Integrand size = 26, antiderivative size = 249 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {b^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )} \] Output:
-1/4*a^5*((b*x^3+a)^2)^(1/2)/x^4/(b*x^3+a)-5*a^4*b*((b*x^3+a)^2)^(1/2)/x/( b*x^3+a)+5*a^3*b^2*x^2*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+2*a^2*b^3*x^5*((b*x^3 +a)^2)^(1/2)/(b*x^3+a)+5*a*b^4*x^8*((b*x^3+a)^2)^(1/2)/(8*b*x^3+8*a)+b^5*x ^11*((b*x^3+a)^2)^(1/2)/(11*b*x^3+11*a)
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (-22 a^5-440 a^4 b x^3+440 a^3 b^2 x^6+176 a^2 b^3 x^9+55 a b^4 x^{12}+8 b^5 x^{15}\right )}{88 x^4 \left (a+b x^3\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^5,x]
Output:
(Sqrt[(a + b*x^3)^2]*(-22*a^5 - 440*a^4*b*x^3 + 440*a^3*b^2*x^6 + 176*a^2* b^3*x^9 + 55*a*b^4*x^12 + 8*b^5*x^15))/(88*x^4*(a + b*x^3))
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.38, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^5}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^5}dx}{a+b x^3}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^5 x^{10}+5 a b^4 x^7+10 a^2 b^3 x^4+10 a^3 b^2 x+\frac {5 a^4 b}{x^2}+\frac {a^5}{x^5}\right )dx}{a+b x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{4 x^4}-\frac {5 a^4 b}{x}+5 a^3 b^2 x^2+2 a^2 b^3 x^5+\frac {5}{8} a b^4 x^8+\frac {b^5 x^{11}}{11}\right )}{a+b x^3}\) |
Input:
Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^5,x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/4*a^5/x^4 - (5*a^4*b)/x + 5*a^3*b^2*x ^2 + 2*a^2*b^3*x^5 + (5*a*b^4*x^8)/8 + (b^5*x^11)/11))/(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 = 2.50 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {\left (-8 b^{5} x^{15}-55 a \,b^{4} x^{12}-176 a^{2} b^{3} x^{9}-440 a^{3} b^{2} x^{6}+440 a^{4} b \,x^{3}+22 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{88 x^{4} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (-8 b^{5} x^{15}-55 a \,b^{4} x^{12}-176 a^{2} b^{3} x^{9}-440 a^{3} b^{2} x^{6}+440 a^{4} b \,x^{3}+22 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{88 x^{4} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (-8 b^{5} x^{15}-55 a \,b^{4} x^{12}-176 a^{2} b^{3} x^{9}-440 a^{3} b^{2} x^{6}+440 a^{4} b \,x^{3}+22 a^{5}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}{88 x^{4} \left (b \,x^{3}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \left (\frac {1}{11} b^{3} x^{11}+\frac {5}{8} a \,b^{2} x^{8}+2 a^{2} b \,x^{5}+5 a^{3} x^{2}\right )}{b \,x^{3}+a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-5 a^{4} b \,x^{3}-\frac {1}{4} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{4}}\) | \(100\) |
Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/88*(-8*b^5*x^15-55*a*b^4*x^12-176*a^2*b^3*x^9-440*a^3*b^2*x^6+440*a^4*b *x^3+22*a^5)*((b*x^3+a)^2)^(5/2)/x^4/(b*x^3+a)^5
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\frac {8 \, b^{5} x^{15} + 55 \, a b^{4} x^{12} + 176 \, a^{2} b^{3} x^{9} + 440 \, a^{3} b^{2} x^{6} - 440 \, a^{4} b x^{3} - 22 \, a^{5}}{88 \, x^{4}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^5,x, algorithm="fricas")
Output:
1/88*(8*b^5*x^15 + 55*a*b^4*x^12 + 176*a^2*b^3*x^9 + 440*a^3*b^2*x^6 - 440 *a^4*b*x^3 - 22*a^5)/x^4
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{5}}\, dx \] Input:
integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**5,x)
Output:
Integral(((a + b*x**3)**2)**(5/2)/x**5, x)
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\frac {8 \, b^{5} x^{15} + 55 \, a b^{4} x^{12} + 176 \, a^{2} b^{3} x^{9} + 440 \, a^{3} b^{2} x^{6} - 440 \, a^{4} b x^{3} - 22 \, a^{5}}{88 \, x^{4}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^5,x, algorithm="maxima")
Output:
1/88*(8*b^5*x^15 + 55*a*b^4*x^12 + 176*a^2*b^3*x^9 + 440*a^3*b^2*x^6 - 440 *a^4*b*x^3 - 22*a^5)/x^4
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{8} \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {20 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{4 \, x^{4}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^5,x, algorithm="giac")
Output:
1/11*b^5*x^11*sgn(b*x^3 + a) + 5/8*a*b^4*x^8*sgn(b*x^3 + a) + 2*a^2*b^3*x^ 5*sgn(b*x^3 + a) + 5*a^3*b^2*x^2*sgn(b*x^3 + a) - 1/4*(20*a^4*b*x^3*sgn(b* x^3 + a) + a^5*sgn(b*x^3 + a))/x^4
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^5} \,d x \] Input:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^5,x)
Output:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^5, x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, dx=\frac {8 b^{5} x^{15}+55 a \,b^{4} x^{12}+176 a^{2} b^{3} x^{9}+440 a^{3} b^{2} x^{6}-440 a^{4} b \,x^{3}-22 a^{5}}{88 x^{4}} \] Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^5,x)
Output:
( - 22*a**5 - 440*a**4*b*x**3 + 440*a**3*b**2*x**6 + 176*a**2*b**3*x**9 + 55*a*b**4*x**12 + 8*b**5*x**15)/(88*x**4)