3.994 \(\int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx\)

Optimal. Leaf size=75 \[ -\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{4 x^4} \]

[Out]

-1/4*(b*x^4+a)^(1/4)/x^4-1/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-1/8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/
a^(3/4)

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Rubi [A]  time = 0.04, antiderivative size = 75, 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, 47, 63, 212, 206, 203} \[ -\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(1/4)/x^5,x]

[Out]

-(a + b*x^4)^(1/4)/(4*x^4) - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(3/4)) - (b*ArcTanh[(a + b*x^4)^(1/4)/
a^(1/4)])/(8*a^(3/4))

Rule 47

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

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt [4]{a+b x}}{x^2} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}+\frac {1}{16} b \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )\\ &=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 \sqrt {a}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 \sqrt {a}}\\ &=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 37, normalized size = 0.49 \[ \frac {b \left (a+b x^4\right )^{5/4} \, _2F_1\left (\frac {5}{4},2;\frac {9}{4};\frac {b x^4}{a}+1\right )}{5 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(1/4)/x^5,x]

[Out]

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

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fricas [B]  time = 0.71, size = 180, normalized size = 2.40 \[ \frac {4 \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}} - \sqrt {\sqrt {b x^{4} + a} b^{2} + a^{2} \sqrt {\frac {b^{4}}{a^{3}}}} a^{2} \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}}}{b^{4}}\right ) - \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b + a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b - a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(b^4/a^3)^(1/4)*x^4*arctan(-((b*x^4 + a)^(1/4)*a^2*b*(b^4/a^3)^(3/4) - sqrt(sqrt(b*x^4 + a)*b^2 + a^2*
sqrt(b^4/a^3))*a^2*(b^4/a^3)^(3/4))/b^4) - (b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b + a*(b^4/a^3)^(1/4)) +
(b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b - a*(b^4/a^3)^(1/4)) - 4*(b*x^4 + a)^(1/4))/x^4

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giac [B]  time = 0.18, size = 207, normalized size = 2.76 \[ \frac {\frac {2 \, \sqrt {2} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x^{4}}}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 2*s
qrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + sqrt(2)*b^2
*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(3/4) + sqrt(2)*(-a)^(1/4)*b^2*lo
g(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a - 8*(b*x^4 + a)^(1/4)*b/x^4)/b

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.02, size = 73, normalized size = 0.97 \[ -\frac {b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{8 \, a^{\frac {3}{4}}} + \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{16 \, a^{\frac {3}{4}}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) + 1/16*b*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4)
+ a^(1/4)))/a^(3/4) - 1/4*(b*x^4 + a)^(1/4)/x^4

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mupad [B]  time = 1.28, size = 55, normalized size = 0.73 \[ -\frac {{\left (b\,x^4+a\right )}^{1/4}}{4\,x^4}-\frac {b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}}-\frac {b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(1/4)/x^5,x)

[Out]

- (a + b*x^4)^(1/4)/(4*x^4) - (b*atan((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(3/4)) - (b*atanh((a + b*x^4)^(1/4)/a^(
1/4)))/(8*a^(3/4))

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sympy [C]  time = 2.86, size = 41, normalized size = 0.55 \[ - \frac {\sqrt [4]{b} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(1/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4))/(4*x**3*gamma(7/4))

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