Integrand size = 30, antiderivative size = 208 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {2 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x)}+\frac {6 b (b d-a e)^2 (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x)}-\frac {2 b^2 (b d-a e) (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^4 (a+b x)} \] Output:
-2/5*(-a*e+b*d)^3*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+6/7*b*(-a*e+ b*d)^2*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)-2/3*b^2*(-a*e+b*d)*(e*x +d)^(9/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/11*b^3*(e*x+d)^(11/2)*((b*x+a)^2 )^(1/2)/e^4/(b*x+a)
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.58 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{5/2} \left (231 a^3 e^3+99 a^2 b e^2 (-2 d+5 e x)+11 a b^2 e \left (8 d^2-20 d e x+35 e^2 x^2\right )+b^3 \left (-16 d^3+40 d^2 e x-70 d e^2 x^2+105 e^3 x^3\right )\right )}{1155 e^4 (a+b x)} \] Input:
Integrate[(d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(231*a^3*e^3 + 99*a^2*b*e^2*(-2*d + 5 *e*x) + 11*a*b^2*e*(8*d^2 - 20*d*e*x + 35*e^2*x^2) + b^3*(-16*d^3 + 40*d^2 *e*x - 70*d*e^2*x^2 + 105*e^3*x^3)))/(1155*e^4*(a + b*x))
Time = 0.45 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1102, 27, 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.
| \(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 (d+e x)^{3/2}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 (d+e x)^{3/2}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3 (d+e x)^{9/2}}{e^3}-\frac {3 b^2 (b d-a e) (d+e x)^{7/2}}{e^3}+\frac {3 b (b d-a e)^2 (d+e x)^{5/2}}{e^3}+\frac {(a e-b d)^3 (d+e x)^{3/2}}{e^3}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 b^2 (d+e x)^{9/2} (b d-a e)}{3 e^4}+\frac {6 b (d+e x)^{7/2} (b d-a e)^2}{7 e^4}-\frac {2 (d+e x)^{5/2} (b d-a e)^3}{5 e^4}+\frac {2 b^3 (d+e x)^{11/2}}{11 e^4}\right )}{a+b x}\) |
Input:
Int[(d + e*x)^(3/2)*(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]*((-2*(b*d - a*e)^3*(d + e*x)^(5/2))/(5*e^4) + (6*b*(b*d - a*e)^2*(d + e*x)^(7/2))/(7*e^4) - (2*b^2*(b*d - a*e)*(d + e *x)^(9/2))/(3*e^4) + (2*b^3*(d + e*x)^(11/2))/(11*e^4)))/(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[((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_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 1.55 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63
| method | result | size |
| gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (105 e^{3} x^{3} b^{3}+385 x^{2} a \,b^{2} e^{3}-70 x^{2} b^{3} d \,e^{2}+495 a^{2} b \,e^{3} x -220 x a \,b^{2} d \,e^{2}+40 b^{3} d^{2} e x +231 e^{3} a^{3}-198 a^{2} b d \,e^{2}+88 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{1155 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
| default | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (105 e^{3} x^{3} b^{3}+385 x^{2} a \,b^{2} e^{3}-70 x^{2} b^{3} d \,e^{2}+495 a^{2} b \,e^{3} x -220 x a \,b^{2} d \,e^{2}+40 b^{3} d^{2} e x +231 e^{3} a^{3}-198 a^{2} b d \,e^{2}+88 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{1155 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
