Integrand size = 23, antiderivative size = 31 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=a x+\frac {b x^2}{2}+\frac {1}{160} \left (2 c+2 a x+b x^2\right )^5 \] Output:
a*x+1/2*b*x^2+1/160*(b*x^2+2*a*x+2*c)^5
Leaf count is larger than twice the leaf count of optimal. \(108\) vs. \(2(31)=62\).
Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.48 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {1}{160} x (2 a+b x) \left (80+80 c^4+16 a^4 x^4+32 a^3 b x^5+24 a^2 b^2 x^6+8 a b^3 x^7+b^4 x^8+80 c^3 x (2 a+b x)+40 c^2 x^2 (2 a+b x)^2+10 c x^3 (2 a+b x)^3\right ) \] Input:
Integrate[(a + b*x)*(1 + (c + a*x + (b*x^2)/2)^4),x]
Output:
(x*(2*a + b*x)*(80 + 80*c^4 + 16*a^4*x^4 + 32*a^3*b*x^5 + 24*a^2*b^2*x^6 + 8*a*b^3*x^7 + b^4*x^8 + 80*c^3*x*(2*a + b*x) + 40*c^2*x^2*(2*a + b*x)^2 + 10*c*x^3*(2*a + b*x)^3))/160
Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2024, 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.
\(\displaystyle \int (a+b x) \left (\left (a x+\frac {b x^2}{2}+c\right )^4+1\right ) \, dx\) |
\(\Big \downarrow \) 2024 |
\(\displaystyle \int \left (\left (a x+\frac {b x^2}{2}+c\right )^4+1\right )d\left (a x+\frac {b x^2}{2}+c\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (a x+\frac {b x^2}{2}+c\right )^5+a x+\frac {b x^2}{2}+c\) |
Input:
Int[(a + b*x)*(1 + (c + a*x + (b*x^2)/2)^4),x]
Output:
c + a*x + (b*x^2)/2 + (c + a*x + (b*x^2)/2)^5/5
Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[P q, x], r = Expon[Qr, x]}, Simp[Coeff[Qr, x, r]/(q*Coeff[Pq, x, q]) Subst[ Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x, r]*D [Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] & & PolyQ[Qr, x]
Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\left (c +x a +\frac {1}{2} b \,x^{2}\right )^{5}}{5}+c +x a +\frac {b \,x^{2}}{2}\) | \(27\) |
norman | \(\left (\frac {1}{4} a^{2} b^{3}+\frac {1}{16} b^{4} c \right ) x^{8}+\left (\frac {1}{2} a^{3} b^{2}+\frac {1}{2} a \,b^{3} c \right ) x^{7}+\left (\frac {1}{5} a^{5}+2 a^{3} b c +\frac {3}{2} a \,b^{2} c^{2}\right ) x^{5}+\left (2 a^{2} c^{3}+\frac {1}{2} b \,c^{4}+\frac {1}{2} b \right ) x^{2}+\left (\frac {1}{2} a^{4} b +\frac {3}{2} c \,a^{2} b^{2}+\frac {1}{4} b^{3} c^{2}\right ) x^{6}+\left (a^{4} c +3 a^{2} b \,c^{2}+\frac {1}{2} b^{2} c^{3}\right ) x^{4}+\left (a \,c^{4}+a \right ) x +\left (2 c^{2} a^{3}+2 a b \,c^{3}\right ) x^{3}+\frac {b^{5} x^{10}}{160}+\frac {a \,b^{4} x^{9}}{16}\) | \(190\) |
