Integrand size = 36, antiderivative size = 98 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(b f-a g) (a c+b c x)^{3+m} \sqrt {a^2+2 a b x+b^2 x^2}}{b^2 c^3 (4+m)}+\frac {g (a c+b c x)^{4+m} \sqrt {a^2+2 a b x+b^2 x^2}}{b^2 c^4 (5+m)} \] Output:
(-a*g+b*f)*(b*c*x+a*c)^(3+m)*((b*x+a)^2)^(1/2)/b^2/c^3/(4+m)+g*(b*c*x+a*c) ^(4+m)*((b*x+a)^2)^(1/2)/b^2/c^4/(5+m)
Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.60 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x)^3 (c (a+b x))^m \sqrt {(a+b x)^2} (-a g+b f (5+m)+b g (4+m) x)}{b^2 (4+m) (5+m)} \] Input:
Integrate[(a*c + b*c*x)^m*(f + g*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
((a + b*x)^3*(c*(a + b*x))^m*Sqrt[(a + b*x)^2]*(-(a*g) + b*f*(5 + m) + b*g *(4 + m)*x))/(b^2*(4 + m)*(5 + m))
Time = 0.42 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {1187, 27, 35, 53, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (f+g x) (a c+b c x)^m \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 (a c+b x c)^m (f+g x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 (a c+b x c)^m (f+g x)dx}{a+b x}\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a c+b x c)^{m+3} (f+g x)dx}{c^3 (a+b x)}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(b f-a g) (a c+b x c)^{m+3}}{b}+\frac {g (a c+b x c)^{m+4}}{b c}\right )dx}{c^3 (a+b x)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {g (a c+b c x)^{m+5}}{b^2 c^2 (m+5)}+\frac {(b f-a g) (a c+b c x)^{m+4}}{b^2 c (m+4)}\right )}{c^3 (a+b x)}\) |
Input:
Int[(a*c + b*c*x)^m*(f + g*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((b*f - a*g)*(a*c + b*c*x)^(4 + m))/(b^2*c *(4 + m)) + (g*(a*c + b*c*x)^(5 + m))/(b^2*c^2*(5 + m))))/(c^3*(a + b*x))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 1.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (c b x +a c \right )^{m} \left (-b g m x -b f m -4 b g x +a g -5 f b \right ) \left (b x +a \right )}{b^{2} \left (m^{2}+9 m +20\right )}\) | \(64\) |
orering | \(-\frac {\left (-b g m x -b f m -4 b g x +a g -5 f b \right ) \left (b x +a \right ) \left (c b x +a c \right )^{m} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{b^{2} \left (m^{2}+9 m +20\right )}\) | \(73\) |
risch | \(-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-b^{5} g m \,x^{5}-4 a \,b^{4} g m \,x^{4}-b^{5} f m \,x^{4}-4 b^{5} g \,x^{5}-6 a^{2} b^{3} g m \,x^{3}-4 a \,b^{4} f m \,x^{3}-15 a \,b^{4} g \,x^{4}-5 b^{5} f \,x^{4}-4 a^{3} b^{2} g m \,x^{2}-6 a^{2} b^{3} f m \,x^{2}-20 a^{2} b^{3} g \,x^{3}-20 a \,b^{4} f \,x^{3}-a^{4} b g m x -4 a^{3} b^{2} f m x -10 a^{3} b^{2} g \,x^{2}-30 a^{2} b^{3} f \,x^{2}-a^{4} b f m -20 a^{3} b^{2} f x +g \,a^{5}-5 a^{4} b f \right ) \left (c \left (b x +a \right )\right )^{m}}{\left (b x +a \right ) \left (4+m \right ) \left (5+m \right ) b^{2}}\) | \(247\) |
Input:
int((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERB OSE)
Output:
-((b*x+a)^2)^(3/2)*(b*c*x+a*c)^m*(-b*g*m*x-b*f*m-4*b*g*x+a*g-5*b*f)*(b*x+a )/b^2/(m^2+9*m+20)
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (76) = 152\).
