Integrand size = 28, antiderivative size = 156 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{11 a (e x)^{21/2}}+\frac {2 (6 b c-11 a d) \left (a x^2+b x^3\right )^{5/2}}{99 a^2 (e x)^{19/2}}-\frac {8 b (6 b c-11 a d) \left (a x^2+b x^3\right )^{5/2}}{693 a^3 e (e x)^{17/2}}+\frac {16 b^2 (6 b c-11 a d) \left (a x^2+b x^3\right )^{5/2}}{3465 a^4 e^2 (e x)^{15/2}} \] Output:
-2/11*c*e*(b*x^3+a*x^2)^(5/2)/a/(e*x)^(21/2)+2/99*(-11*a*d+6*b*c)*(b*x^3+a *x^2)^(5/2)/a^2/(e*x)^(19/2)-8/693*b*(-11*a*d+6*b*c)*(b*x^3+a*x^2)^(5/2)/a ^3/e/(e*x)^(17/2)+16/3465*b^2*(-11*a*d+6*b*c)*(b*x^3+a*x^2)^(5/2)/a^4/e^2/ (e*x)^(15/2)
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.56 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=-\frac {2 \left (x^2 (a+b x)\right )^{5/2} \left (-48 b^3 c x^3+35 a^3 (9 c+11 d x)+8 a b^2 x^2 (15 c+11 d x)-10 a^2 b x (21 c+22 d x)\right )}{3465 a^4 e^8 x^9 (e x)^{3/2}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(19/2),x]
Output:
(-2*(x^2*(a + b*x))^(5/2)*(-48*b^3*c*x^3 + 35*a^3*(9*c + 11*d*x) + 8*a*b^2 *x^2*(15*c + 11*d*x) - 10*a^2*b*x*(21*c + 22*d*x)))/(3465*a^4*e^8*x^9*(e*x )^(3/2))
Time = 0.56 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1944, 1922, 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.
| \(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{(e x)^{19/2}} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle -\frac {(6 b c-11 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{17/2}}dx}{11 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{11 a (e x)^{21/2}}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle -\frac {(6 b c-11 a d) \left (-\frac {4 b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{15/2}}dx}{9 a e}-\frac {2 e \left (a x^2+b x^3\right )^{5/2}}{9 a (e x)^{19/2}}\right )}{11 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{11 a (e x)^{21/2}}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle -\frac {(6 b c-11 a d) \left (-\frac {4 b \left (-\frac {2 b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{13/2}}dx}{7 a e}-\frac {2 e \left (a x^2+b x^3\right )^{5/2}}{7 a (e x)^{17/2}}\right )}{9 a e}-\frac {2 e \left (a x^2+b x^3\right )^{5/2}}{9 a (e x)^{19/2}}\right )}{11 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{11 a (e x)^{21/2}}\) |
| \(\Big \downarrow \) 1920 |
| \(\displaystyle -\frac {(6 b c-11 a d) \left (-\frac {4 b \left (\frac {4 b \left (a x^2+b x^3\right )^{5/2}}{35 a^2 (e x)^{15/2}}-\frac {2 e \left (a x^2+b x^3\right )^{5/2}}{7 a (e x)^{17/2}}\right )}{9 a e}-\frac {2 e \left (a x^2+b x^3\right )^{5/2}}{9 a (e x)^{19/2}}\right )}{11 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{11 a (e x)^{21/2}}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(19/2),x]
Output:
(-2*c*e*(a*x^2 + b*x^3)^(5/2))/(11*a*(e*x)^(21/2)) - ((6*b*c - 11*a*d)*((- 2*e*(a*x^2 + b*x^3)^(5/2))/(9*a*(e*x)^(19/2)) - (4*b*((-2*e*(a*x^2 + b*x^3 )^(5/2))/(7*a*(e*x)^(17/2)) + (4*b*(a*x^2 + b*x^3)^(5/2))/(35*a^2*(e*x)^(1 5/2))))/(9*a*e)))/(11*a*e)
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*(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[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.47 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58
| method | result | size |