gosper | \(\frac {x \left (b^{5} x^{9}+10 a \,b^{4} x^{8}+40 a^{2} b^{3} x^{7}+10 b^{4} c \,x^{7}+80 a^{3} b^{2} x^{6}+80 a \,b^{3} c \,x^{6}+80 a^{4} b \,x^{5}+240 a^{2} b^{2} c \,x^{5}+40 b^{3} c^{2} x^{5}+32 a^{5} x^{4}+320 x^{4} a^{3} b c +240 x^{4} a \,b^{2} c^{2}+160 a^{4} c \,x^{3}+480 a^{2} b \,c^{2} x^{3}+80 b^{2} c^{3} x^{3}+320 a^{3} c^{2} x^{2}+320 a b \,c^{3} x^{2}+320 a^{2} c^{3} x +80 b \,c^{4} x +160 a \,c^{4}+80 b x +160 a \right )}{160}\) | \(206\) |
risch | \(\frac {1}{160} b^{5} x^{10}+\frac {1}{16} a \,b^{4} x^{9}+\frac {1}{4} a^{2} b^{3} x^{8}+\frac {1}{16} x^{8} b^{4} c +\frac {1}{2} a^{3} b^{2} x^{7}+\frac {1}{2} x^{7} a \,b^{3} c +\frac {1}{2} a^{4} b \,x^{6}+\frac {3}{2} x^{6} c \,a^{2} b^{2}+\frac {1}{4} x^{6} b^{3} c^{2}+\frac {1}{5} a^{5} x^{5}+2 x^{5} a^{3} b c +\frac {3}{2} x^{5} a \,b^{2} c^{2}+x^{4} a^{4} c +3 x^{4} a^{2} b \,c^{2}+\frac {1}{2} x^{4} b^{2} c^{3}+2 a^{3} c^{2} x^{3}+2 a b \,c^{3} x^{3}+2 x^{2} a^{2} c^{3}+\frac {1}{2} x^{2} b \,c^{4}+\frac {1}{2} b \,x^{2}+c^{4} x a +x a\) | \(209\) |
parallelrisch | \(\frac {1}{160} b^{5} x^{10}+\frac {1}{16} a \,b^{4} x^{9}+\frac {1}{4} a^{2} b^{3} x^{8}+\frac {1}{16} x^{8} b^{4} c +\frac {1}{2} a^{3} b^{2} x^{7}+\frac {1}{2} x^{7} a \,b^{3} c +\frac {1}{2} a^{4} b \,x^{6}+\frac {3}{2} x^{6} c \,a^{2} b^{2}+\frac {1}{4} x^{6} b^{3} c^{2}+\frac {1}{5} a^{5} x^{5}+2 x^{5} a^{3} b c +\frac {3}{2} x^{5} a \,b^{2} c^{2}+x^{4} a^{4} c +3 x^{4} a^{2} b \,c^{2}+\frac {1}{2} x^{4} b^{2} c^{3}+2 a^{3} c^{2} x^{3}+2 a b \,c^{3} x^{3}+2 x^{2} a^{2} c^{3}+\frac {1}{2} x^{2} b \,c^{4}+\frac {1}{2} b \,x^{2}+c^{4} x a +x a\) | \(209\) |
orering | \(\frac {x \left (b^{5} x^{9}+10 a \,b^{4} x^{8}+40 a^{2} b^{3} x^{7}+10 b^{4} c \,x^{7}+80 a^{3} b^{2} x^{6}+80 a \,b^{3} c \,x^{6}+80 a^{4} b \,x^{5}+240 a^{2} b^{2} c \,x^{5}+40 b^{3} c^{2} x^{5}+32 a^{5} x^{4}+320 x^{4} a^{3} b c +240 x^{4} a \,b^{2} c^{2}+160 a^{4} c \,x^{3}+480 a^{2} b \,c^{2} x^{3}+80 b^{2} c^{3} x^{3}+320 a^{3} c^{2} x^{2}+320 a b \,c^{3} x^{2}+320 a^{2} c^{3} x +80 b \,c^{4} x +160 a \,c^{4}+80 b x +160 a \right ) \left (1+\left (c +x a +\frac {1}{2} b \,x^{2}\right )^{4}\right )}{10 b^{4} x^{8}+80 x^{7} a \,b^{3}+240 b^{2} x^{6} a^{2}+80 c \,b^{3} x^{6}+320 a^{3} b \,x^{5}+480 c \,b^{2} a \,x^{5}+160 x^{4} a^{4}+960 c \,a^{2} b \,x^{4}+240 b^{2} x^{4} c^{2}+640 c \,a^{3} x^{3}+960 a b \,c^{2} x^{3}+960 c^{2} a^{2} x^{2}+320 b \,c^{3} x^{2}+640 a \,c^{3} x +160 c^{4}+160}\) | \(360\) |
Input:
int((b*x+a)*(1+(c+x*a+1/2*b*x^2)^4),x,method=_RETURNVERBOSE)
Output:
1/5*(c+x*a+1/2*b*x^2)^5+c+x*a+1/2*b*x^2
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (27) = 54\).
