Integrand size = 31, antiderivative size = 320 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 a^5 A x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 a^4 (5 A b+a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {10 a^3 b (2 A b+a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {4 a^2 b^2 (A b+a B) x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {10 a b^3 (A b+2 a B) x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)}+\frac {2 b^4 (A b+5 a B) x^{19/2} \sqrt {a^2+2 a b x+b^2 x^2}}{19 (a+b x)}+\frac {2 b^5 B x^{21/2} \sqrt {a^2+2 a b x+b^2 x^2}}{21 (a+b x)} \]
2/9*a^5*A*x^(9/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/11*a^4*(5*A*b+B*a)*x^(11/2)* ((b*x+a)^2)^(1/2)/(b*x+a)+10/13*a^3*b*(2*A*b+B*a)*x^(13/2)*((b*x+a)^2)^(1/ 2)/(b*x+a)+4/3*a^2*b^2*(A*b+B*a)*x^(15/2)*((b*x+a)^2)^(1/2)/(b*x+a)+10/17* a*b^3*(A*b+2*B*a)*x^(17/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/19*b^4*(A*b+5*B*a)* x^(19/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/21*b^5*B*x^(21/2)*((b*x+a)^2)^(1/2)/( b*x+a)
Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.40 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 x^{9/2} \sqrt {(a+b x)^2} \left (29393 a^5 (11 A+9 B x)+101745 a^4 b x (13 A+11 B x)+149226 a^3 b^2 x^2 (15 A+13 B x)+114114 a^2 b^3 x^3 (17 A+15 B x)+45045 a b^4 x^4 (19 A+17 B x)+7293 b^5 x^5 (21 A+19 B x)\right )}{2909907 (a+b x)} \]
(2*x^(9/2)*Sqrt[(a + b*x)^2]*(29393*a^5*(11*A + 9*B*x) + 101745*a^4*b*x*(1 3*A + 11*B*x) + 149226*a^3*b^2*x^2*(15*A + 13*B*x) + 114114*a^2*b^3*x^3*(1 7*A + 15*B*x) + 45045*a*b^4*x^4*(19*A + 17*B*x) + 7293*b^5*x^5*(21*A + 19* B*x)))/(2909907*(a + b*x))
Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.50, 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.
| \(\displaystyle \int x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 x^{7/2} (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^{7/2} (a+b x)^5 (A+B x)dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 B x^{19/2}+b^4 (A b+5 a B) x^{17/2}+5 a b^3 (A b+2 a B) x^{15/2}+10 a^2 b^2 (A b+a B) x^{13/2}+5 a^3 b (2 A b+a B) x^{11/2}+a^4 (5 A b+a B) x^{9/2}+a^5 A x^{7/2}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{9} a^5 A x^{9/2}+\frac {2}{11} a^4 x^{11/2} (a B+5 A b)+\frac {10}{13} a^3 b x^{13/2} (a B+2 A b)+\frac {4}{3} a^2 b^2 x^{15/2} (a B+A b)+\frac {2}{19} b^4 x^{19/2} (5 a B+A b)+\frac {10}{17} a b^3 x^{17/2} (2 a B+A b)+\frac {2}{21} b^5 B x^{21/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^5*A*x^(9/2))/9 + (2*a^4*(5*A*b + a*B) *x^(11/2))/11 + (10*a^3*b*(2*A*b + a*B)*x^(13/2))/13 + (4*a^2*b^2*(A*b + a *B)*x^(15/2))/3 + (10*a*b^3*(A*b + 2*a*B)*x^(17/2))/17 + (2*b^4*(A*b + 5*a *B)*x^(19/2))/19 + (2*b^5*B*x^(21/2))/21))/(a + b*x)
3.8.100.3.1 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 = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44
| method | result | size |
| gosper | \(\frac {2 x^{\frac {9}{2}} \left (138567 B \,b^{5} x^{6}+153153 A \,b^{5} x^{5}+765765 B a \,b^{4} x^{5}+855855 A a \,b^{4} x^{4}+1711710 B \,a^{2} b^{3} x^{4}+1939938 A \,a^{2} b^{3} x^{3}+1939938 B \,a^{3} b^{2} x^{3}+2238390 A \,a^{3} b^{2} x^{2}+1119195 B \,a^{4} b \,x^{2}+1322685 A \,a^{4} b x +264537 a^{5} B x +323323 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2909907 \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(\frac {2 x^{\frac {9}{2}} \left (138567 B \,b^{5} x^{6}+153153 A \,b^{5} x^{5}+765765 B a \,b^{4} x^{5}+855855 A a \,b^{4} x^{4}+1711710 B \,a^{2} b^{3} x^{4}+1939938 A \,a^{2} b^{3} x^{3}+1939938 B \,a^{3} b^{2} x^{3}+2238390 A \,a^{3} b^{2} x^{2}+1119195 B \,a^{4} b \,x^{2}+1322685 A \,a^{4} b x +264537 a^{5} B x +323323 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2909907 \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {9}{2}} \left (138567 B \,b^{5} x^{6}+153153 A \,b^{5} x^{5}+765765 B a \,b^{4} x^{5}+855855 A a \,b^{4} x^{4}+1711710 B \,a^{2} b^{3} x^{4}+1939938 A \,a^{2} b^{3} x^{3}+1939938 B \,a^{3} b^{2} x^{3}+2238390 A \,a^{3} b^{2} x^{2}+1119195 B \,a^{4} b \,x^{2}+1322685 A \,a^{4} b x +264537 a^{5} B x +323323 A \,a^{5}\right )}{2909907 \left (b x +a \right )}\) | \(140\) |