| gosper | \(-\frac {2 x \left (b x +a \right ) \left (88 a \,b^{2} d \,x^{3}-48 b^{3} c \,x^{3}-220 a^{2} b d \,x^{2}+120 a \,b^{2} c \,x^{2}+385 a^{3} d x -210 a^{2} b c x +315 c \,a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{3465 a^{4} \left (e x \right )^{\frac {19}{2}}}\) | \(91\) |
| orering | \(-\frac {2 x \left (b x +a \right ) \left (88 a \,b^{2} d \,x^{3}-48 b^{3} c \,x^{3}-220 a^{2} b d \,x^{2}+120 a \,b^{2} c \,x^{2}+385 a^{3} d x -210 a^{2} b c x +315 c \,a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{3465 a^{4} \left (e x \right )^{\frac {19}{2}}}\) | \(91\) |
| default | \(-\frac {2 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (88 x^{4} a \,b^{3} d -48 x^{4} b^{4} c -132 a^{2} b^{2} d \,x^{3}+72 a \,b^{3} c \,x^{3}+165 a^{3} b d \,x^{2}-90 a^{2} b^{2} c \,x^{2}+385 a^{4} d x +105 a^{3} b c x +315 c \,a^{4}\right )}{3465 x^{8} a^{4} e^{9} \sqrt {e x}}\) | \(115\) |
| risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (88 a \,b^{4} d \,x^{5}-48 b^{5} c \,x^{5}-44 x^{4} a^{2} b^{3} d +24 x^{4} a \,b^{4} c +33 a^{3} b^{2} d \,x^{3}-18 a^{2} b^{3} c \,x^{3}+550 a^{4} b d \,x^{2}+15 a^{3} b^{2} c \,x^{2}+385 a^{5} d x +420 a^{4} b c x +315 c \,a^{5}\right )}{3465 e^{9} x^{6} \sqrt {e x}\, a^{4}}\) | \(137\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(19/2),x,method=_RETURNVERBOSE)
Output:
-2/3465*x*(b*x+a)*(88*a*b^2*d*x^3-48*b^3*c*x^3-220*a^2*b*d*x^2+120*a*b^2*c *x^2+385*a^3*d*x-210*a^2*b*c*x+315*a^3*c)*(b*x^3+a*x^2)^(3/2)/a^4/(e*x)^(1 9/2)
Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=-\frac {2 \, {\left (315 \, a^{5} c - 8 \, {\left (6 \, b^{5} c - 11 \, a b^{4} d\right )} x^{5} + 4 \, {\left (6 \, a b^{4} c - 11 \, a^{2} b^{3} d\right )} x^{4} - 3 \, {\left (6 \, a^{2} b^{3} c - 11 \, a^{3} b^{2} d\right )} x^{3} + 5 \, {\left (3 \, a^{3} b^{2} c + 110 \, a^{4} b d\right )} x^{2} + 35 \, {\left (12 \, a^{4} b c + 11 \, a^{5} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{3465 \, a^{4} e^{10} x^{7}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(19/2),x, algorithm="fricas")
Output:
-2/3465*(315*a^5*c - 8*(6*b^5*c - 11*a*b^4*d)*x^5 + 4*(6*a*b^4*c - 11*a^2* b^3*d)*x^4 - 3*(6*a^2*b^3*c - 11*a^3*b^2*d)*x^3 + 5*(3*a^3*b^2*c + 110*a^4 *b*d)*x^2 + 35*(12*a^4*b*c + 11*a^5*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x)/(a ^4*e^10*x^7)
Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/(e*x)**(19/2),x)
Output:
Timed out
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{\left (e x\right )^{\frac {19}{2}}} \,d x } \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(19/2),x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/(e*x)^(19/2), x)
Time = 0.23 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.21 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=\frac {2 \, {\left ({\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (6 \, a b^{5} c e^{5} \mathrm {sgn}\left (x\right ) - 11 \, a^{2} b^{4} d e^{5} \mathrm {sgn}\left (x\right )\right )} {\left (b x + a\right )}}{a^{5}} - \frac {11 \, {\left (6 \, a^{2} b^{5} c e^{5} \mathrm {sgn}\left (x\right ) - 11 \, a^{3} b^{4} d e^{5} \mathrm {sgn}\left (x\right )\right )}}{a^{5}}\right )} + \frac {99 \, {\left (6 \, a^{3} b^{5} c e^{5} \mathrm {sgn}\left (x\right ) - 11 \, a^{4} b^{4} d e^{5} \mathrm {sgn}\left (x\right )\right )}}{a^{5}}\right )} - \frac {693 \, {\left (a^{4} b^{5} c e^{5} \mathrm {sgn}\left (x\right ) - a^{5} b^{4} d e^{5} \mathrm {sgn}\left (x\right )\right )}}{a^{5}}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b^{7}}{3465 \, {\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {11}{2}} e^{9} {\left | b \right |}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(19/2),x, algorithm="giac")