Time = 0.06 (sec) , antiderivative size = 187, normalized size of antiderivative = 6.03 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {1}{160} \, b^{5} x^{10} + \frac {1}{16} \, a b^{4} x^{9} + \frac {1}{16} \, {\left (4 \, a^{2} b^{3} + b^{4} c\right )} x^{8} + \frac {1}{2} \, {\left (a^{3} b^{2} + a b^{3} c\right )} x^{7} + \frac {1}{4} \, {\left (2 \, a^{4} b + 6 \, a^{2} b^{2} c + b^{3} c^{2}\right )} x^{6} + \frac {1}{10} \, {\left (2 \, a^{5} + 20 \, a^{3} b c + 15 \, a b^{2} c^{2}\right )} x^{5} + \frac {1}{2} \, {\left (2 \, a^{4} c + 6 \, a^{2} b c^{2} + b^{2} c^{3}\right )} x^{4} + 2 \, {\left (a^{3} c^{2} + a b c^{3}\right )} x^{3} + \frac {1}{2} \, {\left (4 \, a^{2} c^{3} + b c^{4} + b\right )} x^{2} + {\left (a c^{4} + a\right )} x \] Input:
integrate((b*x+a)*(1+(c+a*x+1/2*b*x^2)^4),x, algorithm="fricas")
Output:
1/160*b^5*x^10 + 1/16*a*b^4*x^9 + 1/16*(4*a^2*b^3 + b^4*c)*x^8 + 1/2*(a^3* b^2 + a*b^3*c)*x^7 + 1/4*(2*a^4*b + 6*a^2*b^2*c + b^3*c^2)*x^6 + 1/10*(2*a ^5 + 20*a^3*b*c + 15*a*b^2*c^2)*x^5 + 1/2*(2*a^4*c + 6*a^2*b*c^2 + b^2*c^3 )*x^4 + 2*(a^3*c^2 + a*b*c^3)*x^3 + 1/2*(4*a^2*c^3 + b*c^4 + b)*x^2 + (a*c ^4 + a)*x
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (26) = 52\).
