Integrand size = 24, antiderivative size = 223 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \] Output:
-1/5*a^5*((b*x+a)^2)^(1/2)/x^5/(b*x+a)-5/4*a^4*b*((b*x+a)^2)^(1/2)/x^4/(b* x+a)-10/3*a^3*b^2*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-5*a^2*b^3*((b*x+a)^2)^(1/2 )/x^2/(b*x+a)-5*a*b^4*((b*x+a)^2)^(1/2)/x/(b*x+a)+b^5*((b*x+a)^2)^(1/2)*ln (x)/(b*x+a)
Time = 0.48 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\frac {1}{120} \left (-\frac {\sqrt {(a+b x)^2} \left (12 a^4+63 a^3 b x+137 a^2 b^2 x^2+163 a b^3 x^3+137 b^4 x^4\right )}{x^5}+\frac {\sqrt {a^2} \left (12 a^4+75 a^3 b x+200 a^2 b^2 x^2+300 a b^3 x^3+300 b^4 x^4\right )}{x^5}-120 b^5 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )-\frac {120 \sqrt {a^2} b^5 \log (x)}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{a}\right ) \] Input:
Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^6,x]
Output:
(-((Sqrt[(a + b*x)^2]*(12*a^4 + 63*a^3*b*x + 137*a^2*b^2*x^2 + 163*a*b^3*x ^3 + 137*b^4*x^4))/x^5) + (Sqrt[a^2]*(12*a^4 + 75*a^3*b*x + 200*a^2*b^2*x^ 2 + 300*a*b^3*x^3 + 300*b^4*x^4))/x^5 - 120*b^5*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])] - (120*Sqrt[a^2]*b^5*Log[x])/a + (60*Sqrt[a^2]*b^5*Lo g[a*(Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2])])/a + (60*Sqrt[a^2]*b^5*Log[a*(S qrt[a^2] + b*x - Sqrt[(a + b*x)^2])])/a)/120
Time = 0.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1102, 27, 49, 2009}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{x^6}dx}{b^5 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{x^6}dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5}{x^6}+\frac {5 b a^4}{x^5}+\frac {10 b^2 a^3}{x^4}+\frac {10 b^3 a^2}{x^3}+\frac {5 b^4 a}{x^2}+\frac {b^5}{x}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5}{5 x^5}-\frac {5 a^4 b}{4 x^4}-\frac {10 a^3 b^2}{3 x^3}-\frac {5 a^2 b^3}{x^2}-\frac {5 a b^4}{x}+b^5 \log (x)\right )}{a+b x}\) |
Input:
Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^6,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/5*a^5/x^5 - (5*a^4*b)/(4*x^4) - (10*a^3 *b^2)/(3*x^3) - (5*a^2*b^3)/x^2 - (5*a*b^4)/x + b^5*Log[x]))/(a + b*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 0.63 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.34
method | result | size |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (60 \ln \left (x \right ) x^{5} b^{5}-300 a \,b^{4} x^{4}-300 a^{2} b^{3} x^{3}-200 a^{3} b^{2} x^{2}-75 a^{4} b x -12 a^{5}\right )}{60 \left (b x +a \right )^{5} x^{5}}\) | \(76\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-5 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}-\frac {10}{3} a^{3} b^{2} x^{2}-\frac {5}{4} a^{4} b x -\frac {1}{5} a^{5}\right )}{\left (b x +a \right ) x^{5}}+\frac {b^{5} \sqrt {\left (b x +a \right )^{2}}\, \ln \left (x \right )}{b x +a}\) | \(88\) |
Input:
int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x,method=_RETURNVERBOSE)
Output:
1/60*((b*x+a)^2)^(5/2)*(60*ln(x)*x^5*b^5-300*a*b^4*x^4-300*a^2*b^3*x^3-200 *a^3*b^2*x^2-75*a^4*b*x-12*a^5)/(b*x+a)^5/x^5
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.26 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\frac {60 \, b^{5} x^{5} \log \left (x\right ) - 300 \, a b^{4} x^{4} - 300 \, a^{2} b^{3} x^{3} - 200 \, a^{3} b^{2} x^{2} - 75 \, a^{4} b x - 12 \, a^{5}}{60 \, x^{5}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="fricas")
Output:
1/60*(60*b^5*x^5*log(x) - 300*a*b^4*x^4 - 300*a^2*b^3*x^3 - 200*a^3*b^2*x^ 2 - 75*a^4*b*x - 12*a^5)/x^5
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \] Input:
integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**6,x)
Output:
Integral(((a + b*x)**2)**(5/2)/x**6, x)
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (151) = 302\).
Time = 0.04 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} b^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{6} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}}{2 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{6} x}{4 \, a^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{3}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{5}}{15 \, a^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{3 \, a^{4} x} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{15 \, a^{5} x^{2}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{60 \, a^{4} x^{3}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{5 \, a^{2} x^{5}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="maxima")
Output:
(-1)^(2*b^2*x + 2*a*b)*b^5*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*b ^5*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)* b^6*x/a^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5/a + 1/4*(b^2*x^2 + 2*a*b *x + a^2)^(3/2)*b^6*x/a^4 + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^5/a^3 - 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^5/a^5 - 1/3*(b^2*x^2 + 2*a*b*x + a ^2)^(5/2)*b^4/(a^4*x) + 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*x^2) - 11/60*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/(a^4*x^3) + 3/20*(b^2*x^2 + 2 *a*b*x + a^2)^(7/2)*b/(a^3*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2 *x^5)
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=b^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {300 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 300 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 200 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 75 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 12 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{60 \, x^{5}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="giac")
Output:
b^5*log(abs(x))*sgn(b*x + a) - 1/60*(300*a*b^4*x^4*sgn(b*x + a) + 300*a^2* b^3*x^3*sgn(b*x + a) + 200*a^3*b^2*x^2*sgn(b*x + a) + 75*a^4*b*x*sgn(b*x + a) + 12*a^5*sgn(b*x + a))/x^5
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^6} \,d x \] Input:
int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^6,x)
Output:
int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^6, x)
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.26 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) b^{5} x^{5}-12 a^{5}-75 a^{4} b x -200 a^{3} b^{2} x^{2}-300 a^{2} b^{3} x^{3}-300 a \,b^{4} x^{4}}{60 x^{5}} \] Input:
int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x)
Output:
(60*log(x)*b**5*x**5 - 12*a**5 - 75*a**4*b*x - 200*a**3*b**2*x**2 - 300*a* *2*b**3*x**3 - 300*a*b**4*x**4)/(60*x**5)