Time = 0.09 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.23 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (a^{4} b f m + 5 \, a^{4} b f - a^{5} g + {\left (b^{5} g m + 4 \, b^{5} g\right )} x^{5} + {\left (5 \, b^{5} f + 15 \, a b^{4} g + {\left (b^{5} f + 4 \, a b^{4} g\right )} m\right )} x^{4} + 2 \, {\left (10 \, a b^{4} f + 10 \, a^{2} b^{3} g + {\left (2 \, a b^{4} f + 3 \, a^{2} b^{3} g\right )} m\right )} x^{3} + 2 \, {\left (15 \, a^{2} b^{3} f + 5 \, a^{3} b^{2} g + {\left (3 \, a^{2} b^{3} f + 2 \, a^{3} b^{2} g\right )} m\right )} x^{2} + {\left (20 \, a^{3} b^{2} f + {\left (4 \, a^{3} b^{2} f + a^{4} b g\right )} m\right )} x\right )} {\left (b c x + a c\right )}^{m}}{b^{2} m^{2} + 9 \, b^{2} m + 20 \, b^{2}} \] Input:
integrate((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" fricas")
Output:
(a^4*b*f*m + 5*a^4*b*f - a^5*g + (b^5*g*m + 4*b^5*g)*x^5 + (5*b^5*f + 15*a *b^4*g + (b^5*f + 4*a*b^4*g)*m)*x^4 + 2*(10*a*b^4*f + 10*a^2*b^3*g + (2*a* b^4*f + 3*a^2*b^3*g)*m)*x^3 + 2*(15*a^2*b^3*f + 5*a^3*b^2*g + (3*a^2*b^3*f + 2*a^3*b^2*g)*m)*x^2 + (20*a^3*b^2*f + (4*a^3*b^2*f + a^4*b*g)*m)*x)*(b* c*x + a*c)^m/(b^2*m^2 + 9*b^2*m + 20*b^2)
\[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \left (c \left (a + b x\right )\right )^{m} \left (f + g x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((b*c*x+a*c)**m*(g*x+f)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Integral((c*(a + b*x))**m*(f + g*x)*((a + b*x)**2)**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (76) = 152\).
Time = 0.07 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.84 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{4} c^{m} x^{4} + 4 \, a b^{3} c^{m} x^{3} + 6 \, a^{2} b^{2} c^{m} x^{2} + 4 \, a^{3} b c^{m} x + a^{4} c^{m}\right )} {\left (b x + a\right )}^{m} f}{b {\left (m + 4\right )}} + \frac {{\left (b^{5} c^{m} {\left (m + 4\right )} x^{5} + a b^{4} c^{m} {\left (4 \, m + 15\right )} x^{4} + 2 \, a^{2} b^{3} c^{m} {\left (3 \, m + 10\right )} x^{3} + 2 \, a^{3} b^{2} c^{m} {\left (2 \, m + 5\right )} x^{2} + a^{4} b c^{m} m x - a^{5} c^{m}\right )} {\left (b x + a\right )}^{m} g}{{\left (m^{2} + 9 \, m + 20\right )} b^{2}} \] Input:
integrate((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" maxima")
Output:
(b^4*c^m*x^4 + 4*a*b^3*c^m*x^3 + 6*a^2*b^2*c^m*x^2 + 4*a^3*b*c^m*x + a^4*c ^m)*(b*x + a)^m*f/(b*(m + 4)) + (b^5*c^m*(m + 4)*x^5 + a*b^4*c^m*(4*m + 15 )*x^4 + 2*a^2*b^3*c^m*(3*m + 10)*x^3 + 2*a^3*b^2*c^m*(2*m + 5)*x^2 + a^4*b *c^m*m*x - a^5*c^m)*(b*x + a)^m*g/((m^2 + 9*m + 20)*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 545 vs. \(2 (76) = 152\).