Time = 0.05 (sec) , antiderivative size = 194, normalized size of antiderivative = 6.26 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {a b^{4} x^{9}}{16} + \frac {b^{5} x^{10}}{160} + x^{8} \left (\frac {a^{2} b^{3}}{4} + \frac {b^{4} c}{16}\right ) + x^{7} \left (\frac {a^{3} b^{2}}{2} + \frac {a b^{3} c}{2}\right ) + x^{6} \left (\frac {a^{4} b}{2} + \frac {3 a^{2} b^{2} c}{2} + \frac {b^{3} c^{2}}{4}\right ) + x^{5} \left (\frac {a^{5}}{5} + 2 a^{3} b c + \frac {3 a b^{2} c^{2}}{2}\right ) + x^{4} \left (a^{4} c + 3 a^{2} b c^{2} + \frac {b^{2} c^{3}}{2}\right ) + x^{3} \cdot \left (2 a^{3} c^{2} + 2 a b c^{3}\right ) + x^{2} \cdot \left (2 a^{2} c^{3} + \frac {b c^{4}}{2} + \frac {b}{2}\right ) + x \left (a c^{4} + a\right ) \] Input:
integrate((b*x+a)*(1+(c+a*x+1/2*b*x**2)**4),x)
Output:
a*b**4*x**9/16 + b**5*x**10/160 + x**8*(a**2*b**3/4 + b**4*c/16) + x**7*(a **3*b**2/2 + a*b**3*c/2) + x**6*(a**4*b/2 + 3*a**2*b**2*c/2 + b**3*c**2/4) + x**5*(a**5/5 + 2*a**3*b*c + 3*a*b**2*c**2/2) + x**4*(a**4*c + 3*a**2*b* c**2 + b**2*c**3/2) + x**3*(2*a**3*c**2 + 2*a*b*c**3) + x**2*(2*a**2*c**3 + b*c**4/2 + b/2) + x*(a*c**4 + a)
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (27) = 54\).
Time = 0.03 (sec) , antiderivative size = 187, normalized size of antiderivative = 6.03 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {1}{160} \, b^{5} x^{10} + \frac {1}{16} \, a b^{4} x^{9} + \frac {1}{16} \, {\left (4 \, a^{2} b^{3} + b^{4} c\right )} x^{8} + \frac {1}{2} \, {\left (a^{3} b^{2} + a b^{3} c\right )} x^{7} + \frac {1}{4} \, {\left (2 \, a^{4} b + 6 \, a^{2} b^{2} c + b^{3} c^{2}\right )} x^{6} + \frac {1}{10} \, {\left (2 \, a^{5} + 20 \, a^{3} b c + 15 \, a b^{2} c^{2}\right )} x^{5} + \frac {1}{2} \, {\left (2 \, a^{4} c + 6 \, a^{2} b c^{2} + b^{2} c^{3}\right )} x^{4} + 2 \, {\left (a^{3} c^{2} + a b c^{3}\right )} x^{3} + \frac {1}{2} \, {\left (4 \, a^{2} c^{3} + b c^{4} + b\right )} x^{2} + {\left (a c^{4} + a\right )} x \] Input:
integrate((b*x+a)*(1+(c+a*x+1/2*b*x^2)^4),x, algorithm="maxima")
Output:
1/160*b^5*x^10 + 1/16*a*b^4*x^9 + 1/16*(4*a^2*b^3 + b^4*c)*x^8 + 1/2*(a^3* b^2 + a*b^3*c)*x^7 + 1/4*(2*a^4*b + 6*a^2*b^2*c + b^3*c^2)*x^6 + 1/10*(2*a ^5 + 20*a^3*b*c + 15*a*b^2*c^2)*x^5 + 1/2*(2*a^4*c + 6*a^2*b*c^2 + b^2*c^3 )*x^4 + 2*(a^3*c^2 + a*b*c^3)*x^3 + 1/2*(4*a^2*c^3 + b*c^4 + b)*x^2 + (a*c ^4 + a)*x
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (27) = 54\).
