Integrand size = 42, antiderivative size = 160 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2 \left (A b^2-a (b B-a C)\right ) (a+b x) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2}}{11 b^3}+\frac {2 (b B-2 a C) (a+b x)^2 \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2}}{13 b^3}+\frac {2 C (a+b x)^3 \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2}}{15 b^3} \] Output:
2/11*(A*b^2-a*(B*b-C*a))*(b*x+a)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2) /b^3+2/13*(B*b-2*C*a)*(b*x+a)^2*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2)/ b^3+2/15*C*(b*x+a)^3*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2)/b^3
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.40 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2 (a+b x)^4 \sqrt {(a+b x)^3} \left (195 A b^2+8 a^2 C+11 b^2 x (15 B+13 C x)-2 a b (15 B+22 C x)\right )}{2145 b^3} \] Input:
Integrate[(A + B*x + C*x^2)*(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^(3/2 ),x]
Output:
(2*(a + b*x)^4*Sqrt[(a + b*x)^3]*(195*A*b^2 + 8*a^2*C + 11*b^2*x*(15*B + 1 3*C*x) - 2*a*b*(15*B + 22*C*x)))/(2145*b^3)
Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2008, 1140, 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.
\(\displaystyle \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2008 |
\(\displaystyle \frac {\left ((a+b x)^3\right )^{3/2} \int (a+b x)^{9/2} \left (C x^2+B x+A\right )dx}{(a+b x)^{9/2}}\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \frac {\left ((a+b x)^3\right )^{3/2} \int \left (\frac {C (a+b x)^{13/2}}{b^2}+\frac {(b B-2 a C) (a+b x)^{11/2}}{b^2}+\frac {\left (A b^2-a (b B-a C)\right ) (a+b x)^{9/2}}{b^2}\right )dx}{(a+b x)^{9/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left ((a+b x)^3\right )^{3/2} \left (\frac {2 (a+b x)^{11/2} \left (A b^2-a (b B-a C)\right )}{11 b^3}+\frac {2 (a+b x)^{13/2} (b B-2 a C)}{13 b^3}+\frac {2 C (a+b x)^{15/2}}{15 b^3}\right )}{(a+b x)^{9/2}}\) |
Input:
Int[(A + B*x + C*x^2)*(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^(3/2),x]
Output:
(((a + b*x)^3)^(3/2)*((2*(A*b^2 - a*(b*B - a*C))*(a + b*x)^(11/2))/(11*b^3 ) + (2*(b*B - 2*a*C)*(a + b*x)^(13/2))/(13*b^3) + (2*C*(a + b*x)^(15/2))/( 15*b^3)))/(a + b*x)^(9/2)
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p) Int[u*(a + b*x)^(Expon[Px, x]*p), x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; !IntegerQ[p] && PolyQ[Px, x ] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (143 C \,b^{2} x^{2}+165 B \,b^{2} x -44 C a b x +195 A \,b^{2}-30 a b B +8 C \,a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{\frac {3}{2}}}{2145 b^{3}}\) | \(80\) |
default | \(\frac {2 \left (b x +a \right ) \left (143 C \,b^{2} x^{2}+165 B \,b^{2} x -44 C a b x +195 A \,b^{2}-30 a b B +8 C \,a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{\frac {3}{2}}}{2145 b^{3}}\) | \(80\) |
orering | \(\frac {2 \left (b x +a \right ) \left (143 C \,b^{2} x^{2}+165 B \,b^{2} x -44 C a b x +195 A \,b^{2}-30 a b B +8 C \,a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{\frac {3}{2}}}{2145 b^{3}}\) | \(80\) |