Time = 0.20 (sec) , antiderivative size = 545, normalized size of antiderivative = 5.56 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b c x + a c\right )}^{m} b^{5} g m x^{5} \mathrm {sgn}\left (b x + a\right ) + {\left (b c x + a c\right )}^{m} b^{5} f m x^{4} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (b c x + a c\right )}^{m} a b^{4} g m x^{4} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (b c x + a c\right )}^{m} b^{5} g x^{5} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (b c x + a c\right )}^{m} a b^{4} f m x^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, {\left (b c x + a c\right )}^{m} a^{2} b^{3} g m x^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (b c x + a c\right )}^{m} b^{5} f x^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (b c x + a c\right )}^{m} a b^{4} g x^{4} \mathrm {sgn}\left (b x + a\right ) + 6 \, {\left (b c x + a c\right )}^{m} a^{2} b^{3} f m x^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (b c x + a c\right )}^{m} a^{3} b^{2} g m x^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (b c x + a c\right )}^{m} a b^{4} f x^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (b c x + a c\right )}^{m} a^{2} b^{3} g x^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (b c x + a c\right )}^{m} a^{3} b^{2} f m x \mathrm {sgn}\left (b x + a\right ) + {\left (b c x + a c\right )}^{m} a^{4} b g m x \mathrm {sgn}\left (b x + a\right ) + 30 \, {\left (b c x + a c\right )}^{m} a^{2} b^{3} f x^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (b c x + a c\right )}^{m} a^{3} b^{2} g x^{2} \mathrm {sgn}\left (b x + a\right ) + {\left (b c x + a c\right )}^{m} a^{4} b f m \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (b c x + a c\right )}^{m} a^{3} b^{2} f x \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (b c x + a c\right )}^{m} a^{4} b f \mathrm {sgn}\left (b x + a\right ) - {\left (b c x + a c\right )}^{m} a^{5} g \mathrm {sgn}\left (b x + a\right )}{b^{2} m^{2} + 9 \, b^{2} m + 20 \, b^{2}} \] Input:
integrate((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" giac")
Output:
((b*c*x + a*c)^m*b^5*g*m*x^5*sgn(b*x + a) + (b*c*x + a*c)^m*b^5*f*m*x^4*sg n(b*x + a) + 4*(b*c*x + a*c)^m*a*b^4*g*m*x^4*sgn(b*x + a) + 4*(b*c*x + a*c )^m*b^5*g*x^5*sgn(b*x + a) + 4*(b*c*x + a*c)^m*a*b^4*f*m*x^3*sgn(b*x + a) + 6*(b*c*x + a*c)^m*a^2*b^3*g*m*x^3*sgn(b*x + a) + 5*(b*c*x + a*c)^m*b^5*f *x^4*sgn(b*x + a) + 15*(b*c*x + a*c)^m*a*b^4*g*x^4*sgn(b*x + a) + 6*(b*c*x + a*c)^m*a^2*b^3*f*m*x^2*sgn(b*x + a) + 4*(b*c*x + a*c)^m*a^3*b^2*g*m*x^2 *sgn(b*x + a) + 20*(b*c*x + a*c)^m*a*b^4*f*x^3*sgn(b*x + a) + 20*(b*c*x + a*c)^m*a^2*b^3*g*x^3*sgn(b*x + a) + 4*(b*c*x + a*c)^m*a^3*b^2*f*m*x*sgn(b* x + a) + (b*c*x + a*c)^m*a^4*b*g*m*x*sgn(b*x + a) + 30*(b*c*x + a*c)^m*a^2 *b^3*f*x^2*sgn(b*x + a) + 10*(b*c*x + a*c)^m*a^3*b^2*g*x^2*sgn(b*x + a) + (b*c*x + a*c)^m*a^4*b*f*m*sgn(b*x + a) + 20*(b*c*x + a*c)^m*a^3*b^2*f*x*sg n(b*x + a) + 5*(b*c*x + a*c)^m*a^4*b*f*sgn(b*x + a) - (b*c*x + a*c)^m*a^5* g*sgn(b*x + a))/(b^2*m^2 + 9*b^2*m + 20*b^2)