2/2909907*x^(9/2)*(138567*B*b^5*x^6+153153*A*b^5*x^5+765765*B*a*b^4*x^5+85 5855*A*a*b^4*x^4+1711710*B*a^2*b^3*x^4+1939938*A*a^2*b^3*x^3+1939938*B*a^3 *b^2*x^3+2238390*A*a^3*b^2*x^2+1119195*B*a^4*b*x^2+1322685*A*a^4*b*x+26453 7*B*a^5*x+323323*A*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5
Time = 0.35 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.39 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{2909907} \, {\left (138567 \, B b^{5} x^{10} + 323323 \, A a^{5} x^{4} + 153153 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{9} + 855855 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + 1939938 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{7} + 1119195 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + 264537 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{5}\right )} \sqrt {x} \]
2/2909907*(138567*B*b^5*x^10 + 323323*A*a^5*x^4 + 153153*(5*B*a*b^4 + A*b^ 5)*x^9 + 855855*(2*B*a^2*b^3 + A*a*b^4)*x^8 + 1939938*(B*a^3*b^2 + A*a^2*b ^3)*x^7 + 1119195*(B*a^4*b + 2*A*a^3*b^2)*x^6 + 264537*(B*a^5 + 5*A*a^4*b) *x^5)*sqrt(x)
Timed out. \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.75 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{2078505} \, {\left (6435 \, {\left (17 \, b^{5} x^{2} + 19 \, a b^{4} x\right )} x^{\frac {15}{2}} + 32604 \, {\left (15 \, a b^{4} x^{2} + 17 \, a^{2} b^{3} x\right )} x^{\frac {13}{2}} + 63954 \, {\left (13 \, a^{2} b^{3} x^{2} + 15 \, a^{3} b^{2} x\right )} x^{\frac {11}{2}} + 58140 \, {\left (11 \, a^{3} b^{2} x^{2} + 13 \, a^{4} b x\right )} x^{\frac {9}{2}} + 20995 \, {\left (9 \, a^{4} b x^{2} + 11 \, a^{5} x\right )} x^{\frac {7}{2}}\right )} A + \frac {2}{4849845} \, {\left (12155 \, {\left (19 \, b^{5} x^{2} + 21 \, a b^{4} x\right )} x^{\frac {17}{2}} + 60060 \, {\left (17 \, a b^{4} x^{2} + 19 \, a^{2} b^{3} x\right )} x^{\frac {15}{2}} + 114114 \, {\left (15 \, a^{2} b^{3} x^{2} + 17 \, a^{3} b^{2} x\right )} x^{\frac {13}{2}} + 99484 \, {\left (13 \, a^{3} b^{2} x^{2} + 15 \, a^{4} b x\right )} x^{\frac {11}{2}} + 33915 \, {\left (11 \, a^{4} b x^{2} + 13 \, a^{5} x\right )} x^{\frac {9}{2}}\right )} B \]
2/2078505*(6435*(17*b^5*x^2 + 19*a*b^4*x)*x^(15/2) + 32604*(15*a*b^4*x^2 + 17*a^2*b^3*x)*x^(13/2) + 63954*(13*a^2*b^3*x^2 + 15*a^3*b^2*x)*x^(11/2) + 58140*(11*a^3*b^2*x^2 + 13*a^4*b*x)*x^(9/2) + 20995*(9*a^4*b*x^2 + 11*a^5 *x)*x^(7/2))*A + 2/4849845*(12155*(19*b^5*x^2 + 21*a*b^4*x)*x^(17/2) + 600 60*(17*a*b^4*x^2 + 19*a^2*b^3*x)*x^(15/2) + 114114*(15*a^2*b^3*x^2 + 17*a^ 3*b^2*x)*x^(13/2) + 99484*(13*a^3*b^2*x^2 + 15*a^4*b*x)*x^(11/2) + 33915*( 11*a^4*b*x^2 + 13*a^5*x)*x^(9/2))*B
Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.62 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{21} \, B b^{5} x^{\frac {21}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{19} \, B a b^{4} x^{\frac {19}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{19} \, A b^{5} x^{\frac {19}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{17} \, B a^{2} b^{3} x^{\frac {17}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{17} \, A a b^{4} x^{\frac {17}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{3} \, B a^{3} b^{2} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{3} \, A a^{2} b^{3} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{13} \, B a^{4} b x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{13} \, A a^{3} b^{2} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, B a^{5} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{11} \, A a^{4} b x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, A a^{5} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/21*B*b^5*x^(21/2)*sgn(b*x + a) + 10/19*B*a*b^4*x^(19/2)*sgn(b*x + a) + 2 /19*A*b^5*x^(19/2)*sgn(b*x + a) + 20/17*B*a^2*b^3*x^(17/2)*sgn(b*x + a) + 10/17*A*a*b^4*x^(17/2)*sgn(b*x + a) + 4/3*B*a^3*b^2*x^(15/2)*sgn(b*x + a) + 4/3*A*a^2*b^3*x^(15/2)*sgn(b*x + a) + 10/13*B*a^4*b*x^(13/2)*sgn(b*x + a ) + 20/13*A*a^3*b^2*x^(13/2)*sgn(b*x + a) + 2/11*B*a^5*x^(11/2)*sgn(b*x + a) + 10/11*A*a^4*b*x^(11/2)*sgn(b*x + a) + 2/9*A*a^5*x^(9/2)*sgn(b*x + a)
Timed out. \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x^{7/2}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]