| orering | \(\frac {2 \left (105 e^{3} x^{3} b^{3}+385 x^{2} a \,b^{2} e^{3}-70 x^{2} b^{3} d \,e^{2}+495 a^{2} b \,e^{3} x -220 x a \,b^{2} d \,e^{2}+40 b^{3} d^{2} e x +231 e^{3} a^{3}-198 a^{2} b d \,e^{2}+88 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (e x +d \right )^{\frac {5}{2}} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{1155 e^{4} \left (b x +a \right )^{3}}\) | \(141\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (105 b^{3} x^{5} e^{5}+385 a \,b^{2} e^{5} x^{4}+140 b^{3} d \,e^{4} x^{4}+495 a^{2} b \,e^{5} x^{3}+550 a \,b^{2} d \,e^{4} x^{3}+5 b^{3} d^{2} e^{3} x^{3}+231 a^{3} e^{5} x^{2}+792 a^{2} b d \,e^{4} x^{2}+33 a \,b^{2} d^{2} e^{3} x^{2}-6 b^{3} d^{3} e^{2} x^{2}+462 a^{3} d \,e^{4} x +99 a^{2} b \,d^{2} e^{3} x -44 a \,b^{2} d^{3} e^{2} x +8 b^{3} d^{4} e x +231 a^{3} d^{2} e^{3}-198 a^{2} b \,d^{3} e^{2}+88 a \,b^{2} d^{4} e -16 b^{3} d^{5}\right ) \sqrt {e x +d}}{1155 \left (b x +a \right ) e^{4}}\) | \(244\) |
Input:
int((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/1155*(e*x+d)^(5/2)*(105*b^3*e^3*x^3+385*a*b^2*e^3*x^2-70*b^3*d*e^2*x^2+4 95*a^2*b*e^3*x-220*a*b^2*d*e^2*x+40*b^3*d^2*e*x+231*a^3*e^3-198*a^2*b*d*e^ 2+88*a*b^2*d^2*e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3
Time = 0.09 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.04 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{3} e^{5} x^{5} - 16 \, b^{3} d^{5} + 88 \, a b^{2} d^{4} e - 198 \, a^{2} b d^{3} e^{2} + 231 \, a^{3} d^{2} e^{3} + 35 \, {\left (4 \, b^{3} d e^{4} + 11 \, a b^{2} e^{5}\right )} x^{4} + 5 \, {\left (b^{3} d^{2} e^{3} + 110 \, a b^{2} d e^{4} + 99 \, a^{2} b e^{5}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{3} e^{2} - 11 \, a b^{2} d^{2} e^{3} - 264 \, a^{2} b d e^{4} - 77 \, a^{3} e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{4} e - 44 \, a b^{2} d^{3} e^{2} + 99 \, a^{2} b d^{2} e^{3} + 462 \, a^{3} d e^{4}\right )} x\right )} \sqrt {e x + d}}{1155 \, e^{4}} \] Input:
integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
Output:
2/1155*(105*b^3*e^5*x^5 - 16*b^3*d^5 + 88*a*b^2*d^4*e - 198*a^2*b*d^3*e^2 + 231*a^3*d^2*e^3 + 35*(4*b^3*d*e^4 + 11*a*b^2*e^5)*x^4 + 5*(b^3*d^2*e^3 + 110*a*b^2*d*e^4 + 99*a^2*b*e^5)*x^3 - 3*(2*b^3*d^3*e^2 - 11*a*b^2*d^2*e^3 - 264*a^2*b*d*e^4 - 77*a^3*e^5)*x^2 + (8*b^3*d^4*e - 44*a*b^2*d^3*e^2 + 9 9*a^2*b*d^2*e^3 + 462*a^3*d*e^4)*x)*sqrt(e*x + d)/e^4
\[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \left (d + e x\right )^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((e*x+d)**(3/2)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Integral((d + e*x)**(3/2)*((a + b*x)**2)**(3/2), x)
Time = 0.05 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.04 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{3} e^{5} x^{5} - 16 \, b^{3} d^{5} + 88 \, a b^{2} d^{4} e - 198 \, a^{2} b d^{3} e^{2} + 231 \, a^{3} d^{2} e^{3} + 35 \, {\left (4 \, b^{3} d e^{4} + 11 \, a b^{2} e^{5}\right )} x^{4} + 5 \, {\left (b^{3} d^{2} e^{3} + 110 \, a b^{2} d e^{4} + 99 \, a^{2} b e^{5}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{3} e^{2} - 11 \, a b^{2} d^{2} e^{3} - 264 \, a^{2} b d e^{4} - 77 \, a^{3} e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{4} e - 44 \, a b^{2} d^{3} e^{2} + 99 \, a^{2} b d^{2} e^{3} + 462 \, a^{3} d e^{4}\right )} x\right )} \sqrt {e x + d}}{1155 \, e^{4}} \] Input:
integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
Output:
2/1155*(105*b^3*e^5*x^5 - 16*b^3*d^5 + 88*a*b^2*d^4*e - 198*a^2*b*d^3*e^2 + 231*a^3*d^2*e^3 + 35*(4*b^3*d*e^4 + 11*a*b^2*e^5)*x^4 + 5*(b^3*d^2*e^3 + 110*a*b^2*d*e^4 + 99*a^2*b*e^5)*x^3 - 3*(2*b^3*d^3*e^2 - 11*a*b^2*d^2*e^3 - 264*a^2*b*d*e^4 - 77*a^3*e^5)*x^2 + (8*b^3*d^4*e - 44*a*b^2*d^3*e^2 + 9 9*a^2*b*d^2*e^3 + 462*a^3*d*e^4)*x)*sqrt(e*x + d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (148) = 296\).
