Integrand size = 31, antiderivative size = 220 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 a^3 A x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {6 a b (A b+a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 b^3 B x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{15 (a+b x)} \]
2/7*a^3*A*x^(7/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/9*a^2*(3*A*b+B*a)*x^(9/2)*(( b*x+a)^2)^(1/2)/(b*x+a)+6/11*a*b*(A*b+B*a)*x^(11/2)*((b*x+a)^2)^(1/2)/(b*x +a)+2/13*b^2*(A*b+3*B*a)*x^(13/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/15*b^3*B*x^( 15/2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.40 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 x^{7/2} \sqrt {(a+b x)^2} \left (715 a^3 (9 A+7 B x)+1365 a^2 b x (11 A+9 B x)+945 a b^2 x^2 (13 A+11 B x)+231 b^3 x^3 (15 A+13 B x)\right )}{45045 (a+b x)} \]
(2*x^(7/2)*Sqrt[(a + b*x)^2]*(715*a^3*(9*A + 7*B*x) + 1365*a^2*b*x*(11*A + 9*B*x) + 945*a*b^2*x^2*(13*A + 11*B*x) + 231*b^3*x^3*(15*A + 13*B*x)))/(4 5045*(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.
\(\displaystyle \int x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x^{5/2} (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^{5/2} (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^{13/2}+b^2 (A b+3 a B) x^{11/2}+3 a b (A b+a B) x^{9/2}+a^2 (3 A b+a B) x^{7/2}+a^3 A x^{5/2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{7} a^3 A x^{7/2}+\frac {2}{9} a^2 x^{9/2} (a B+3 A b)+\frac {2}{13} b^2 x^{13/2} (3 a B+A b)+\frac {6}{11} a b x^{11/2} (a B+A b)+\frac {2}{15} b^3 B x^{15/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^3*A*x^(7/2))/7 + (2*a^2*(3*A*b + a*B) *x^(9/2))/9 + (6*a*b*(A*b + a*B)*x^(11/2))/11 + (2*b^2*(A*b + 3*a*B)*x^(13 /2))/13 + (2*b^3*B*x^(15/2))/15))/(a + b*x)
3.8.92.3.1 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 = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(\frac {2 x^{\frac {7}{2}} \left (3003 x^{4} B \,b^{3}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 A a \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 A \,a^{2} b x +5005 a^{3} B x +6435 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{45045 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {2 x^{\frac {7}{2}} \left (3003 x^{4} B \,b^{3}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 A a \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 A \,a^{2} b x +5005 a^{3} B x +6435 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{45045 \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {7}{2}} \left (3003 x^{4} B \,b^{3}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 A a \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 A \,a^{2} b x +5005 a^{3} B x +6435 A \,a^{3}\right )}{45045 \left (b x +a \right )}\) | \(92\) |
2/45045*x^(7/2)*(3003*B*b^3*x^4+3465*A*b^3*x^3+10395*B*a*b^2*x^3+12285*A*a *b^2*x^2+12285*B*a^2*b*x^2+15015*A*a^2*b*x+5005*B*a^3*x+6435*A*a^3)*((b*x+ a)^2)^(3/2)/(b*x+a)^3
Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{45045} \, {\left (3003 \, B b^{3} x^{7} + 6435 \, A a^{3} x^{3} + 3465 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 12285 \, {\left (B a^{2} b + A a b^{2}\right )} x^{5} + 5005 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4}\right )} \sqrt {x} \]
2/45045*(3003*B*b^3*x^7 + 6435*A*a^3*x^3 + 3465*(3*B*a*b^2 + A*b^3)*x^6 + 12285*(B*a^2*b + A*a*b^2)*x^5 + 5005*(B*a^3 + 3*A*a^2*b)*x^4)*sqrt(x)
\[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^{\frac {5}{2}} \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.62 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{9009} \, {\left (63 \, {\left (11 \, b^{3} x^{2} + 13 \, a b^{2} x\right )} x^{\frac {9}{2}} + 182 \, {\left (9 \, a b^{2} x^{2} + 11 \, a^{2} b x\right )} x^{\frac {7}{2}} + 143 \, {\left (7 \, a^{2} b x^{2} + 9 \, a^{3} x\right )} x^{\frac {5}{2}}\right )} A + \frac {2}{6435} \, {\left (33 \, {\left (13 \, b^{3} x^{2} + 15 \, a b^{2} x\right )} x^{\frac {11}{2}} + 90 \, {\left (11 \, a b^{2} x^{2} + 13 \, a^{2} b x\right )} x^{\frac {9}{2}} + 65 \, {\left (9 \, a^{2} b x^{2} + 11 \, a^{3} x\right )} x^{\frac {7}{2}}\right )} B \]
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))*A + 2/6435*(33*(13*b^3* x^2 + 15*a*b^2*x)*x^(11/2) + 90*(11*a*b^2*x^2 + 13*a^2*b*x)*x^(9/2) + 65*( 9*a^2*b*x^2 + 11*a^3*x)*x^(7/2))*B
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{15} \, B b^{3} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{13} \, B a b^{2} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{13} \, A b^{3} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, B a^{2} b x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, A a b^{2} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, B a^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A a^{2} b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, A a^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/15*B*b^3*x^(15/2)*sgn(b*x + a) + 6/13*B*a*b^2*x^(13/2)*sgn(b*x + a) + 2/ 13*A*b^3*x^(13/2)*sgn(b*x + a) + 6/11*B*a^2*b*x^(11/2)*sgn(b*x + a) + 6/11 *A*a*b^2*x^(11/2)*sgn(b*x + a) + 2/9*B*a^3*x^(9/2)*sgn(b*x + a) + 2/3*A*a^ 2*b*x^(9/2)*sgn(b*x + a) + 2/7*A*a^3*x^(7/2)*sgn(b*x + a)
Timed out. \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^{5/2}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]