\(\int \frac {(a+b x^4)^{5/4}}{x^5} \, dx\) [1053]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=\frac {5}{4} b \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}-\frac {5}{8} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {5}{8} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]

[Out]

5/4*b*(b*x^4+a)^(1/4)-1/4*(b*x^4+a)^(5/4)/x^4-5/8*a^(1/4)*b*arctan((b*x^4+a)^(1/4)/a^(1/4))-5/8*a^(1/4)*b*arct
anh((b*x^4+a)^(1/4)/a^(1/4))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {272, 43, 52, 65, 218, 212, 209} \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=-\frac {5}{8} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {5}{8} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}+\frac {5}{4} b \sqrt [4]{a+b x^4} \]

[In]

Int[(a + b*x^4)^(5/4)/x^5,x]

[Out]

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

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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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^2} \, dx,x,x^4\right ) \\ & = -\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}+\frac {1}{16} (5 b) \text {Subst}\left (\int \frac {\sqrt [4]{a+b x}}{x} \, dx,x,x^4\right ) \\ & = \frac {5}{4} b \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}+\frac {1}{16} (5 a b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right ) \\ & = \frac {5}{4} b \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}+\frac {1}{4} (5 a) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right ) \\ & = \frac {5}{4} b \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}-\frac {1}{8} \left (5 \sqrt {a} b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )-\frac {1}{8} \left (5 \sqrt {a} b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right ) \\ & = \frac {5}{4} b \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{4 x^4}-\frac {5}{8} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {5}{8} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=\frac {1}{8} \left (-\frac {2 \left (a-4 b x^4\right ) \sqrt [4]{a+b x^4}}{x^4}-5 \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-5 \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )\right ) \]

[In]

Integrate[(a + b*x^4)^(5/4)/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 4.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {-5 b \,x^{4} \left (\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right ) a^{\frac {1}{4}}-4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-4 b \,x^{4}+a \right )}{16 x^{4}}\)	\(88\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=-\frac {5 \, \left (a b^{4}\right )^{\frac {1}{4}} x^{4} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b + 5 \, \left (a b^{4}\right )^{\frac {1}{4}}\right ) + 5 i \, \left (a b^{4}\right )^{\frac {1}{4}} x^{4} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b + 5 i \, \left (a b^{4}\right )^{\frac {1}{4}}\right ) - 5 i \, \left (a b^{4}\right )^{\frac {1}{4}} x^{4} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b - 5 i \, \left (a b^{4}\right )^{\frac {1}{4}}\right ) - 5 \, \left (a b^{4}\right )^{\frac {1}{4}} x^{4} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b - 5 \, \left (a b^{4}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (4 \, b x^{4} - a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, x^{4}} \]

[In]

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

[Out]

-1/16*(5*(a*b^4)^(1/4)*x^4*log(5*(b*x^4 + a)^(1/4)*b + 5*(a*b^4)^(1/4)) + 5*I*(a*b^4)^(1/4)*x^4*log(5*(b*x^4 +
a)^(1/4)*b + 5*I*(a*b^4)^(1/4)) - 5*I*(a*b^4)^(1/4)*x^4*log(5*(b*x^4 + a)^(1/4)*b - 5*I*(a*b^4)^(1/4)) - 5*(a
*b^4)^(1/4)*x^4*log(5*(b*x^4 + a)^(1/4)*b - 5*(a*b^4)^(1/4)) - 4*(4*b*x^4 - a)*(b*x^4 + a)^(1/4))/x^4

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=- \frac {b^{\frac {5}{4}} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

-b**(5/4)*x*gamma(-1/4)*hyper((-5/4, -1/4), (3/4,), a*exp_polar(I*pi)/(b*x**4))/(4*gamma(3/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=-\frac {5}{16} \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )} a + {\left (b x^{4} + a\right )}^{\frac {1}{4}} b - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (67) = 134\).

Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.43 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=-\frac {10 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + 10 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + 5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) - 5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) - 32 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} + \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b}{x^{4}}}{32 \, b} \]

[In]

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

[Out]

-1/32*(10*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4)) + 1
0*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4)) + 5*sqrt(2
)*(-a)^(1/4)*b^2*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) - 5*sqrt(2)*(-a)^(1/4)
*b^2*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) - 32*(b*x^4 + a)^(1/4)*b^2 + 8*(b
*x^4 + a)^(1/4)*a*b/x^4)/b

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^5} \, dx=b\,{\left (b\,x^4+a\right )}^{1/4}-\frac {a\,{\left (b\,x^4+a\right )}^{1/4}}{4\,x^4}-\frac {5\,a^{1/4}\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8}+\frac {a^{1/4}\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{8} \]

[In]

int((a + b*x^4)^(5/4)/x^5,x)

[Out]

b*(a + b*x^4)^(1/4) - (a*(a + b*x^4)^(1/4))/(4*x^4) - (5*a^(1/4)*b*atan((a + b*x^4)^(1/4)/a^(1/4)))/8 + (a^(1/
4)*b*atan(((a + b*x^4)^(1/4)*1i)/a^(1/4))*5i)/8