3.12.45 \(\int \frac {1}{x^5 (a+b x^4)^{5/4}} \, dx\) [1145]

Optimal. Leaf size=97 \[ -\frac {5 b}{4 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{4 a x^4 \sqrt [4]{a+b x^4}}-\frac {5 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}+\frac {5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \]

[Out]

-5/4*b/a^2/(b*x^4+a)^(1/4)-1/4/a/x^4/(b*x^4+a)^(1/4)-5/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)+5/8*b*arcta
nh((b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {272, 44, 53, 65, 304, 209, 212} \begin {gather*} -\frac {5 b \text {ArcTan}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}+\frac {5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}-\frac {5 b}{4 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{4 a x^4 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^4)^(5/4)),x]

[Out]

(-5*b)/(4*a^2*(a + b*x^4)^(1/4)) - 1/(4*a*x^4*(a + b*x^4)^(1/4)) - (5*b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*
a^(9/4)) + (5*b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(9/4))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 272

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 304

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^5 \left (a+b x^4\right )^{5/4}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{5/4}} \, dx,x,x^4\right )\\ &=\frac {1}{a x^4 \sqrt [4]{a+b x^4}}+\frac {5 \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{4 a}\\ &=\frac {1}{a x^4 \sqrt [4]{a+b x^4}}-\frac {5 \left (a+b x^4\right )^{3/4}}{4 a^2 x^4}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{x \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{16 a^2}\\ &=\frac {1}{a x^4 \sqrt [4]{a+b x^4}}-\frac {5 \left (a+b x^4\right )^{3/4}}{4 a^2 x^4}-\frac {5 \text {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )}{4 a^2}\\ &=\frac {1}{a x^4 \sqrt [4]{a+b x^4}}-\frac {5 \left (a+b x^4\right )^{3/4}}{4 a^2 x^4}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 a^2}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 a^2}\\ &=\frac {1}{a x^4 \sqrt [4]{a+b x^4}}-\frac {5 \left (a+b x^4\right )^{3/4}}{4 a^2 x^4}-\frac {5 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}+\frac {5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 81, normalized size = 0.84 \begin {gather*} \frac {-\frac {2 \sqrt [4]{a} \left (a+5 b x^4\right )}{x^4 \sqrt [4]{a+b x^4}}-5 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a^(1/4)*(a + 5*b*x^4))/(x^4*(a + b*x^4)^(1/4)) - 5*b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)] + 5*b*ArcTanh[(a +
b*x^4)^(1/4)/a^(1/4)])/(8*a^(9/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{5} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 110, normalized size = 1.13 \begin {gather*} -\frac {5 \, {\left (b x^{4} + a\right )} b - 4 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}} - \frac {5 \, b {\left (\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}}}\right )}}{16 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*(b*x^4 + a)*b - 4*a*b)/((b*x^4 + a)^(5/4)*a^2 - (b*x^4 + a)^(1/4)*a^3) - 5/16*b*(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^2

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (73) = 146\).
time = 0.38, size = 259, normalized size = 2.67 \begin {gather*} \frac {20 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b^{3} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} - \sqrt {a^{5} b^{4} \sqrt {\frac {b^{4}}{a^{9}}} + \sqrt {b x^{4} + a} b^{6}} a^{2} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}}}{b^{4}}\right ) + 5 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 5 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 4 \, {\left (5 \, b x^{4} + a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(20*(a^2*b*x^8 + a^3*x^4)*(b^4/a^9)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*a^2*b^3*(b^4/a^9)^(1/4) - sqrt(a^5*b
^4*sqrt(b^4/a^9) + sqrt(b*x^4 + a)*b^6)*a^2*(b^4/a^9)^(1/4))/b^4) + 5*(a^2*b*x^8 + a^3*x^4)*(b^4/a^9)^(1/4)*lo
g(125*a^7*(b^4/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^3) - 5*(a^2*b*x^8 + a^3*x^4)*(b^4/a^9)^(1/4)*log(-125*a^7*
(b^4/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^3) - 4*(5*b*x^4 + a)*(b*x^4 + a)^(3/4))/(a^2*b*x^8 + a^3*x^4)

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Sympy [C] Result contains complex when optimal does not.
time = 0.96, size = 39, normalized size = 0.40 \begin {gather*} - \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{9} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(9/4)*hyper((5/4, 9/4), (13/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(5/4)*x**9*gamma(13/4))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (73) = 146\).
time = 1.51, size = 234, normalized size = 2.41 \begin {gather*} \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \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 )}{16 \, a^{3}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \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 )}{16 \, a^{3}} + \frac {5 \, \sqrt {2} b \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 )}{32 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \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 )}{32 \, a^{3}} - \frac {5 \, {\left (b x^{4} + a\right )} b - 4 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*sqrt(2)*(-a)^(3/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^3 + 5/16
*sqrt(2)*(-a)^(3/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^3 + 5/32*sq
rt(2)*b*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) + 5/32*sqrt(2)
*(-a)^(3/4)*b*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^3 - 1/4*(5*(b*x^4 + a)
*b - 4*a*b)/(((b*x^4 + a)^(5/4) - (b*x^4 + a)^(1/4)*a)*a^2)

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Mupad [B]
time = 1.39, size = 87, normalized size = 0.90 \begin {gather*} \frac {5\,b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {5\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {\frac {b}{a}-\frac {5\,b\,\left (b\,x^4+a\right )}{4\,a^2}}{a\,{\left (b\,x^4+a\right )}^{1/4}-{\left (b\,x^4+a\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^4)^(5/4)),x)

[Out]

(5*b*atanh((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(9/4)) - (5*b*atan((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(9/4)) - (b/a
- (5*b*(a + b*x^4))/(4*a^2))/(a*(a + b*x^4)^(1/4) - (a + b*x^4)^(5/4))

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