Output:
2/3465*((b*x + a)*(4*(b*x + a)*(2*(6*a*b^5*c*e^5*sgn(x) - 11*a^2*b^4*d*e^5 *sgn(x))*(b*x + a)/a^5 - 11*(6*a^2*b^5*c*e^5*sgn(x) - 11*a^3*b^4*d*e^5*sgn (x))/a^5) + 99*(6*a^3*b^5*c*e^5*sgn(x) - 11*a^4*b^4*d*e^5*sgn(x))/a^5) - 6 93*(a^4*b^5*c*e^5*sgn(x) - a^5*b^4*d*e^5*sgn(x))/a^5)*(b*x + a)^(5/2)*b^7/ (((b*x + a)*b*e - a*b*e)^(11/2)*e^9*abs(b))
Time = 9.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,a\,c}{11\,e^9}+\frac {x\,\left (770\,d\,a^5+840\,b\,c\,a^4\right )}{3465\,a^4\,e^9}-\frac {x^5\,\left (96\,b^5\,c-176\,a\,b^4\,d\right )}{3465\,a^4\,e^9}+\frac {2\,b^2\,x^3\,\left (11\,a\,d-6\,b\,c\right )}{1155\,a^2\,e^9}-\frac {8\,b^3\,x^4\,\left (11\,a\,d-6\,b\,c\right )}{3465\,a^3\,e^9}+\frac {2\,b\,x^2\,\left (110\,a\,d+3\,b\,c\right )}{693\,a\,e^9}\right )}{x^6\,\sqrt {e\,x}} \] Input:
int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/(e*x)^(19/2),x)
Output:
-((a*x^2 + b*x^3)^(1/2)*((2*a*c)/(11*e^9) + (x*(770*a^5*d + 840*a^4*b*c))/ (3465*a^4*e^9) - (x^5*(96*b^5*c - 176*a*b^4*d))/(3465*a^4*e^9) + (2*b^2*x^ 3*(11*a*d - 6*b*c))/(1155*a^2*e^9) - (8*b^3*x^4*(11*a*d - 6*b*c))/(3465*a^ 3*e^9) + (2*b*x^2*(110*a*d + 3*b*c))/(693*a*e^9)))/(x^6*(e*x)^(1/2))
Time = 0.22 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.49 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{19/2}} \, dx=\frac {2 \sqrt {e}\, \left (-315 \sqrt {x}\, \sqrt {b x +a}\, a^{5} c -385 \sqrt {x}\, \sqrt {b x +a}\, a^{5} d x -420 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b c x -550 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b d \,x^{2}-15 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} c \,x^{2}-33 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} d \,x^{3}+18 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} c \,x^{3}+44 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} d \,x^{4}-24 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} c \,x^{4}-88 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} d \,x^{5}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{5} c \,x^{5}+88 \sqrt {b}\, a \,b^{4} d \,x^{6}-48 \sqrt {b}\, b^{5} c \,x^{6}\right )}{3465 a^{4} e^{10} x^{6}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(19/2),x)
Output:
(2*sqrt(e)*( - 315*sqrt(x)*sqrt(a + b*x)*a**5*c - 385*sqrt(x)*sqrt(a + b*x )*a**5*d*x - 420*sqrt(x)*sqrt(a + b*x)*a**4*b*c*x - 550*sqrt(x)*sqrt(a + b *x)*a**4*b*d*x**2 - 15*sqrt(x)*sqrt(a + b*x)*a**3*b**2*c*x**2 - 33*sqrt(x) *sqrt(a + b*x)*a**3*b**2*d*x**3 + 18*sqrt(x)*sqrt(a + b*x)*a**2*b**3*c*x** 3 + 44*sqrt(x)*sqrt(a + b*x)*a**2*b**3*d*x**4 - 24*sqrt(x)*sqrt(a + b*x)*a *b**4*c*x**4 - 88*sqrt(x)*sqrt(a + b*x)*a*b**4*d*x**5 + 48*sqrt(x)*sqrt(a + b*x)*b**5*c*x**5 + 88*sqrt(b)*a*b**4*d*x**6 - 48*sqrt(b)*b**5*c*x**6))/( 3465*a**4*e**10*x**6)