Integrand size = 15, antiderivative size = 88 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=-\frac {a^2 \sqrt {a+b x^2}}{5 x^5}-\frac {11 a b \sqrt {a+b x^2}}{15 x^3}-\frac {23 b^2 \sqrt {a+b x^2}}{15 x}+b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \] Output:
-1/5*a^2*(b*x^2+a)^(1/2)/x^5-11/15*a*b*(b*x^2+a)^(1/2)/x^3-23/15*b^2*(b*x^ 2+a)^(1/2)/x+b^(5/2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {\sqrt {a+b x^2} \left (-3 a^2-11 a b x^2-23 b^2 x^4\right )}{15 x^5}-b^{5/2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \] Input:
Integrate[(a + b*x^2)^(5/2)/x^6,x]
Output:
(Sqrt[a + b*x^2]*(-3*a^2 - 11*a*b*x^2 - 23*b^2*x^4))/(15*x^5) - b^(5/2)*Lo g[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {247, 247, 247, 224, 219}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle b \int \frac {\left (b x^2+a\right )^{3/2}}{x^4}dx-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle b \left (b \int \frac {\sqrt {b x^2+a}}{x^2}dx-\frac {\left (a+b x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle b \left (b \left (b \int \frac {1}{\sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2}}{x}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle b \left (b \left (b \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-\frac {\sqrt {a+b x^2}}{x}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b \left (b \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2}}{x}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}\) |
Input:
Int[(a + b*x^2)^(5/2)/x^6,x]
Output:
-1/5*(a + b*x^2)^(5/2)/x^5 + b*(-1/3*(a + b*x^2)^(3/2)/x^3 + b*(-(Sqrt[a + b*x^2]/x) + Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]))
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 0.41 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.65
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (23 b^{2} x^{4}+11 a b \,x^{2}+3 a^{2}\right )}{15 x^{5}}+b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )\) | \(57\) |
pseudoelliptic | \(\frac {15 b^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) x^{5}-\sqrt {b \,x^{2}+a}\, \left (23 b^{2} x^{4}+11 a b \,x^{2}+3 a^{2}\right )}{15 x^{5}}\) | \(64\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}}+\frac {2 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )}{5 a}\) | \(140\) |
Input:
int((b*x^2+a)^(5/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/15*(b*x^2+a)^(1/2)*(23*b^2*x^4+11*a*b*x^2+3*a^2)/x^5+b^(5/2)*ln(b^(1/2) *x+(b*x^2+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=\left [\frac {15 \, b^{\frac {5}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (23 \, b^{2} x^{4} + 11 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{5}}, -\frac {15 \, \sqrt {-b} b^{2} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (23 \, b^{2} x^{4} + 11 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{15 \, x^{5}}\right ] \] Input:
integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="fricas")
Output:
[1/30*(15*b^(5/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2* (23*b^2*x^4 + 11*a*b*x^2 + 3*a^2)*sqrt(b*x^2 + a))/x^5, -1/15*(15*sqrt(-b) *b^2*x^5*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (23*b^2*x^4 + 11*a*b*x^2 + 3 *a^2)*sqrt(b*x^2 + a))/x^5]
Time = 2.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=- \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {23 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} - \frac {b^{\frac {5}{2}} \log {\left (\frac {a}{b x^{2}} \right )}}{2} + b^{\frac {5}{2}} \log {\left (\sqrt {\frac {a}{b x^{2}} + 1} + 1 \right )} \] Input:
integrate((b*x**2+a)**(5/2)/x**6,x)
Output:
-a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 11*a*b**(3/2)*sqrt(a/(b*x**2 ) + 1)/(15*x**2) - 23*b**(5/2)*sqrt(a/(b*x**2) + 1)/15 - b**(5/2)*log(a/(b *x**2))/2 + b**(5/2)*log(sqrt(a/(b*x**2) + 1) + 1)
Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} b^{3} x}{a} + b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{15 \, a^{2} x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{5 \, a x^{5}} \] Input:
integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="maxima")
Output:
2/3*(b*x^2 + a)^(3/2)*b^3*x/a^2 + sqrt(b*x^2 + a)*b^3*x/a + b^(5/2)*arcsin h(b*x/sqrt(a*b)) - 8/15*(b*x^2 + a)^(5/2)*b^2/(a^2*x) - 2/15*(b*x^2 + a)^( 7/2)*b/(a^2*x^3) - 1/5*(b*x^2 + a)^(7/2)/(a*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (70) = 140\).
Time = 0.14 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.91 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=-\frac {1}{2} \, b^{\frac {5}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a b^{\frac {5}{2}} - 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} b^{\frac {5}{2}} + 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} b^{\frac {5}{2}} - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} b^{\frac {5}{2}} + 23 \, a^{5} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \] Input:
integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="giac")
Output:
-1/2*b^(5/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(5/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^2*b^(5 /2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*b^(5/2) - 70*(sqrt(b)*x - sq rt(b*x^2 + a))^2*a^4*b^(5/2) + 23*a^5*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^6} \,d x \] Input:
int((a + b*x^2)^(5/2)/x^6,x)
Output:
int((a + b*x^2)^(5/2)/x^6, x)
Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{2}-11 \sqrt {b \,x^{2}+a}\, a b \,x^{2}-23 \sqrt {b \,x^{2}+a}\, b^{2} x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{5}+5 \sqrt {b}\, b^{2} x^{5}}{15 x^{5}} \] Input:
int((b*x^2+a)^(5/2)/x^6,x)
Output:
( - 3*sqrt(a + b*x**2)*a**2 - 11*sqrt(a + b*x**2)*a*b*x**2 - 23*sqrt(a + b *x**2)*b**2*x**4 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))* b**2*x**5 + 5*sqrt(b)*b**2*x**5)/(15*x**5)