trager | \(\frac {2 \left (143 C \,b^{6} x^{6}+165 B \,b^{6} x^{5}+528 C \,b^{5} a \,x^{5}+195 A \,x^{4} b^{6}+630 B \,x^{4} a \,b^{5}+690 C \,x^{4} a^{2} b^{4}+780 A \,x^{3} a \,b^{5}+870 B \,x^{3} a^{2} b^{4}+340 C \,x^{3} a^{3} b^{3}+1170 A \,x^{2} a^{2} b^{4}+480 B \,x^{2} a^{3} b^{3}+15 C \,x^{2} a^{4} b^{2}+780 A x \,a^{3} b^{3}+45 B x \,a^{4} b^{2}-12 C x \,a^{5} b +195 A \,a^{4} b^{2}-30 B \,a^{5} b +8 C \,a^{6}\right ) \sqrt {b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}}}{2145 b^{3}}\) | \(215\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{3}}\, \left (143 C \,b^{7} x^{7}+165 B \,b^{7} x^{6}+671 C a \,b^{6} x^{6}+195 A \,b^{7} x^{5}+795 B a \,b^{6} x^{5}+1218 C \,a^{2} b^{5} x^{5}+975 A a \,b^{6} x^{4}+1500 B \,a^{2} b^{5} x^{4}+1030 C \,a^{3} b^{4} x^{4}+1950 A \,a^{2} b^{5} x^{3}+1350 B \,a^{3} b^{4} x^{3}+355 C \,a^{4} b^{3} x^{3}+1950 A \,a^{3} b^{4} x^{2}+525 B \,a^{4} b^{3} x^{2}+3 C \,a^{5} b^{2} x^{2}+975 A \,a^{4} b^{3} x +15 B \,a^{5} b^{2} x -4 C \,a^{6} b x +195 A \,a^{5} b^{2}-30 B \,a^{6} b +8 C \,a^{7}\right )}{2145 \left (b x +a \right ) b^{3}}\) | \(238\) |
Input:
int((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2),x,method=_RETU RNVERBOSE)
Output:
2/2145*(b*x+a)*(143*C*b^2*x^2+165*B*b^2*x-44*C*a*b*x+195*A*b^2-30*B*a*b+8* C*a^2)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2)/b^3
Time = 0.08 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.28 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2 \, {\left (143 \, C b^{6} x^{6} + 8 \, C a^{6} - 30 \, B a^{5} b + 195 \, A a^{4} b^{2} + 33 \, {\left (16 \, C a b^{5} + 5 \, B b^{6}\right )} x^{5} + 15 \, {\left (46 \, C a^{2} b^{4} + 42 \, B a b^{5} + 13 \, A b^{6}\right )} x^{4} + 10 \, {\left (34 \, C a^{3} b^{3} + 87 \, B a^{2} b^{4} + 78 \, A a b^{5}\right )} x^{3} + 15 \, {\left (C a^{4} b^{2} + 32 \, B a^{3} b^{3} + 78 \, A a^{2} b^{4}\right )} x^{2} - 3 \, {\left (4 \, C a^{5} b - 15 \, B a^{4} b^{2} - 260 \, A a^{3} b^{3}\right )} x\right )} \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}{2145 \, b^{3}} \] Input:
integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2),x, algor ithm="fricas")
Output:
2/2145*(143*C*b^6*x^6 + 8*C*a^6 - 30*B*a^5*b + 195*A*a^4*b^2 + 33*(16*C*a* b^5 + 5*B*b^6)*x^5 + 15*(46*C*a^2*b^4 + 42*B*a*b^5 + 13*A*b^6)*x^4 + 10*(3 4*C*a^3*b^3 + 87*B*a^2*b^4 + 78*A*a*b^5)*x^3 + 15*(C*a^4*b^2 + 32*B*a^3*b^ 3 + 78*A*a^2*b^4)*x^2 - 3*(4*C*a^5*b - 15*B*a^4*b^2 - 260*A*a^3*b^3)*x)*sq rt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)/b^3
\[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\int \left (A + B x + C x^{2}\right ) \left (\left (a + b x\right )^{3}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((C*x**2+B*x+A)*(b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3)**(3/2), x)
Output:
Integral((A + B*x + C*x**2)*((a + b*x)**3)**(3/2), x)
Time = 0.05 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.21 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} {\left (b x + a\right )}^{\frac {3}{2}} A}{11 \, b} + \frac {2 \, {\left (11 \, b^{5} x^{5} + 42 \, a b^{4} x^{4} + 58 \, a^{2} b^{3} x^{3} + 32 \, a^{3} b^{2} x^{2} + 3 \, a^{4} b x - 2 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {3}{2}} B}{143 \, b^{2}} + \frac {2 \, {\left (143 \, b^{6} x^{6} + 528 \, a b^{5} x^{5} + 690 \, a^{2} b^{4} x^{4} + 340 \, a^{3} b^{3} x^{3} + 15 \, a^{4} b^{2} x^{2} - 12 \, a^{5} b x + 8 \, a^{6}\right )} {\left (b x + a\right )}^{\frac {3}{2}} C}{2145 \, b^{3}} \] Input:
integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2),x, algor ithm="maxima")
Output:
2/11*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*(b*x + a)^( 3/2)*A/b + 2/143*(11*b^5*x^5 + 42*a*b^4*x^4 + 58*a^2*b^3*x^3 + 32*a^3*b^2* x^2 + 3*a^4*b*x - 2*a^5)*(b*x + a)^(3/2)*B/b^2 + 2/2145*(143*b^6*x^6 + 528 *a*b^5*x^5 + 690*a^2*b^4*x^4 + 340*a^3*b^3*x^3 + 15*a^4*b^2*x^2 - 12*a^5*b *x + 8*a^6)*(b*x + a)^(3/2)*C/b^3
Leaf count of result is larger than twice the leaf count of optimal. 1138 vs. \(2 (148) = 296\).
