Integrand size = 31, antiderivative size = 220 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 a^3 A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 a b (A b+a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 b^3 B x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)} \]
2/5*a^3*A*x^(5/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/7*a^2*(3*A*b+B*a)*x^(7/2)*(( b*x+a)^2)^(1/2)/(b*x+a)+2/3*a*b*(A*b+B*a)*x^(9/2)*((b*x+a)^2)^(1/2)/(b*x+a )+2/11*b^2*(A*b+3*B*a)*x^(11/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/13*b^3*B*x^(13 /2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.40 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 x^{5/2} \sqrt {(a+b x)^2} \left (429 a^3 (7 A+5 B x)+715 a^2 b x (9 A+7 B x)+455 a b^2 x^2 (11 A+9 B x)+105 b^3 x^3 (13 A+11 B x)\right )}{15015 (a+b x)} \]
(2*x^(5/2)*Sqrt[(a + b*x)^2]*(429*a^3*(7*A + 5*B*x) + 715*a^2*b*x*(9*A + 7 *B*x) + 455*a*b^2*x^2*(11*A + 9*B*x) + 105*b^3*x^3*(13*A + 11*B*x)))/(1501 5*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.51, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1187, 27, 85, 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.
\(\displaystyle \int x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x^{3/2} (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^{3/2} (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^{11/2}+b^2 (A b+3 a B) x^{9/2}+3 a b (A b+a B) x^{7/2}+a^2 (3 A b+a B) x^{5/2}+a^3 A x^{3/2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{5} a^3 A x^{5/2}+\frac {2}{7} a^2 x^{7/2} (a B+3 A b)+\frac {2}{11} b^2 x^{11/2} (3 a B+A b)+\frac {2}{3} a b x^{9/2} (a B+A b)+\frac {2}{13} b^3 B x^{13/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^3*A*x^(5/2))/5 + (2*a^2*(3*A*b + a*B) *x^(7/2))/7 + (2*a*b*(A*b + a*B)*x^(9/2))/3 + (2*b^2*(A*b + 3*a*B)*x^(11/2 ))/11 + (2*b^3*B*x^(13/2))/13))/(a + b*x)
3.8.93.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
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 = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(\frac {2 x^{\frac {5}{2}} \left (1155 x^{4} B \,b^{3}+1365 A \,b^{3} x^{3}+4095 B a \,b^{2} x^{3}+5005 A a \,b^{2} x^{2}+5005 B \,a^{2} b \,x^{2}+6435 A \,a^{2} b x +2145 a^{3} B x +3003 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15015 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {2 x^{\frac {5}{2}} \left (1155 x^{4} B \,b^{3}+1365 A \,b^{3} x^{3}+4095 B a \,b^{2} x^{3}+5005 A a \,b^{2} x^{2}+5005 B \,a^{2} b \,x^{2}+6435 A \,a^{2} b x +2145 a^{3} B x +3003 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15015 \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {5}{2}} \left (1155 x^{4} B \,b^{3}+1365 A \,b^{3} x^{3}+4095 B a \,b^{2} x^{3}+5005 A a \,b^{2} x^{2}+5005 B \,a^{2} b \,x^{2}+6435 A \,a^{2} b x +2145 a^{3} B x +3003 A \,a^{3}\right )}{15015 \left (b x +a \right )}\) | \(92\) |
2/15015*x^(5/2)*(1155*B*b^3*x^4+1365*A*b^3*x^3+4095*B*a*b^2*x^3+5005*A*a*b ^2*x^2+5005*B*a^2*b*x^2+6435*A*a^2*b*x+2145*B*a^3*x+3003*A*a^3)*((b*x+a)^2 )^(3/2)/(b*x+a)^3
Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{15015} \, {\left (1155 \, B b^{3} x^{6} + 3003 \, A a^{3} x^{2} + 1365 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{5} + 5005 \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} + 2145 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )} \sqrt {x} \]
2/15015*(1155*B*b^3*x^6 + 3003*A*a^3*x^2 + 1365*(3*B*a*b^2 + A*b^3)*x^5 + 5005*(B*a^2*b + A*a*b^2)*x^4 + 2145*(B*a^3 + 3*A*a^2*b)*x^3)*sqrt(x)
\[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^{\frac {3}{2}} \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.62 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{3465} \, {\left (35 \, {\left (9 \, b^{3} x^{2} + 11 \, a b^{2} x\right )} x^{\frac {7}{2}} + 110 \, {\left (7 \, a b^{2} x^{2} + 9 \, a^{2} b x\right )} x^{\frac {5}{2}} + 99 \, {\left (5 \, a^{2} b x^{2} + 7 \, a^{3} x\right )} x^{\frac {3}{2}}\right )} A + \frac {2}{9009} \, {\left (63 \, {\left (11 \, b^{3} x^{2} + 13 \, a b^{2} x\right )} x^{\frac {9}{2}} + 182 \, {\left (9 \, a b^{2} x^{2} + 11 \, a^{2} b x\right )} x^{\frac {7}{2}} + 143 \, {\left (7 \, a^{2} b x^{2} + 9 \, a^{3} x\right )} x^{\frac {5}{2}}\right )} B \]
2/3465*(35*(9*b^3*x^2 + 11*a*b^2*x)*x^(7/2) + 110*(7*a*b^2*x^2 + 9*a^2*b*x )*x^(5/2) + 99*(5*a^2*b*x^2 + 7*a^3*x)*x^(3/2))*A + 2/9009*(63*(11*b^3*x^2 + 13*a*b^2*x)*x^(9/2) + 182*(9*a*b^2*x^2 + 11*a^2*b*x)*x^(7/2) + 143*(7*a ^2*b*x^2 + 9*a^3*x)*x^(5/2))*B
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{13} \, B b^{3} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, B a b^{2} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a^{2} b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A a b^{2} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, B a^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, A a^{2} b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/13*B*b^3*x^(13/2)*sgn(b*x + a) + 6/11*B*a*b^2*x^(11/2)*sgn(b*x + a) + 2/ 11*A*b^3*x^(11/2)*sgn(b*x + a) + 2/3*B*a^2*b*x^(9/2)*sgn(b*x + a) + 2/3*A* a*b^2*x^(9/2)*sgn(b*x + a) + 2/7*B*a^3*x^(7/2)*sgn(b*x + a) + 6/7*A*a^2*b* x^(7/2)*sgn(b*x + a) + 2/5*A*a^3*x^(5/2)*sgn(b*x + a)
Timed out. \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^{3/2}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]