\(\int \frac {(a+b x^4)^{5/4}}{x^9} \, dx\) [1054]

Optimal result

Integrand size = 15, antiderivative size = 98 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=-\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}-\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}} \]

[Out]

-5/32*b*(b*x^4+a)^(1/4)/x^4-1/8*(b*x^4+a)^(5/4)/x^8-5/64*b^2*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-5/64*b^2*
arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 43, 65, 218, 212, 209} \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=-\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8} \]

[In]

Int[(a + b*x^4)^(5/4)/x^9,x]

[Out]

(-5*b*(a + b*x^4)^(1/4))/(32*x^4) - (a + b*x^4)^(5/4)/(8*x^8) - (5*b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(64*
a^(3/4)) - (5*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(3/4))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 218

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] &&  !Gt
Q[a/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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {(a+b x)^{5/4}}{x^3} \, dx,x,x^4\right ) \\ & = -\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}+\frac {1}{32} (5 b) \text {Subst}\left (\int \frac {\sqrt [4]{a+b x}}{x^2} \, dx,x,x^4\right ) \\ & = -\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}+\frac {1}{128} \left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right ) \\ & = -\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}+\frac {1}{32} (5 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right ) \\ & = -\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 \sqrt {a}}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 \sqrt {a}} \\ & = -\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8}-\frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=\frac {\left (-4 a-9 b x^4\right ) \sqrt [4]{a+b x^4}}{32 x^8}-\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}} \]

[In]

Integrate[(a + b*x^4)^(5/4)/x^9,x]

[Out]

((-4*a - 9*b*x^4)*(a + b*x^4)^(1/4))/(32*x^8) - (5*b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(3/4)) - (5*b^
2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(3/4))

Maple [A] (verified)

Time = 4.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {-10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b^{2} x^{8}-5 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b^{2} x^{8}-36 b \,x^{4} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {3}{4}}-16 a^{\frac {7}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{128 x^{8} a^{\frac {3}{4}}}\)	\(108\)

[In]

int((b*x^4+a)^(5/4)/x^9,x,method=_RETURNVERBOSE)

[Out]

1/128*(-10*arctan((b*x^4+a)^(1/4)/a^(1/4))*b^2*x^8-5*ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))
*b^2*x^8-36*b*x^4*(b*x^4+a)^(1/4)*a^(3/4)-16*a^(7/4)*(b*x^4+a)^(1/4))/x^8/a^(3/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=-\frac {5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} + 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) + 5 i \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} + 5 i \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) - 5 i \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} - 5 i \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) - 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} - 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) + 4 \, {\left (9 \, b x^{4} + 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, x^{8}} \]

[In]

integrate((b*x^4+a)^(5/4)/x^9,x, algorithm="fricas")

[Out]

-1/128*(5*(b^8/a^3)^(1/4)*x^8*log(5*(b*x^4 + a)^(1/4)*b^2 + 5*(b^8/a^3)^(1/4)*a) + 5*I*(b^8/a^3)^(1/4)*x^8*log
(5*(b*x^4 + a)^(1/4)*b^2 + 5*I*(b^8/a^3)^(1/4)*a) - 5*I*(b^8/a^3)^(1/4)*x^8*log(5*(b*x^4 + a)^(1/4)*b^2 - 5*I*
(b^8/a^3)^(1/4)*a) - 5*(b^8/a^3)^(1/4)*x^8*log(5*(b*x^4 + a)^(1/4)*b^2 - 5*(b^8/a^3)^(1/4)*a) + 4*(9*b*x^4 + 4
*a)*(b*x^4 + a)^(1/4))/x^8

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=- \frac {b^{\frac {5}{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((b*x**4+a)**(5/4)/x**9,x)

[Out]

-b**(5/4)*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4))/(4*x**3*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=-\frac {5 \, b^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{64 \, a^{\frac {3}{4}}} + \frac {5 \, b^{2} \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{128 \, a^{\frac {3}{4}}} - \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2} - 5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} - 2 \, {\left (b x^{4} + a\right )} a + a^{2}\right )}} \]

[In]

integrate((b*x^4+a)^(5/4)/x^9,x, algorithm="maxima")

[Out]

-5/64*b^2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) + 5/128*b^2*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)
^(1/4) + a^(1/4)))/a^(3/4) - 1/32*(9*(b*x^4 + a)^(5/4)*b^2 - 5*(b*x^4 + a)^(1/4)*a*b^2)/((b*x^4 + a)^2 - 2*(b*
x^4 + a)*a + a^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (74) = 148\).

Time = 0.30 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.37 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=\frac {\frac {10 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {10 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} - \frac {8 \, {\left (9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{3} - 5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{3}\right )}}{b^{2} x^{8}}}{256 \, b} \]

[In]

integrate((b*x^4+a)^(5/4)/x^9,x, algorithm="giac")

[Out]

1/256*(10*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 1
0*sqrt(2)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 5*sqrt(2
)*b^3*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(3/4) + 5*sqrt(2)*(-a)^(1/4)
*b^3*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a - 8*(9*(b*x^4 + a)^(5/4)*b^3 -
5*(b*x^4 + a)^(1/4)*a*b^3)/(b^2*x^8))/b

Mupad [B] (verification not implemented)

Time = 6.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^9} \, dx=\frac {5\,a\,{\left (b\,x^4+a\right )}^{1/4}}{32\,x^8}-\frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{3/4}}-\frac {9\,{\left (b\,x^4+a\right )}^{5/4}}{32\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{3/4}} \]

[In]

int((a + b*x^4)^(5/4)/x^9,x)

[Out]

(b^2*atan(((a + b*x^4)^(1/4)*1i)/a^(1/4))*5i)/(64*a^(3/4)) - (5*b^2*atan((a + b*x^4)^(1/4)/a^(1/4)))/(64*a^(3/
4)) - (9*(a + b*x^4)^(5/4))/(32*x^8) + (5*a*(a + b*x^4)^(1/4))/(32*x^8)