Integrand size = 15, antiderivative size = 93 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=-\frac {a^2 \sqrt {a+b x^2}}{6 x^6}-\frac {13 a b \sqrt {a+b x^2}}{24 x^4}-\frac {11 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}} \] Output:
-1/6*a^2*(b*x^2+a)^(1/2)/x^6-13/24*a*b*(b*x^2+a)^(1/2)/x^4-11/16*b^2*(b*x^ 2+a)^(1/2)/x^2-5/16*b^3*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a^2-26 a b x^2-33 b^2 x^4\right )}{48 x^6}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}} \] Input:
Integrate[(a + b*x^2)^(5/2)/x^7,x]
Output:
(Sqrt[a + b*x^2]*(-8*a^2 - 26*a*b*x^2 - 33*b^2*x^4))/(48*x^6) - (5*b^3*Arc Tanh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*Sqrt[a])
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {243, 51, 51, 51, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2}}{x^8}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} b \int \frac {\left (b x^2+a\right )^{3/2}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{5/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} b \left (\frac {3}{4} b \int \frac {\sqrt {b x^2+a}}{x^4}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{5/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} b \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{5/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} b \left (\frac {3}{4} b \left (\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{5/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} b \left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{5/2}}{3 x^6}\right )\) |
Input:
Int[(a + b*x^2)^(5/2)/x^7,x]
Output:
(-1/3*(a + b*x^2)^(5/2)/x^6 + (5*b*(-1/2*(a + b*x^2)^(3/2)/x^4 + (3*b*(-(S qrt[a + b*x^2]/x^2) - (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/4))/6 )/2
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] :> 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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (33 b^{2} x^{4}+26 a b \,x^{2}+8 a^{2}\right )}{48 x^{6}}-\frac {5 b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{16 \sqrt {a}}\) | \(68\) |
pseudoelliptic | \(\frac {-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) b^{3} x^{6}-33 b^{2} x^{4} \sqrt {b \,x^{2}+a}\, \sqrt {a}-26 a^{\frac {3}{2}} b \,x^{2} \sqrt {b \,x^{2}+a}-8 a^{\frac {5}{2}} \sqrt {b \,x^{2}+a}}{48 x^{6} \sqrt {a}}\) | \(84\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\) | \(139\) |
Input:
int((b*x^2+a)^(5/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/48*(b*x^2+a)^(1/2)*(33*b^2*x^4+26*a*b*x^2+8*a^2)/x^6-5/16*b^3/a^(1/2)*l n((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} x^{6} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (33 \, a b^{2} x^{4} + 26 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{96 \, a x^{6}}, \frac {15 \, \sqrt {-a} b^{3} x^{6} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (33 \, a b^{2} x^{4} + 26 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{48 \, a x^{6}}\right ] \] Input:
integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="fricas")
Output:
[1/96*(15*sqrt(a)*b^3*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x ^2) - 2*(33*a*b^2*x^4 + 26*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a*x^6), 1/ 48*(15*sqrt(-a)*b^3*x^6*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - (33*a*b^2*x^4 + 26*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a*x^6)]
Time = 2.68 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=- \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{6 x^{5}} - \frac {13 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{24 x^{3}} - \frac {11 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{16 x} - \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 \sqrt {a}} \] Input:
integrate((b*x**2+a)**(5/2)/x**7,x)
Output:
-a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(6*x**5) - 13*a*b**(3/2)*sqrt(a/(b*x**2 ) + 1)/(24*x**3) - 11*b**(5/2)*sqrt(a/(b*x**2) + 1)/(16*x) - 5*b**3*asinh( sqrt(a)/(sqrt(b)*x))/(16*sqrt(a))
Time = 0.05 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=-\frac {5 \, b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, \sqrt {a}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{16 \, a^{3}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}}{48 \, a^{2}} + \frac {5 \, \sqrt {b x^{2} + a} b^{3}}{16 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{16 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{6 \, a x^{6}} \] Input:
integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="maxima")
Output:
-5/16*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/16*(b*x^2 + a)^(5/2)*b ^3/a^3 + 5/48*(b*x^2 + a)^(3/2)*b^3/a^2 + 5/16*sqrt(b*x^2 + a)*b^3/a - 1/1 6*(b*x^2 + a)^(7/2)*b^2/(a^3*x^2) - 1/24*(b*x^2 + a)^(7/2)*b/(a^2*x^4) - 1 /6*(b*x^2 + a)^(7/2)/(a*x^6)
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=\frac {1}{48} \, b^{3} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {33 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} - 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x^{2} + a} a^{2}}{b^{3} x^{6}}\right )} \] Input:
integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="giac")
Output:
1/48*b^3*(15*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (33*(b*x^2 + a)^( 5/2) - 40*(b*x^2 + a)^(3/2)*a + 15*sqrt(b*x^2 + a)*a^2)/(b^3*x^6))
Time = 0.88 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=\frac {5\,a\,{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {11\,{\left (b\,x^2+a\right )}^{5/2}}{16\,x^6}-\frac {5\,a^2\,\sqrt {b\,x^2+a}}{16\,x^6}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,\sqrt {a}} \] Input:
int((a + b*x^2)^(5/2)/x^7,x)
Output:
(b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(16*a^(1/2)) - (11*(a + b*x^ 2)^(5/2))/(16*x^6) + (5*a*(a + b*x^2)^(3/2))/(6*x^6) - (5*a^2*(a + b*x^2)^ (1/2))/(16*x^6)
Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^7} \, dx=\frac {-8 \sqrt {b \,x^{2}+a}\, a^{3}-26 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}-33 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}}{48 a \,x^{6}} \] Input:
int((b*x^2+a)^(5/2)/x^7,x)
Output:
( - 8*sqrt(a + b*x**2)*a**3 - 26*sqrt(a + b*x**2)*a**2*b*x**2 - 33*sqrt(a + b*x**2)*a*b**2*x**4 + 15*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt( b)*x)/sqrt(a))*b**3*x**6 - 15*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sq rt(b)*x)/sqrt(a))*b**3*x**6)/(48*a*x**6)