Integrand size = 24, antiderivative size = 167 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=-\frac {2 a^3 (b c-a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^5 x^7}+\frac {2 a^2 (3 b c-4 a d) \left (a x^2+b x^3\right )^{9/2}}{9 b^5 x^9}-\frac {6 a (b c-2 a d) \left (a x^2+b x^3\right )^{11/2}}{11 b^5 x^{11}}+\frac {2 (b c-4 a d) \left (a x^2+b x^3\right )^{13/2}}{13 b^5 x^{13}}+\frac {2 d \left (a x^2+b x^3\right )^{15/2}}{15 b^5 x^{15}} \] Output:
-2/7*a^3*(-a*d+b*c)*(b*x^3+a*x^2)^(7/2)/b^5/x^7+2/9*a^2*(-4*a*d+3*b*c)*(b* x^3+a*x^2)^(9/2)/b^5/x^9-6/11*a*(-2*a*d+b*c)*(b*x^3+a*x^2)^(11/2)/b^5/x^11 +2/13*(-4*a*d+b*c)*(b*x^3+a*x^2)^(13/2)/b^5/x^13+2/15*d*(b*x^3+a*x^2)^(15/ 2)/b^5/x^15
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.59 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\frac {2 x (a+b x)^4 \left (128 a^4 d+168 a^2 b^2 x (5 c+6 d x)+231 b^4 x^3 (15 c+13 d x)-16 a^3 b (15 c+28 d x)-42 a b^3 x^2 (45 c+44 d x)\right )}{45045 b^5 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^2,x]
Output:
(2*x*(a + b*x)^4*(128*a^4*d + 168*a^2*b^2*x*(5*c + 6*d*x) + 231*b^4*x^3*(1 5*c + 13*d*x) - 16*a^3*b*(15*c + 28*d*x) - 42*a*b^3*x^2*(45*c + 44*d*x)))/ (45045*b^5*Sqrt[x^2*(a + b*x)])
Time = 0.61 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1945, 1922, 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.
| \(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2} (c+d x)}{x^2} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {(15 b c-8 a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^2}dx}{15 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle \frac {(15 b c-8 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^3}dx}{13 b}\right )}{15 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle \frac {(15 b c-8 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^4}dx}{11 b}\right )}{13 b}\right )}{15 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle \frac {(15 b c-8 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^5}dx}{9 b}\right )}{11 b}\right )}{13 b}\right )}{15 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}\) |
| \(\Big \downarrow \) 1920 |
| \(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {4 a \left (a x^2+b x^3\right )^{7/2}}{63 b^2 x^7}\right )}{11 b}\right )}{13 b}\right ) (15 b c-8 a d)}{15 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^2,x]
Output:
(2*d*(a*x^2 + b*x^3)^(7/2))/(15*b*x^3) + ((15*b*c - 8*a*d)*((2*(a*x^2 + b* x^3)^(7/2))/(13*b*x^4) - (6*a*((2*(a*x^2 + b*x^3)^(7/2))/(11*b*x^5) - (4*a *((-4*a*(a*x^2 + b*x^3)^(7/2))/(63*b^2*x^7) + (2*(a*x^2 + b*x^3)^(7/2))/(9 *b*x^6)))/(11*b)))/(13*b)))/(15*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*(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[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Result contains higher order function than in optimal. Order 3 vs. order 2.
Time = 0.55 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.51
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (a^{2} x \left (a d +\frac {5 b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {\left (7 x b \left (\frac {11 d x}{35}+c \right ) a^{\frac {3}{2}}+\left (\frac {23 d x}{5}-\frac {3 c}{2}\right ) a^{\frac {5}{2}}+b^{2} x^{2} \sqrt {a}\, \left (\frac {3 d x}{5}+c \right )\right ) \sqrt {b x +a}}{3}\right )}{\sqrt {a}\, x}\) | \(86\) |
| gosper | \(\frac {2 \left (b x +a \right ) \left (3003 d \,x^{4} b^{4}-1848 a \,b^{3} d \,x^{3}+3465 b^{4} c \,x^{3}+1008 a^{2} b^{2} d \,x^{2}-1890 a \,b^{3} c \,x^{2}-448 a^{3} b d x +840 a^{2} b^{2} c x +128 a^{4} d -240 a^{3} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(109\) |
| default | \(\frac {2 \left (b x +a \right ) \left (3003 d \,x^{4} b^{4}-1848 a \,b^{3} d \,x^{3}+3465 b^{4} c \,x^{3}+1008 a^{2} b^{2} d \,x^{2}-1890 a \,b^{3} c \,x^{2}-448 a^{3} b d x +840 a^{2} b^{2} c x +128 a^{4} d -240 a^{3} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(109\) |
