Integrand size = 24, antiderivative size = 222 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {5 a b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \] Output:
-1/3*a^5*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-5/2*a^4*b*((b*x+a)^2)^(1/2)/x^2/(b* x+a)-10*a^3*b^2*((b*x+a)^2)^(1/2)/x/(b*x+a)+5*a*b^4*x*((b*x+a)^2)^(1/2)/(b *x+a)+b^5*x^2*((b*x+a)^2)^(1/2)/(2*b*x+2*a)+10*a^2*b^3*((b*x+a)^2)^(1/2)*l n(x)/(b*x+a)
Time = 0.76 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {\left (4 a^5+30 a^4 b x+120 a^3 b^2 x^2+53 a^2 b^3 x^3-60 a b^4 x^4-6 b^5 x^5\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{12 x^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}-10 a^2 b^3 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )-10 a \sqrt {a^2} b^3 \log (x)+5 a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+5 a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right ) \] Input:
Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^4,x]
Output:
((4*a^5 + 30*a^4*b*x + 120*a^3*b^2*x^2 + 53*a^2*b^3*x^3 - 60*a*b^4*x^4 - 6 *b^5*x^5)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(12*x^3*(a^ 2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) - 10*a^2*b^3*ArcTanh[(b*x)/(Sqrt [a^2] - Sqrt[(a + b*x)^2])] - 10*a*Sqrt[a^2]*b^3*Log[x] + 5*a*Sqrt[a^2]*b^ 3*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] + 5*a*Sqrt[a^2]*b^3*Log[Sqrt[a^ 2] + b*x - Sqrt[(a + b*x)^2]]
Time = 0.38 (sec) , antiderivative size = 88, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{x^4}dx}{b^5 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{x^4}dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5}{x^4}+\frac {5 b a^4}{x^3}+\frac {10 b^2 a^3}{x^2}+\frac {10 b^3 a^2}{x}+5 b^4 a+b^5 x\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5}{3 x^3}-\frac {5 a^4 b}{2 x^2}-\frac {10 a^3 b^2}{x}+10 a^2 b^3 \log (x)+5 a b^4 x+\frac {b^5 x^2}{2}\right )}{a+b x}\) |
Input:
Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^4,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/3*a^5/x^3 - (5*a^4*b)/(2*x^2) - (10*a^3 *b^2)/x + 5*a*b^4*x + (b^5*x^2)/2 + 10*a^2*b^3*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 = 0.62 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.34
method | result | size |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (3 b^{5} x^{5}+60 \ln \left (x \right ) x^{3} a^{2} b^{3}+30 a \,b^{4} x^{4}-60 a^{3} b^{2} x^{2}-15 a^{4} b x -2 a^{5}\right )}{6 \left (b x +a \right )^{5} x^{3}}\) | \(76\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{4} \left (\frac {1}{2} b \,x^{2}+5 a x \right )}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-10 a^{3} b^{2} x^{2}-\frac {5}{2} a^{4} b x -\frac {1}{3} a^{5}\right )}{\left (b x +a \right ) x^{3}}+\frac {10 a^{2} b^{3} \sqrt {\left (b x +a \right )^{2}}\, \ln \left (x \right )}{b x +a}\) | \(103\) |
Input:
int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^4,x,method=_RETURNVERBOSE)
Output:
1/6*((b*x+a)^2)^(5/2)*(3*b^5*x^5+60*ln(x)*x^3*a^2*b^3+30*a*b^4*x^4-60*a^3* b^2*x^2-15*a^4*b*x-2*a^5)/(b*x+a)^5/x^3
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {3 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} \log \left (x\right ) - 60 \, a^{3} b^{2} x^{2} - 15 \, a^{4} b x - 2 \, a^{5}}{6 \, x^{3}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^4,x, algorithm="fricas")
Output:
1/6*(3*b^5*x^5 + 30*a*b^4*x^4 + 60*a^2*b^3*x^3*log(x) - 60*a^3*b^2*x^2 - 1 5*a^4*b*x - 2*a^5)/x^3
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{4}}\, dx \] Input:
integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**4,x)
Output:
Integral(((a + b*x)**2)**(5/2)/x**4, x)
Time = 0.04 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a^{2} b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a^{2} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4} x + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a b^{3} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4} x}{2 \, a^{2}} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{3}}{6 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{6 \, a^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{6 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{3 \, a^{2} x^{3}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^4,x, algorithm="maxima")
Output:
10*(-1)^(2*b^2*x + 2*a*b)*a^2*b^3*log(2*b^2*x + 2*a*b) - 10*(-1)^(2*a*b*x + 2*a^2)*a^2*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) + 5*sqrt(b^2*x^2 + 2*a *b*x + a^2)*b^4*x + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*b^3 + 5/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4*x/a^2 + 35/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^ 3/a + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^3/a^3 - 11/6*(b^2*x^2 + 2*a*b* x + a^2)^(5/2)*b^2/(a^2*x) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^ 2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^3)
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} \, b^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{4} x \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {60 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{6 \, x^{3}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^4,x, algorithm="giac")
Output:
1/2*b^5*x^2*sgn(b*x + a) + 5*a*b^4*x*sgn(b*x + a) + 10*a^2*b^3*log(abs(x)) *sgn(b*x + a) - 1/6*(60*a^3*b^2*x^2*sgn(b*x + a) + 15*a^4*b*x*sgn(b*x + a) + 2*a^5*sgn(b*x + a))/x^3
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^4} \,d x \] Input:
int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^4,x)
Output:
int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^4, x)
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{2} b^{3} x^{3}-2 a^{5}-15 a^{4} b x -60 a^{3} b^{2} x^{2}+30 a \,b^{4} x^{4}+3 b^{5} x^{5}}{6 x^{3}} \] Input:
int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^4,x)
Output:
(60*log(x)*a**2*b**3*x**3 - 2*a**5 - 15*a**4*b*x - 60*a**3*b**2*x**2 + 30* a*b**4*x**4 + 3*b**5*x**5)/(6*x**3)