Time = 12.08 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.59 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx={\left (a\,c+b\,c\,x\right )}^m\,\left (\frac {a^3\,\left (5\,b\,f-a\,g+b\,f\,m\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^2\,\left (m^2+9\,m+20\right )}+\frac {3\,a\,x^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (3\,a\,g+5\,b\,f+a\,g\,m+b\,f\,m\right )}{m^2+9\,m+20}+\frac {b\,x^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (11\,a\,g+5\,b\,f+3\,a\,g\,m+b\,f\,m\right )}{m^2+9\,m+20}+\frac {b^2\,g\,x^4\,\left (m+4\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{m^2+9\,m+20}+\frac {a^2\,x\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a\,g+15\,b\,f+a\,g\,m+3\,b\,f\,m\right )}{b\,\left (m^2+9\,m+20\right )}\right ) \] Input:
int((f + g*x)*(a*c + b*c*x)^m*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
Output:
(a*c + b*c*x)^m*((a^3*(5*b*f - a*g + b*f*m)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2 ))/(b^2*(9*m + m^2 + 20)) + (3*a*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(3*a* g + 5*b*f + a*g*m + b*f*m))/(9*m + m^2 + 20) + (b*x^3*(a^2 + b^2*x^2 + 2*a *b*x)^(1/2)*(11*a*g + 5*b*f + 3*a*g*m + b*f*m))/(9*m + m^2 + 20) + (b^2*g* x^4*(m + 4)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(9*m + m^2 + 20) + (a^2*x*(a^ 2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a*g + 15*b*f + a*g*m + 3*b*f*m))/(b*(9*m + m ^2 + 20)))
Time = 0.27 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.32 \[ \int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (b c x +a c \right )^{m} \left (b^{5} g m \,x^{5}+4 a \,b^{4} g m \,x^{4}+b^{5} f m \,x^{4}+4 b^{5} g \,x^{5}+6 a^{2} b^{3} g m \,x^{3}+4 a \,b^{4} f m \,x^{3}+15 a \,b^{4} g \,x^{4}+5 b^{5} f \,x^{4}+4 a^{3} b^{2} g m \,x^{2}+6 a^{2} b^{3} f m \,x^{2}+20 a^{2} b^{3} g \,x^{3}+20 a \,b^{4} f \,x^{3}+a^{4} b g m x +4 a^{3} b^{2} f m x +10 a^{3} b^{2} g \,x^{2}+30 a^{2} b^{3} f \,x^{2}+a^{4} b f m +20 a^{3} b^{2} f x -a^{5} g +5 a^{4} b f \right )}{b^{2} \left (m^{2}+9 m +20\right )} \] Input:
int((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
((a*c + b*c*x)**m*( - a**5*g + a**4*b*f*m + 5*a**4*b*f + a**4*b*g*m*x + 4* a**3*b**2*f*m*x + 20*a**3*b**2*f*x + 4*a**3*b**2*g*m*x**2 + 10*a**3*b**2*g *x**2 + 6*a**2*b**3*f*m*x**2 + 30*a**2*b**3*f*x**2 + 6*a**2*b**3*g*m*x**3 + 20*a**2*b**3*g*x**3 + 4*a*b**4*f*m*x**3 + 20*a*b**4*f*x**3 + 4*a*b**4*g* m*x**4 + 15*a*b**4*g*x**4 + b**5*f*m*x**4 + 5*b**5*f*x**4 + b**5*g*m*x**5 + 4*b**5*g*x**5))/(b**2*(m**2 + 9*m + 20))