Integrand size = 15, antiderivative size = 81 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=-a b \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} b \left (a+\frac {b}{x^4}\right )^{3/2}+\frac {1}{4} a^2 \sqrt {a+\frac {b}{x^4}} x^4+\frac {5}{4} a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \] Output:
-a*b*(a+b/x^4)^(1/2)-1/6*b*(a+b/x^4)^(3/2)+1/4*a^2*(a+b/x^4)^(1/2)*x^4+5/4 *a^(3/2)*b*arctanh((a+b/x^4)^(1/2)/a^(1/2))
Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {\sqrt {a+\frac {b}{x^4}} \left (\sqrt {b+a x^4} \left (-2 b^2-14 a b x^4+3 a^2 x^8\right )+15 a^{3/2} b x^6 \log \left (\sqrt {a} x^2+\sqrt {b+a x^4}\right )\right )}{12 x^4 \sqrt {b+a x^4}} \] Input:
Integrate[(a + b/x^4)^(5/2)*x^3,x]
Output:
(Sqrt[a + b/x^4]*(Sqrt[b + a*x^4]*(-2*b^2 - 14*a*b*x^4 + 3*a^2*x^8) + 15*a ^(3/2)*b*x^6*Log[Sqrt[a]*x^2 + Sqrt[b + a*x^4]]))/(12*x^4*Sqrt[b + a*x^4])
Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 51, 60, 60, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (a+\frac {b}{x^4}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{4} \int \left (a+\frac {b}{x^4}\right )^{5/2} x^8d\frac {1}{x^4}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{2} b \int \left (a+\frac {b}{x^4}\right )^{3/2} x^4d\frac {1}{x^4}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{2} b \left (a \int \sqrt {a+\frac {b}{x^4}} x^4d\frac {1}{x^4}+\frac {2}{3} \left (a+\frac {b}{x^4}\right )^{3/2}\right )\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{2} b \left (a \left (a \int \frac {x^4}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x^4}+2 \sqrt {a+\frac {b}{x^4}}\right )+\frac {2}{3} \left (a+\frac {b}{x^4}\right )^{3/2}\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{2} b \left (a \left (\frac {2 a \int \frac {1}{\frac {1}{b x^8}-\frac {a}{b}}d\sqrt {a+\frac {b}{x^4}}}{b}+2 \sqrt {a+\frac {b}{x^4}}\right )+\frac {2}{3} \left (a+\frac {b}{x^4}\right )^{3/2}\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{2} b \left (a \left (2 \sqrt {a+\frac {b}{x^4}}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+\frac {b}{x^4}\right )^{3/2}\right )\right )\) |
Input:
Int[(a + b/x^4)^(5/2)*x^3,x]
Output:
((a + b/x^4)^(5/2)*x^4 - (5*b*((2*(a + b/x^4)^(3/2))/3 + a*(2*Sqrt[a + b/x ^4] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b/x^4]/Sqrt[a]])))/2)/4
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11
method | result | size |
risch | \(\frac {\left (3 a^{2} x^{8}-14 a b \,x^{4}-2 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{12 x^{4}}+\frac {5 a^{\frac {3}{2}} b \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{4 \sqrt {a \,x^{4}+b}}\) | \(90\) |
default | \(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{4} \left (3 a^{2} x^{8} \sqrt {a \,x^{4}+b}+15 a^{\frac {3}{2}} b \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right ) x^{6}-14 a b \sqrt {a \,x^{4}+b}\, x^{4}-2 b^{2} \sqrt {a \,x^{4}+b}\right )}{12 \left (a \,x^{4}+b \right )^{\frac {5}{2}}}\) | \(103\) |
Input:
int((a+b/x^4)^(5/2)*x^3,x,method=_RETURNVERBOSE)
Output:
1/12*(3*a^2*x^8-14*a*b*x^4-2*b^2)/x^4*((a*x^4+b)/x^4)^(1/2)+5/4*a^(3/2)*b* ln(a^(1/2)*x^2+(a*x^4+b)^(1/2))*((a*x^4+b)/x^4)^(1/2)*x^2/(a*x^4+b)^(1/2)
Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.99 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b x^{4} \log \left (-2 \, a x^{4} - 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + 2 \, {\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{24 \, x^{4}}, -\frac {15 \, \sqrt {-a} a b x^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a}\right ) - {\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, x^{4}}\right ] \] Input:
integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="fricas")
Output:
[1/24*(15*a^(3/2)*b*x^4*log(-2*a*x^4 - 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) + 2*(3*a^2*x^8 - 14*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4, -1/ 12*(15*sqrt(-a)*a*b*x^4*arctan(sqrt(-a)*sqrt((a*x^4 + b)/x^4)/a) - (3*a^2* x^8 - 14*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4]
Time = 3.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {a^{\frac {5}{2}} x^{4} \sqrt {1 + \frac {b}{a x^{4}}}}{4} - \frac {7 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{4}}}}{6} - \frac {5 a^{\frac {3}{2}} b \log {\left (\frac {b}{a x^{4}} \right )}}{8} + \frac {5 a^{\frac {3}{2}} b \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{4} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{4}}}}{6 x^{4}} \] Input:
integrate((a+b/x**4)**(5/2)*x**3,x)
Output:
a**(5/2)*x**4*sqrt(1 + b/(a*x**4))/4 - 7*a**(3/2)*b*sqrt(1 + b/(a*x**4))/6 - 5*a**(3/2)*b*log(b/(a*x**4))/8 + 5*a**(3/2)*b*log(sqrt(1 + b/(a*x**4)) + 1)/4 - sqrt(a)*b**2*sqrt(1 + b/(a*x**4))/(6*x**4)
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {1}{4} \, \sqrt {a + \frac {b}{x^{4}}} a^{2} x^{4} - \frac {5}{8} \, a^{\frac {3}{2}} b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right ) - \frac {1}{6} \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} b - \sqrt {a + \frac {b}{x^{4}}} a b \] Input:
integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="maxima")
Output:
1/4*sqrt(a + b/x^4)*a^2*x^4 - 5/8*a^(3/2)*b*log((sqrt(a + b/x^4) - sqrt(a) )/(sqrt(a + b/x^4) + sqrt(a))) - 1/6*(a + b/x^4)^(3/2)*b - sqrt(a + b/x^4) *a*b
Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (63) = 126\).
Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.75 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {1}{4} \, \sqrt {a x^{4} + b} a^{2} x^{2} - \frac {5}{8} \, a^{\frac {3}{2}} b \log \left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2}\right ) + \frac {9 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{4} a^{\frac {3}{2}} b^{2} - 12 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} a^{\frac {3}{2}} b^{3} + 7 \, a^{\frac {3}{2}} b^{4}}{3 \, {\left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} - b\right )}^{3}} \] Input:
integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="giac")
Output:
1/4*sqrt(a*x^4 + b)*a^2*x^2 - 5/8*a^(3/2)*b*log((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2) + 1/3*(9*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^4*a^(3/2)*b^2 - 12*(sqrt (a)*x^2 - sqrt(a*x^4 + b))^2*a^(3/2)*b^3 + 7*a^(3/2)*b^4)/((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2 - b)^3
Time = 1.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {a^2\,x^4\,\sqrt {a+\frac {b}{x^4}}}{4}-\frac {b\,{\left (a+\frac {b}{x^4}\right )}^{3/2}}{6}-a\,b\,\sqrt {a+\frac {b}{x^4}}-\frac {a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x^4}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{4} \] Input:
int(x^3*(a + b/x^4)^(5/2),x)
Output:
(a^2*x^4*(a + b/x^4)^(1/2))/4 - (b*(a + b/x^4)^(3/2))/6 - (a^(3/2)*b*atan( ((a + b/x^4)^(1/2)*1i)/a^(1/2))*5i)/4 - a*b*(a + b/x^4)^(1/2)
Time = 0.21 (sec) , antiderivative size = 445, normalized size of antiderivative = 5.49 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {480 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a^{3} b \,x^{14}+360 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a^{2} b^{2} x^{10}+30 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a \,b^{3} x^{6}+96 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, a^{4} x^{18}-248 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, a^{3} b \,x^{14}-534 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, a^{2} b^{2} x^{10}-215 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, a \,b^{3} x^{6}-20 \sqrt {a}\, \sqrt {a \,x^{4}+b}\, b^{4} x^{2}+480 \,\mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a^{4} b \,x^{16}+600 \,\mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a^{3} b^{2} x^{12}+150 \,\mathrm {log}\left (\frac {\sqrt {a \,x^{4}+b}+\sqrt {a}\, x^{2}}{\sqrt {b}}\right ) a^{2} b^{3} x^{8}+96 a^{5} x^{20}-200 a^{4} b \,x^{16}-670 a^{3} b^{2} x^{12}-445 a^{2} b^{3} x^{8}-80 a \,b^{4} x^{4}-4 b^{5}}{24 x^{6} \left (16 \sqrt {a \,x^{4}+b}\, a^{2} x^{8}+12 \sqrt {a \,x^{4}+b}\, a b \,x^{4}+\sqrt {a \,x^{4}+b}\, b^{2}+16 \sqrt {a}\, a^{2} x^{10}+20 \sqrt {a}\, a b \,x^{6}+5 \sqrt {a}\, b^{2} x^{2}\right )} \] Input:
int((a+b/x^4)^(5/2)*x^3,x)
Output:
(480*sqrt(a)*sqrt(a*x**4 + b)*log((sqrt(a*x**4 + b) + sqrt(a)*x**2)/sqrt(b ))*a**3*b*x**14 + 360*sqrt(a)*sqrt(a*x**4 + b)*log((sqrt(a*x**4 + b) + sqr t(a)*x**2)/sqrt(b))*a**2*b**2*x**10 + 30*sqrt(a)*sqrt(a*x**4 + b)*log((sqr t(a*x**4 + b) + sqrt(a)*x**2)/sqrt(b))*a*b**3*x**6 + 96*sqrt(a)*sqrt(a*x** 4 + b)*a**4*x**18 - 248*sqrt(a)*sqrt(a*x**4 + b)*a**3*b*x**14 - 534*sqrt(a )*sqrt(a*x**4 + b)*a**2*b**2*x**10 - 215*sqrt(a)*sqrt(a*x**4 + b)*a*b**3*x **6 - 20*sqrt(a)*sqrt(a*x**4 + b)*b**4*x**2 + 480*log((sqrt(a*x**4 + b) + sqrt(a)*x**2)/sqrt(b))*a**4*b*x**16 + 600*log((sqrt(a*x**4 + b) + sqrt(a)* x**2)/sqrt(b))*a**3*b**2*x**12 + 150*log((sqrt(a*x**4 + b) + sqrt(a)*x**2) /sqrt(b))*a**2*b**3*x**8 + 96*a**5*x**20 - 200*a**4*b*x**16 - 670*a**3*b** 2*x**12 - 445*a**2*b**3*x**8 - 80*a*b**4*x**4 - 4*b**5)/(24*x**6*(16*sqrt( a*x**4 + b)*a**2*x**8 + 12*sqrt(a*x**4 + b)*a*b*x**4 + sqrt(a*x**4 + b)*b* *2 + 16*sqrt(a)*a**2*x**10 + 20*sqrt(a)*a*b*x**6 + 5*sqrt(a)*b**2*x**2))