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.84 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {1}{160} \, {\left (b x^{2} + 2 \, a x\right )}^{5} + \frac {1}{16} \, {\left (b x^{2} + 2 \, a x\right )}^{4} c + \frac {1}{4} \, {\left (b x^{2} + 2 \, a x\right )}^{3} c^{2} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )}^{2} c^{3} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} c^{4} + \frac {1}{2} \, b x^{2} + a x \] Input:
integrate((b*x+a)*(1+(c+a*x+1/2*b*x^2)^4),x, algorithm="giac")
Output:
1/160*(b*x^2 + 2*a*x)^5 + 1/16*(b*x^2 + 2*a*x)^4*c + 1/4*(b*x^2 + 2*a*x)^3 *c^2 + 1/2*(b*x^2 + 2*a*x)^2*c^3 + 1/2*(b*x^2 + 2*a*x)*c^4 + 1/2*b*x^2 + a *x
Time = 21.81 (sec) , antiderivative size = 180, normalized size of antiderivative = 5.81 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=x^6\,\left (\frac {a^4\,b}{2}+\frac {3\,a^2\,b^2\,c}{2}+\frac {b^3\,c^2}{4}\right )+x^4\,\left (a^4\,c+3\,a^2\,b\,c^2+\frac {b^2\,c^3}{2}\right )+x^2\,\left (2\,a^2\,c^3+\frac {b\,c^4}{2}+\frac {b}{2}\right )+x^5\,\left (\frac {a^5}{5}+2\,a^3\,b\,c+\frac {3\,a\,b^2\,c^2}{2}\right )+\frac {b^5\,x^{10}}{160}+x^8\,\left (\frac {a^2\,b^3}{4}+\frac {c\,b^4}{16}\right )+\frac {a\,b^4\,x^9}{16}+a\,x\,\left (c^4+1\right )+\frac {a\,b^2\,x^7\,\left (a^2+b\,c\right )}{2}+2\,a\,c^2\,x^3\,\left (a^2+b\,c\right ) \] Input:
int(((c + a*x + (b*x^2)/2)^4 + 1)*(a + b*x),x)
Output:
x^6*((a^4*b)/2 + (b^3*c^2)/4 + (3*a^2*b^2*c)/2) + x^4*(a^4*c + (b^2*c^3)/2 + 3*a^2*b*c^2) + x^2*(b/2 + (b*c^4)/2 + 2*a^2*c^3) + x^5*(a^5/5 + (3*a*b^ 2*c^2)/2 + 2*a^3*b*c) + (b^5*x^10)/160 + x^8*((b^4*c)/16 + (a^2*b^3)/4) + (a*b^4*x^9)/16 + a*x*(c^4 + 1) + (a*b^2*x^7*(b*c + a^2))/2 + 2*a*c^2*x^3*( b*c + a^2)
Time = 0.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 6.61 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^4\right ) \, dx=\frac {x \left (b^{5} x^{9}+10 a \,b^{4} x^{8}+40 a^{2} b^{3} x^{7}+10 b^{4} c \,x^{7}+80 a^{3} b^{2} x^{6}+80 a \,b^{3} c \,x^{6}+80 a^{4} b \,x^{5}+240 a^{2} b^{2} c \,x^{5}+40 b^{3} c^{2} x^{5}+32 a^{5} x^{4}+320 a^{3} b c \,x^{4}+240 a \,b^{2} c^{2} x^{4}+160 a^{4} c \,x^{3}+480 a^{2} b \,c^{2} x^{3}+80 b^{2} c^{3} x^{3}+320 a^{3} c^{2} x^{2}+320 a b \,c^{3} x^{2}+320 a^{2} c^{3} x +80 b \,c^{4} x +160 a \,c^{4}+80 b x +160 a \right )}{160} \] Input:
int((b*x+a)*(1+(c+a*x+1/2*b*x^2)^4),x)
Output:
(x*(32*a**5*x**4 + 80*a**4*b*x**5 + 160*a**4*c*x**3 + 80*a**3*b**2*x**6 + 320*a**3*b*c*x**4 + 320*a**3*c**2*x**2 + 40*a**2*b**3*x**7 + 240*a**2*b**2 *c*x**5 + 480*a**2*b*c**2*x**3 + 320*a**2*c**3*x + 10*a*b**4*x**8 + 80*a*b **3*c*x**6 + 240*a*b**2*c**2*x**4 + 320*a*b*c**3*x**2 + 160*a*c**4 + 160*a + b**5*x**9 + 10*b**4*c*x**7 + 40*b**3*c**2*x**5 + 80*b**2*c**3*x**3 + 80 *b*c**4*x + 80*b*x))/160