Integrand size = 24, antiderivative size = 88 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 a (b c-a d) x^3}{3 b^3 \left (a x^2+b x^3\right )^{3/2}}-\frac {2 (b c-2 a d) x}{b^3 \sqrt {a x^2+b x^3}}+\frac {2 d \sqrt {a x^2+b x^3}}{b^3 x} \] Output:
2/3*a*(-a*d+b*c)*x^3/b^3/(b*x^3+a*x^2)^(3/2)-2*(-2*a*d+b*c)*x/b^3/(b*x^3+a *x^2)^(1/2)+2*d*(b*x^3+a*x^2)^(1/2)/b^3/x
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 x^3 \left (8 a^2 d-2 a b (c-6 d x)+3 b^2 x (-c+d x)\right )}{3 b^3 \left (x^2 (a+b x)\right )^{3/2}} \] Input:
Integrate[(x^6*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*x^3*(8*a^2*d - 2*a*b*(c - 6*d*x) + 3*b^2*x*(-c + d*x)))/(3*b^3*(x^2*(a + b*x))^(3/2))
Time = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1943, 1921, 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.
\(\displaystyle \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {(b c-4 a d) \int \frac {x^4}{\left (b x^3+a x^2\right )^{3/2}}dx}{3 a b}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {(b c-4 a d) \left (\frac {2 \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{b}-\frac {2 x^2}{b \sqrt {a x^2+b x^3}}\right )}{3 a b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {\left (\frac {4 \sqrt {a x^2+b x^3}}{b^2 x}-\frac {2 x^2}{b \sqrt {a x^2+b x^3}}\right ) (b c-4 a d)}{3 a b}\) |
Input:
Int[(x^6*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*(b*c - a*d)*x^5)/(3*a*b*(a*x^2 + b*x^3)^(3/2)) - ((b*c - 4*a*d)*((-2*x^ 2)/(b*Sqrt[a*x^2 + b*x^3]) + (4*Sqrt[a*x^2 + b*x^3])/(b^2*x)))/(3*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[((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 = 0.90 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}+12 a b d x -3 b^{2} c x +8 a^{2} d -2 a b c \right ) x^{5}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
default | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}+12 a b d x -3 b^{2} c x +8 a^{2} d -2 a b c \right ) x^{5}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
orering | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}+12 a b d x -3 b^{2} c x +8 a^{2} d -2 a b c \right ) x^{5}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
trager | \(\frac {2 \left (3 b^{2} d \,x^{2}+12 a b d x -3 b^{2} c x +8 a^{2} d -2 a b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 x \,b^{3} \left (b x +a \right )^{2}}\) | \(63\) |
risch | \(\frac {2 d \left (b x +a \right ) x}{b^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {2 \left (6 a b d x -3 b^{2} c x +5 a^{2} d -2 a b c \right ) x}{3 b^{3} \left (b x +a \right ) \sqrt {x^{2} \left (b x +a \right )}}\) | \(74\) |
pseudoelliptic | \(-\frac {4096 \left (-\frac {11 x^{6} \left (\frac {9 d x}{11}+c \right ) b^{7}}{2048}+\frac {33 x^{5} a \left (\frac {49 d x}{66}+c \right ) b^{6}}{3584}-\frac {33 x^{4} \left (\frac {7 d x}{11}+c \right ) a^{2} b^{5}}{1792}+\frac {11 x^{3} a^{3} \left (\frac {21 d x}{44}+c \right ) b^{4}}{224}-\frac {33 \left (\frac {7 d x}{33}+c \right ) x^{2} a^{4} b^{3}}{112}-\frac {33 x \,a^{5} \left (-\frac {7 d x}{22}+c \right ) b^{2}}{28}-\frac {11 a^{6} \left (-\frac {21 d x}{11}+c \right ) b}{14}+a^{7} d \right )}{99 \left (b x +a \right )^{\frac {3}{2}} b^{8}}\) | \(126\) |
Input:
int(x^6*(d*x+c)/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/3*(b*x+a)*(3*b^2*d*x^2+12*a*b*d*x-3*b^2*c*x+8*a^2*d-2*a*b*c)*x^5/b^3/(b* x^3+a*x^2)^(5/2)
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} d x^{2} - 2 \, a b c + 8 \, a^{2} d - 3 \, {\left (b^{2} c - 4 \, a b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, {\left (b^{5} x^{3} + 2 \, a b^{4} x^{2} + a^{2} b^{3} x\right )}} \] Input:
integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
Output:
2/3*(3*b^2*d*x^2 - 2*a*b*c + 8*a^2*d - 3*(b^2*c - 4*a*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^5*x^3 + 2*a*b^4*x^2 + a^2*b^3*x)
\[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {x^{6} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**6*(d*x+c)/(b*x**3+a*x**2)**(5/2),x)
Output:
Integral(x**6*(c + d*x)/(x**2*(a + b*x))**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b x + 2 \, a\right )} c}{3 \, {\left (b^{3} x + a b^{2}\right )} \sqrt {b x + a}} + \frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} d}{3 \, {\left (b^{4} x + a b^{3}\right )} \sqrt {b x + a}} \] Input:
integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
Output:
-2/3*(3*b*x + 2*a)*c/((b^3*x + a*b^2)*sqrt(b*x + a)) + 2/3*(3*b^2*x^2 + 12 *a*b*x + 8*a^2)*d/((b^4*x + a*b^3)*sqrt(b*x + a))
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {4 \, {\left (b c - 4 \, a d\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {a} b^{3}} + \frac {2 \, \sqrt {b x + a} d}{b^{3} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (3 \, {\left (b x + a\right )} b c - a b c - 6 \, {\left (b x + a\right )} a d + a^{2} d\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
Output:
4/3*(b*c - 4*a*d)*sgn(x)/(sqrt(a)*b^3) + 2*sqrt(b*x + a)*d/(b^3*sgn(x)) - 2/3*(3*(b*x + a)*b*c - a*b*c - 6*(b*x + a)*a*d + a^2*d)/((b*x + a)^(3/2)*b ^3*sgn(x))
Time = 9.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (8\,d\,a^2+12\,d\,a\,b\,x-2\,c\,a\,b+3\,d\,b^2\,x^2-3\,c\,b^2\,x\right )}{3\,b^3\,x\,{\left (a+b\,x\right )}^2} \] Input:
int((x^6*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(8*a^2*d + 3*b^2*d*x^2 - 2*a*b*c - 3*b^2*c*x + 12 *a*b*d*x))/(3*b^3*x*(a + b*x)^2)
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 b^{2} d \,x^{2}+8 a b d x -2 b^{2} c x +\frac {16}{3} a^{2} d -\frac {4}{3} a b c}{\sqrt {b x +a}\, b^{3} \left (b x +a \right )} \] Input:
int(x^6*(d*x+c)/(b*x^3+a*x^2)^(5/2),x)
Output:
(2*(8*a**2*d - 2*a*b*c + 12*a*b*d*x - 3*b**2*c*x + 3*b**2*d*x**2))/(3*sqrt (a + b*x)*b**3*(a + b*x))