Integrand size = 24, antiderivative size = 125 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 a^2 (b c-a d) x^3}{3 b^4 \left (a x^2+b x^3\right )^{3/2}}+\frac {2 a (2 b c-3 a d) x}{b^4 \sqrt {a x^2+b x^3}}+\frac {2 (b c-3 a d) \sqrt {a x^2+b x^3}}{b^4 x}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b^4 x^3} \] Output:
-2/3*a^2*(-a*d+b*c)*x^3/b^4/(b*x^3+a*x^2)^(3/2)+2*a*(-3*a*d+2*b*c)*x/b^4/( b*x^3+a*x^2)^(1/2)+2*(-3*a*d+b*c)*(b*x^3+a*x^2)^(1/2)/b^4/x+2/3*d*(b*x^3+a *x^2)^(3/2)/b^4/x^3
Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 x^3 \left (-16 a^3 d+8 a^2 b (c-3 d x)-6 a b^2 x (-2 c+d x)+b^3 x^2 (3 c+d x)\right )}{3 b^4 \left (x^2 (a+b x)\right )^{3/2}} \] Input:
Integrate[(x^7*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*x^3*(-16*a^3*d + 8*a^2*b*(c - 3*d*x) - 6*a*b^2*x*(-2*c + d*x) + b^3*x^2 *(3*c + d*x)))/(3*b^4*(x^2*(a + b*x))^(3/2))
Time = 0.51 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1943, 1921, 1922, 1920}
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.
\(\displaystyle \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {2 x^6 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {(b c-2 a d) \int \frac {x^5}{\left (b x^3+a x^2\right )^{3/2}}dx}{a b}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {2 x^6 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {(b c-2 a d) \left (\frac {4 \int \frac {x^2}{\sqrt {b x^3+a x^2}}dx}{b}-\frac {2 x^3}{b \sqrt {a x^2+b x^3}}\right )}{a b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 x^6 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {(b c-2 a d) \left (\frac {4 \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{3 b}\right )}{b}-\frac {2 x^3}{b \sqrt {a x^2+b x^3}}\right )}{a b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 x^6 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {\left (\frac {4 \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {4 a \sqrt {a x^2+b x^3}}{3 b^2 x}\right )}{b}-\frac {2 x^3}{b \sqrt {a x^2+b x^3}}\right ) (b c-2 a d)}{a b}\) |
Input:
Int[(x^7*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*(b*c - a*d)*x^6)/(3*a*b*(a*x^2 + b*x^3)^(3/2)) - ((b*c - 2*a*d)*((-2*x^ 3)/(b*Sqrt[a*x^2 + b*x^3]) + (4*((2*Sqrt[a*x^2 + b*x^3])/(3*b) - (4*a*Sqrt [a*x^2 + b*x^3])/(3*b^2*x)))/b))/(a*b)
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 1.01 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-b^{3} d \,x^{3}+6 a \,b^{2} d \,x^{2}-3 b^{3} c \,x^{2}+24 a^{2} b d x -12 a \,b^{2} c x +16 a^{3} d -8 c \,a^{2} b \right ) x^{5}}{3 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(85\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-b^{3} d \,x^{3}+6 a \,b^{2} d \,x^{2}-3 b^{3} c \,x^{2}+24 a^{2} b d x -12 a \,b^{2} c x +16 a^{3} d -8 c \,a^{2} b \right ) x^{5}}{3 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(85\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-b^{3} d \,x^{3}+6 a \,b^{2} d \,x^{2}-3 b^{3} c \,x^{2}+24 a^{2} b d x -12 a \,b^{2} c x +16 a^{3} d -8 c \,a^{2} b \right ) x^{5}}{3 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(85\) |
trager | \(-\frac {2 \left (-b^{3} d \,x^{3}+6 a \,b^{2} d \,x^{2}-3 b^{3} c \,x^{2}+24 a^{2} b d x -12 a \,b^{2} c x +16 a^{3} d -8 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 x \,b^{4} \left (b x +a \right )^{2}}\) | \(87\) |
risch | \(-\frac {2 \left (-b d x +8 a d -3 b c \right ) \left (b x +a \right ) x}{3 b^{4} \sqrt {x^{2} \left (b x +a \right )}}-\frac {2 a \left (9 a b d x -6 b^{2} c x +8 a^{2} d -5 a b c \right ) x}{3 b^{4} \left (b x +a \right ) \sqrt {x^{2} \left (b x +a \right )}}\) | \(88\) |