Time = 0.14 (sec) , antiderivative size = 1138, normalized size of antiderivative = 7.11 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2),x, algor ithm="giac")
Output:
2/45045*(45045*sqrt(b*x + a)*A*a^5*sgn(b*x + a) + 75075*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A*a^4*sgn(b*x + a) + 15015*((b*x + a)^(3/2) - 3*sqrt(b *x + a)*a)*B*a^5*sgn(b*x + a)/b + 30030*(3*(b*x + a)^(5/2) - 10*(b*x + a)^ (3/2)*a + 15*sqrt(b*x + a)*a^2)*A*a^3*sgn(b*x + a) + 3003*(3*(b*x + a)^(5/ 2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*C*a^5*sgn(b*x + a)/b^2 + 15015*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*B *a^4*sgn(b*x + a)/b + 12870*(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*a^2*sgn(b*x + a) + 6435*(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)*C*a^4*sgn(b*x + a)/b^2 + 12870*(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)*B*a^3*sgn(b* x + a)/b + 715*(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*a*sgn(b*x + a) + 1430*(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)*C*a^3*sgn(b*x + a)/b^2 + 1430*(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)*B*a^2*sgn(b* x + a)/b + 65*(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*sqr t(b*x + a)*a^5)*A*sgn(b*x + a) + 650*(63*(b*x + a)^(11/2) - 385*(b*x + ...
Time = 12.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.51 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2\,{\left (a+b\,x\right )}^4\,\sqrt {a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3}\,\left (8\,C\,a^2-44\,C\,a\,b\,x-30\,B\,a\,b+143\,C\,b^2\,x^2+165\,B\,b^2\,x+195\,A\,b^2\right )}{2145\,b^3} \] Input:
int((A + B*x + C*x^2)*(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^(3/2),x)
Output:
(2*(a + b*x)^4*(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^(1/2)*(195*A*b^2 + 8*C*a^2 + 143*C*b^2*x^2 - 30*B*a*b + 165*B*b^2*x - 44*C*a*b*x))/(2145*b^ 3)
Time = 0.15 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^{3/2} \, dx=\frac {2 \sqrt {b x +a}\, \left (143 b^{7} c \,x^{7}+671 a \,b^{6} c \,x^{6}+165 b^{8} x^{6}+1218 a^{2} b^{5} c \,x^{5}+990 a \,b^{7} x^{5}+1030 a^{3} b^{4} c \,x^{4}+2475 a^{2} b^{6} x^{4}+355 a^{4} b^{3} c \,x^{3}+3300 a^{3} b^{5} x^{3}+3 a^{5} b^{2} c \,x^{2}+2475 a^{4} b^{4} x^{2}-4 a^{6} b c x +990 a^{5} b^{3} x +8 a^{7} c +165 a^{6} b^{2}\right )}{2145 b^{3}} \] Input:
int((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(3/2),x)
Output:
(2*sqrt(a + b*x)*(8*a**7*c + 165*a**6*b**2 - 4*a**6*b*c*x + 990*a**5*b**3* x + 3*a**5*b**2*c*x**2 + 2475*a**4*b**4*x**2 + 355*a**4*b**3*c*x**3 + 3300 *a**3*b**5*x**3 + 1030*a**3*b**4*c*x**4 + 2475*a**2*b**6*x**4 + 1218*a**2* b**5*c*x**5 + 990*a*b**7*x**5 + 671*a*b**6*c*x**6 + 165*b**8*x**6 + 143*b* *7*c*x**7))/(2145*b**3)