| orering | \(\frac {2 \left (b x +a \right ) \left (3003 d \,x^{4} b^{4}-1848 a \,b^{3} d \,x^{3}+3465 b^{4} c \,x^{3}+1008 a^{2} b^{2} d \,x^{2}-1890 a \,b^{3} c \,x^{2}-448 a^{3} b d x +840 a^{2} b^{2} c x +128 a^{4} d -240 a^{3} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(109\) |
| risch | \(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (3003 b^{7} d \,x^{7}+7161 a \,b^{6} d \,x^{6}+3465 b^{7} c \,x^{6}+4473 a^{2} b^{5} d \,x^{5}+8505 a \,b^{6} c \,x^{5}+35 a^{3} b^{4} d \,x^{4}+5565 a^{2} b^{5} c \,x^{4}-40 a^{4} b^{3} d \,x^{3}+75 a^{3} b^{4} c \,x^{3}+48 a^{5} b^{2} d \,x^{2}-90 a^{4} b^{3} c \,x^{2}-64 a^{6} b d x +120 a^{5} b^{2} c x +128 a^{7} d -240 a^{6} b c \right )}{45045 x \,b^{5}}\) | \(174\) |
| trager | \(\frac {2 \left (3003 b^{7} d \,x^{7}+7161 a \,b^{6} d \,x^{6}+3465 b^{7} c \,x^{6}+4473 a^{2} b^{5} d \,x^{5}+8505 a \,b^{6} c \,x^{5}+35 a^{3} b^{4} d \,x^{4}+5565 a^{2} b^{5} c \,x^{4}-40 a^{4} b^{3} d \,x^{3}+75 a^{3} b^{4} c \,x^{3}+48 a^{5} b^{2} d \,x^{2}-90 a^{4} b^{3} c \,x^{2}-64 a^{6} b d x +120 a^{5} b^{2} c x +128 a^{7} d -240 a^{6} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{45045 b^{5} x}\) | \(176\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-2*(a^2*x*(a*d+5/2*b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))-1/3*(7*x*b*(11/35*d *x+c)*a^(3/2)+(23/5*d*x-3/2*c)*a^(5/2)+b^2*x^2*a^(1/2)*(3/5*d*x+c))*(b*x+a )^(1/2))/a^(1/2)/x
Time = 0.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.05 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\frac {2 \, {\left (3003 \, b^{7} d x^{7} - 240 \, a^{6} b c + 128 \, a^{7} d + 231 \, {\left (15 \, b^{7} c + 31 \, a b^{6} d\right )} x^{6} + 63 \, {\left (135 \, a b^{6} c + 71 \, a^{2} b^{5} d\right )} x^{5} + 35 \, {\left (159 \, a^{2} b^{5} c + a^{3} b^{4} d\right )} x^{4} + 5 \, {\left (15 \, a^{3} b^{4} c - 8 \, a^{4} b^{3} d\right )} x^{3} - 6 \, {\left (15 \, a^{4} b^{3} c - 8 \, a^{5} b^{2} d\right )} x^{2} + 8 \, {\left (15 \, a^{5} b^{2} c - 8 \, a^{6} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{45045 \, b^{5} x} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^2,x, algorithm="fricas")
Output:
2/45045*(3003*b^7*d*x^7 - 240*a^6*b*c + 128*a^7*d + 231*(15*b^7*c + 31*a*b ^6*d)*x^6 + 63*(135*a*b^6*c + 71*a^2*b^5*d)*x^5 + 35*(159*a^2*b^5*c + a^3* b^4*d)*x^4 + 5*(15*a^3*b^4*c - 8*a^4*b^3*d)*x^3 - 6*(15*a^4*b^3*c - 8*a^5* b^2*d)*x^2 + 8*(15*a^5*b^2*c - 8*a^6*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^5*x)
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{2}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**2,x)
Output:
Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**2, x)
Time = 0.04 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\frac {2 \, {\left (231 \, b^{6} x^{6} + 567 \, a b^{5} x^{5} + 371 \, a^{2} b^{4} x^{4} + 5 \, a^{3} b^{3} x^{3} - 6 \, a^{4} b^{2} x^{2} + 8 \, a^{5} b x - 16 \, a^{6}\right )} \sqrt {b x + a} c}{3003 \, b^{4}} + \frac {2 \, {\left (3003 \, b^{7} x^{7} + 7161 \, a b^{6} x^{6} + 4473 \, a^{2} b^{5} x^{5} + 35 \, a^{3} b^{4} x^{4} - 40 \, a^{4} b^{3} x^{3} + 48 \, a^{5} b^{2} x^{2} - 64 \, a^{6} b x + 128 \, a^{7}\right )} \sqrt {b x + a} d}{45045 \, b^{5}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^2,x, algorithm="maxima")
Output:
2/3003*(231*b^6*x^6 + 567*a*b^5*x^5 + 371*a^2*b^4*x^4 + 5*a^3*b^3*x^3 - 6* a^4*b^2*x^2 + 8*a^5*b*x - 16*a^6)*sqrt(b*x + a)*c/b^4 + 2/45045*(3003*b^7* x^7 + 7161*a*b^6*x^6 + 4473*a^2*b^5*x^5 + 35*a^3*b^4*x^4 - 40*a^4*b^3*x^3 + 48*a^5*b^2*x^2 - 64*a^6*b*x + 128*a^7)*sqrt(b*x + a)*d/b^5
Leaf count of result is larger than twice the leaf count of optimal. 650 vs. \(2 (147) = 294\).