Time = 0.15 (sec) , antiderivative size = 638, normalized size of antiderivative = 3.07 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:
integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
Output:
2/3465*(3465*sqrt(e*x + d)*a^3*d^2*sgn(b*x + a) + 2310*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^3*d*sgn(b*x + a) + 3465*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2*b*d^2*sgn(b*x + a)/e + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^ (3/2)*d + 15*sqrt(e*x + d)*d^2)*a^3*sgn(b*x + a) + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b^2*d^2*sgn(b*x + a)/e^2 + 1386*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a ^2*b*d*sgn(b*x + a)/e + 99*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35* (e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*b^3*d^2*sgn(b*x + a)/e^3 + 594 *(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*s qrt(e*x + d)*d^3)*a*b^2*d*sgn(b*x + a)/e^2 + 297*(5*(e*x + d)^(7/2) - 21*( e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a^2*b*sg n(b*x + a)/e + 22*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^3*d*sgn (b*x + a)/e^3 + 33*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*a*b^2*sg n(b*x + a)/e^2 + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2)*d^4 - 69 3*sqrt(e*x + d)*d^5)*b^3*sgn(b*x + a)/e^3)/e
Timed out. \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int {\left (d+e\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \] Input:
int((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
Output:
int((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
Time = 0.26 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.09 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {e x +d}\, \left (105 b^{3} e^{5} x^{5}+385 a \,b^{2} e^{5} x^{4}+140 b^{3} d \,e^{4} x^{4}+495 a^{2} b \,e^{5} x^{3}+550 a \,b^{2} d \,e^{4} x^{3}+5 b^{3} d^{2} e^{3} x^{3}+231 a^{3} e^{5} x^{2}+792 a^{2} b d \,e^{4} x^{2}+33 a \,b^{2} d^{2} e^{3} x^{2}-6 b^{3} d^{3} e^{2} x^{2}+462 a^{3} d \,e^{4} x +99 a^{2} b \,d^{2} e^{3} x -44 a \,b^{2} d^{3} e^{2} x +8 b^{3} d^{4} e x +231 a^{3} d^{2} e^{3}-198 a^{2} b \,d^{3} e^{2}+88 a \,b^{2} d^{4} e -16 b^{3} d^{5}\right )}{1155 e^{4}} \] Input:
int((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
(2*sqrt(d + e*x)*(231*a**3*d**2*e**3 + 462*a**3*d*e**4*x + 231*a**3*e**5*x **2 - 198*a**2*b*d**3*e**2 + 99*a**2*b*d**2*e**3*x + 792*a**2*b*d*e**4*x** 2 + 495*a**2*b*e**5*x**3 + 88*a*b**2*d**4*e - 44*a*b**2*d**3*e**2*x + 33*a *b**2*d**2*e**3*x**2 + 550*a*b**2*d*e**4*x**3 + 385*a*b**2*e**5*x**4 - 16* b**3*d**5 + 8*b**3*d**4*e*x - 6*b**3*d**3*e**2*x**2 + 5*b**3*d**2*e**3*x** 3 + 140*b**3*d*e**4*x**4 + 105*b**3*e**5*x**5))/(1155*e**4)