^{2}\right )} \frac {1}{a^{5}}^{\frac {1}{4}} \log \left (a^{4} \frac {1}{a^{5}}^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right ) - {\left (a b x^{4} + a^{2}\right )} \frac {1}{a^{5}}^{\frac {1}{4}} \log \left (-a^{4} \frac {1}{a^{5}}^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{4 \, {\left (a b x^{4} + a^{2}\right )}} \]

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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giac [B]  time = 0.17, size = 199, normalized size = 2.84 \[ -\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a^{2}} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a^{2}} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a^{2}} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a^{2}} + \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a} \]

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)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {5}{4}} x}\, dx \]

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 [A]  time = 3.09, size = 75, normalized size = 1.07 \[ \frac {\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}}{4 \, a} + \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.23, size = 52, normalized size = 0.74 \[ \frac {\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}}+\frac {1}{a\,{\left (b\,x^4+a\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^4)^(5/4)),x)

[Out]

atan((a + b*x^4)^(1/4)/a^(1/4))/(2*a^(5/4)) - atanh((a + b*x^4)^(1/4)/a^(1/4))/(2*a^(5/4)) + 1/(a*(a + b*x^4)^
(1/4))

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sympy [C]  time = 1.82, size = 39, normalized size = 0.56 \[ - \frac {\Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{5} \Gamma \left (\frac {9}{4}\right )} \]

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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