Time = 0.14 (sec) , antiderivative size = 650, normalized size of antiderivative = 3.89 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^2,x, algorithm="giac")
Output:
2/45045*(1287*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/ 2)*a^2 - 35*sqrt(b*x + a)*a^3)*a^3*c*sgn(x)/b^3 + 429*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^ 3 + 315*sqrt(b*x + a)*a^4)*a^2*c*sgn(x)/b^3 + 143*(35*(b*x + a)^(9/2) - 18 0*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a^3*d*sgn(x)/b^4 + 195*(63*(b*x + a)^(11/2) - 385*( b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 11 55*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a*c*sgn(x)/b^3 + 195*(63*( b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*( b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a^2 *d*sgn(x)/b^4 + 15*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005* (b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)*c*sgn(x)/b^3 + 45*(23 1*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2) *a^5 + 3003*sqrt(b*x + a)*a^6)*a*d*sgn(x)/b^4 + 7*(429*(b*x + a)^(15/2) - 3465*(b*x + a)^(13/2)*a + 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/ 2)*a^3 + 32175*(b*x + a)^(7/2)*a^4 - 27027*(b*x + a)^(5/2)*a^5 + 15015*(b* x + a)^(3/2)*a^6 - 6435*sqrt(b*x + a)*a^7)*d*sgn(x)/b^4)/b + 32/45045*(15* a^(13/2)*b*c - 8*a^(15/2)*d)*sgn(x)/b^5
Time = 9.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {256\,a^7\,d-480\,a^6\,b\,c}{45045\,b^5}+\frac {2\,a\,x^5\,\left (71\,a\,d+135\,b\,c\right )}{715}+\frac {2\,b\,x^6\,\left (31\,a\,d+15\,b\,c\right )}{195}+\frac {2\,b^2\,d\,x^7}{15}-\frac {16\,a^5\,x\,\left (8\,a\,d-15\,b\,c\right )}{45045\,b^4}-\frac {2\,a^3\,x^3\,\left (8\,a\,d-15\,b\,c\right )}{9009\,b^2}+\frac {4\,a^4\,x^2\,\left (8\,a\,d-15\,b\,c\right )}{15015\,b^3}+\frac {2\,a^2\,x^4\,\left (a\,d+159\,b\,c\right )}{1287\,b}\right )}{x} \] Input:
int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^2,x)
Output:
((a*x^2 + b*x^3)^(1/2)*((256*a^7*d - 480*a^6*b*c)/(45045*b^5) + (2*a*x^5*( 71*a*d + 135*b*c))/715 + (2*b*x^6*(31*a*d + 15*b*c))/195 + (2*b^2*d*x^7)/1 5 - (16*a^5*x*(8*a*d - 15*b*c))/(45045*b^4) - (2*a^3*x^3*(8*a*d - 15*b*c)) /(9009*b^2) + (4*a^4*x^2*(8*a*d - 15*b*c))/(15015*b^3) + (2*a^2*x^4*(a*d + 159*b*c))/(1287*b)))/x
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^2} \, dx=\frac {2 \sqrt {b x +a}\, \left (3003 b^{7} d \,x^{7}+7161 a \,b^{6} d \,x^{6}+3465 b^{7} c \,x^{6}+4473 a^{2} b^{5} d \,x^{5}+8505 a \,b^{6} c \,x^{5}+35 a^{3} b^{4} d \,x^{4}+5565 a^{2} b^{5} c \,x^{4}-40 a^{4} b^{3} d \,x^{3}+75 a^{3} b^{4} c \,x^{3}+48 a^{5} b^{2} d \,x^{2}-90 a^{4} b^{3} c \,x^{2}-64 a^{6} b d x +120 a^{5} b^{2} c x +128 a^{7} d -240 a^{6} b c \right )}{45045 b^{5}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^2,x)
Output:
(2*sqrt(a + b*x)*(128*a**7*d - 240*a**6*b*c - 64*a**6*b*d*x + 120*a**5*b** 2*c*x + 48*a**5*b**2*d*x**2 - 90*a**4*b**3*c*x**2 - 40*a**4*b**3*d*x**3 + 75*a**3*b**4*c*x**3 + 35*a**3*b**4*d*x**4 + 5565*a**2*b**5*c*x**4 + 4473*a **2*b**5*d*x**5 + 8505*a*b**6*c*x**5 + 7161*a*b**6*d*x**6 + 3465*b**7*c*x* *6 + 3003*b**7*d*x**7))/(45045*b**5)