Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^2 b^3 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {5 a b^4 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {b^5 x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 \left (a+b x^3\right )} \] Output:
-1/5*a^5*((b*x^3+a)^2)^(1/2)/x^5/(b*x^3+a)-5/2*a^4*b*((b*x^3+a)^2)^(1/2)/x ^2/(b*x^3+a)+10*a^3*b^2*x*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5*a^2*b^3*x^4*((b* x^3+a)^2)^(1/2)/(2*b*x^3+2*a)+5*a*b^4*x^7*((b*x^3+a)^2)^(1/2)/(7*b*x^3+7*a )+b^5*x^10*((b*x^3+a)^2)^(1/2)/(10*b*x^3+10*a)
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (-14 a^5-175 a^4 b x^3+700 a^3 b^2 x^6+175 a^2 b^3 x^9+50 a b^4 x^{12}+7 b^5 x^{15}\right )}{70 x^5 \left (a+b x^3\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^6,x]
Output:
(Sqrt[(a + b*x^3)^2]*(-14*a^5 - 175*a^4*b*x^3 + 700*a^3*b^2*x^6 + 175*a^2* b^3*x^9 + 50*a*b^4*x^12 + 7*b^5*x^15))/(70*x^5*(a + b*x^3))
Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 802, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^6}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^6}dx}{a+b x^3}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^5 x^9+5 a b^4 x^6+10 a^2 b^3 x^3+10 a^3 b^2+\frac {5 a^4 b}{x^3}+\frac {a^5}{x^6}\right )dx}{a+b x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{5 x^5}-\frac {5 a^4 b}{2 x^2}+10 a^3 b^2 x+\frac {5}{2} a^2 b^3 x^4+\frac {5}{7} a b^4 x^7+\frac {b^5 x^{10}}{10}\right )}{a+b x^3}\) |
Input:
Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^6,x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/5*a^5/x^5 - (5*a^4*b)/(2*x^2) + 10*a^ 3*b^2*x + (5*a^2*b^3*x^4)/2 + (5*a*b^4*x^7)/7 + (b^5*x^10)/10))/(a + b*x^3 )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 3.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {\left (-7 b^{5} x^{15}-50 a \,b^{4} x^{12}-175 a^{2} b^{3} x^{9}-700 a^{3} b^{2} x^{6}+175 a^{4} b \,x^{3}+14 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{70 x^{5} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (-7 b^{5} x^{15}-50 a \,b^{4} x^{12}-175 a^{2} b^{3} x^{9}-700 a^{3} b^{2} x^{6}+175 a^{4} b \,x^{3}+14 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{70 x^{5} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (-7 b^{5} x^{15}-50 a \,b^{4} x^{12}-175 a^{2} b^{3} x^{9}-700 a^{3} b^{2} x^{6}+175 a^{4} b \,x^{3}+14 a^{5}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}{70 x^{5} \left (b \,x^{3}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \left (\frac {1}{10} b^{3} x^{10}+\frac {5}{7} a \,b^{2} x^{7}+\frac {5}{2} a^{2} b \,x^{4}+10 a^{3} x \right )}{b \,x^{3}+a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {5}{2} a^{4} b \,x^{3}-\frac {1}{5} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{5}}\) | \(98\) |
Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/70*(-7*b^5*x^15-50*a*b^4*x^12-175*a^2*b^3*x^9-700*a^3*b^2*x^6+175*a^4*b *x^3+14*a^5)*((b*x^3+a)^2)^(5/2)/x^5/(b*x^3+a)^5
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\frac {7 \, b^{5} x^{15} + 50 \, a b^{4} x^{12} + 175 \, a^{2} b^{3} x^{9} + 700 \, a^{3} b^{2} x^{6} - 175 \, a^{4} b x^{3} - 14 \, a^{5}}{70 \, x^{5}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^6,x, algorithm="fricas")
Output:
1/70*(7*b^5*x^15 + 50*a*b^4*x^12 + 175*a^2*b^3*x^9 + 700*a^3*b^2*x^6 - 175 *a^4*b*x^3 - 14*a^5)/x^5
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \] Input:
integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**6,x)
Output:
Integral(((a + b*x**3)**2)**(5/2)/x**6, x)
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\frac {7 \, b^{5} x^{15} + 50 \, a b^{4} x^{12} + 175 \, a^{2} b^{3} x^{9} + 700 \, a^{3} b^{2} x^{6} - 175 \, a^{4} b x^{3} - 14 \, a^{5}}{70 \, x^{5}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^6,x, algorithm="maxima")
Output:
1/70*(7*b^5*x^15 + 50*a*b^4*x^12 + 175*a^2*b^3*x^9 + 700*a^3*b^2*x^6 - 175 *a^4*b*x^3 - 14*a^5)/x^5
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\frac {1}{10} \, b^{5} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{7} \, a b^{4} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{2} \, a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 10 \, a^{3} b^{2} x \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {25 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{10 \, x^{5}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^6,x, algorithm="giac")
Output:
1/10*b^5*x^10*sgn(b*x^3 + a) + 5/7*a*b^4*x^7*sgn(b*x^3 + a) + 5/2*a^2*b^3* x^4*sgn(b*x^3 + a) + 10*a^3*b^2*x*sgn(b*x^3 + a) - 1/10*(25*a^4*b*x^3*sgn( b*x^3 + a) + 2*a^5*sgn(b*x^3 + a))/x^5
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^6} \,d x \] Input:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^6,x)
Output:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^6, x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^6} \, dx=\frac {7 b^{5} x^{15}+50 a \,b^{4} x^{12}+175 a^{2} b^{3} x^{9}+700 a^{3} b^{2} x^{6}-175 a^{4} b \,x^{3}-14 a^{5}}{70 x^{5}} \] Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^6,x)
Output:
( - 14*a**5 - 175*a**4*b*x**3 + 700*a**3*b**2*x**6 + 175*a**2*b**3*x**9 + 50*a*b**4*x**12 + 7*b**5*x**15)/(70*x**5)