pseudoelliptic | \(\frac {\frac {2 \left (\frac {11 d x}{13}+c \right ) x^{7} b^{8}}{11}-\frac {28 x^{6} a \left (\frac {72 d x}{91}+c \right ) b^{7}}{99}+\frac {16 x^{5} \left (\frac {28 d x}{39}+c \right ) a^{2} b^{6}}{33}-\frac {32 x^{4} \left (\frac {8 d x}{13}+c \right ) a^{3} b^{5}}{33}+\frac {256 x^{3} \left (\frac {6 d x}{13}+c \right ) a^{4} b^{4}}{99}-\frac {512 \left (\frac {8 d x}{39}+c \right ) x^{2} a^{5} b^{3}}{33}-\frac {2048 x \,a^{6} \left (-\frac {4 d x}{13}+c \right ) b^{2}}{33}-\frac {4096 \left (-\frac {24 d x}{13}+c \right ) a^{7} b}{99}+\frac {65536 a^{8} d}{1287}}{b^{9} \left (b x +a \right )^{\frac {3}{2}}}\) | \(143\) |
Input:
int(x^7*(d*x+c)/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(b*x+a)*(-b^3*d*x^3+6*a*b^2*d*x^2-3*b^3*c*x^2+24*a^2*b*d*x-12*a*b^2*c *x+16*a^3*d-8*a^2*b*c)*x^5/b^4/(b*x^3+a*x^2)^(5/2)
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (b^{3} d x^{3} + 8 \, a^{2} b c - 16 \, a^{3} d + 3 \, {\left (b^{3} c - 2 \, a b^{2} d\right )} x^{2} + 12 \, {\left (a b^{2} c - 2 \, a^{2} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, {\left (b^{6} x^{3} + 2 \, a b^{5} x^{2} + a^{2} b^{4} x\right )}} \] Input:
integrate(x^7*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
Output:
2/3*(b^3*d*x^3 + 8*a^2*b*c - 16*a^3*d + 3*(b^3*c - 2*a*b^2*d)*x^2 + 12*(a* b^2*c - 2*a^2*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^6*x^3 + 2*a*b^5*x^2 + a^2*b^4 *x)
\[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {x^{7} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**7*(d*x+c)/(b*x**3+a*x**2)**(5/2),x)
Output:
Integral(x**7*(c + d*x)/(x**2*(a + b*x))**(5/2), x)
Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.76 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} c}{3 \, {\left (b^{4} x + a b^{3}\right )} \sqrt {b x + a}} + \frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} d}{3 \, {\left (b^{5} x + a b^{4}\right )} \sqrt {b x + a}} \] Input:
integrate(x^7*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
Output:
2/3*(3*b^2*x^2 + 12*a*b*x + 8*a^2)*c/((b^4*x + a*b^3)*sqrt(b*x + a)) + 2/3 *(b^3*x^3 - 6*a*b^2*x^2 - 24*a^2*b*x - 16*a^3)*d/((b^5*x + a*b^4)*sqrt(b*x + a))
Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.97 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {16 \, {\left (a b c - 2 \, a^{2} d\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {a} b^{4}} + \frac {2 \, {\left (6 \, {\left (b x + a\right )} a b c - a^{2} b c - 9 \, {\left (b x + a\right )} a^{2} d + a^{3} d\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (3 \, \sqrt {b x + a} b^{9} c + {\left (b x + a\right )}^{\frac {3}{2}} b^{8} d - 9 \, \sqrt {b x + a} a b^{8} d\right )}}{3 \, b^{12} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^7*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
Output:
-16/3*(a*b*c - 2*a^2*d)*sgn(x)/(sqrt(a)*b^4) + 2/3*(6*(b*x + a)*a*b*c - a^ 2*b*c - 9*(b*x + a)*a^2*d + a^3*d)/((b*x + a)^(3/2)*b^4*sgn(x)) + 2/3*(3*s qrt(b*x + a)*b^9*c + (b*x + a)^(3/2)*b^8*d - 9*sqrt(b*x + a)*a*b^8*d)/(b^1 2*sgn(x))
Time = 9.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.68 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-16\,d\,a^3-24\,d\,a^2\,b\,x+8\,c\,a^2\,b-6\,d\,a\,b^2\,x^2+12\,c\,a\,b^2\,x+d\,b^3\,x^3+3\,c\,b^3\,x^2\right )}{3\,b^4\,x\,{\left (a+b\,x\right )}^2} \] Input:
int((x^7*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(3*b^3*c*x^2 - 16*a^3*d + b^3*d*x^3 + 8*a^2*b*c + 12*a*b^2*c*x - 24*a^2*b*d*x - 6*a*b^2*d*x^2))/(3*b^4*x*(a + b*x)^2)
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int \frac {x^7 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\frac {2}{3} b^{3} d \,x^{3}-4 a \,b^{2} d \,x^{2}+2 b^{3} c \,x^{2}-16 a^{2} b d x +8 a \,b^{2} c x -\frac {32}{3} a^{3} d +\frac {16}{3} a^{2} b c}{\sqrt {b x +a}\, b^{4} \left (b x +a \right )} \] Input:
int(x^7*(d*x+c)/(b*x^3+a*x^2)^(5/2),x)
Output:
(2*( - 16*a**3*d + 8*a**2*b*c - 24*a**2*b*d*x + 12*a*b**2*c*x - 6*a*b**2*d *x**2 + 3*b**3*c*x**2 + b**3*d*x**3))/(3*sqrt(a + b*x)*b**4*(a + b*x))