Integrand size = 31, antiderivative size = 220 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 a^3 A x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {6 a b (A b+a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{15 (a+b x)}+\frac {2 b^3 B x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)} \]
2/9*a^3*A*x^(9/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/11*a^2*(3*A*b+B*a)*x^(11/2)* ((b*x+a)^2)^(1/2)/(b*x+a)+6/13*a*b*(A*b+B*a)*x^(13/2)*((b*x+a)^2)^(1/2)/(b *x+a)+2/15*b^2*(A*b+3*B*a)*x^(15/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/17*b^3*B*x ^(17/2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.40 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 x^{9/2} \sqrt {(a+b x)^2} \left (1105 a^3 (11 A+9 B x)+2295 a^2 b x (13 A+11 B x)+1683 a b^2 x^2 (15 A+13 B x)+429 b^3 x^3 (17 A+15 B x)\right )}{109395 (a+b x)} \]
(2*x^(9/2)*Sqrt[(a + b*x)^2]*(1105*a^3*(11*A + 9*B*x) + 2295*a^2*b*x*(13*A + 11*B*x) + 1683*a*b^2*x^2*(15*A + 13*B*x) + 429*b^3*x^3*(17*A + 15*B*x)) )/(109395*(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.
\(\displaystyle \int x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x^{7/2} (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^{7/2} (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^{15/2}+b^2 (A b+3 a B) x^{13/2}+3 a b (A b+a B) x^{11/2}+a^2 (3 A b+a B) x^{9/2}+a^3 A x^{7/2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{9} a^3 A x^{9/2}+\frac {2}{11} a^2 x^{11/2} (a B+3 A b)+\frac {2}{15} b^2 x^{15/2} (3 a B+A b)+\frac {6}{13} a b x^{13/2} (a B+A b)+\frac {2}{17} b^3 B x^{17/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^3*A*x^(9/2))/9 + (2*a^2*(3*A*b + a*B) *x^(11/2))/11 + (6*a*b*(A*b + a*B)*x^(13/2))/13 + (2*b^2*(A*b + 3*a*B)*x^( 15/2))/15 + (2*b^3*B*x^(17/2))/17))/(a + b*x)
3.8.91.3.1 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 = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(\frac {2 x^{\frac {9}{2}} \left (6435 x^{4} B \,b^{3}+7293 A \,b^{3} x^{3}+21879 B a \,b^{2} x^{3}+25245 A a \,b^{2} x^{2}+25245 B \,a^{2} b \,x^{2}+29835 A \,a^{2} b x +9945 a^{3} B x +12155 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{109395 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {2 x^{\frac {9}{2}} \left (6435 x^{4} B \,b^{3}+7293 A \,b^{3} x^{3}+21879 B a \,b^{2} x^{3}+25245 A a \,b^{2} x^{2}+25245 B \,a^{2} b \,x^{2}+29835 A \,a^{2} b x +9945 a^{3} B x +12155 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{109395 \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {9}{2}} \left (6435 x^{4} B \,b^{3}+7293 A \,b^{3} x^{3}+21879 B a \,b^{2} x^{3}+25245 A a \,b^{2} x^{2}+25245 B \,a^{2} b \,x^{2}+29835 A \,a^{2} b x +9945 a^{3} B x +12155 A \,a^{3}\right )}{109395 \left (b x +a \right )}\) | \(92\) |
2/109395*x^(9/2)*(6435*B*b^3*x^4+7293*A*b^3*x^3+21879*B*a*b^2*x^3+25245*A* a*b^2*x^2+25245*B*a^2*b*x^2+29835*A*a^2*b*x+9945*B*a^3*x+12155*A*a^3)*((b* x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{109395} \, {\left (6435 \, B b^{3} x^{8} + 12155 \, A a^{3} x^{4} + 7293 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{7} + 25245 \, {\left (B a^{2} b + A a b^{2}\right )} x^{6} + 9945 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5}\right )} \sqrt {x} \]
2/109395*(6435*B*b^3*x^8 + 12155*A*a^3*x^4 + 7293*(3*B*a*b^2 + A*b^3)*x^7 + 25245*(B*a^2*b + A*a*b^2)*x^6 + 9945*(B*a^3 + 3*A*a^2*b)*x^5)*sqrt(x)
Timed out. \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.62 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{6435} \, {\left (33 \, {\left (13 \, b^{3} x^{2} + 15 \, a b^{2} x\right )} x^{\frac {11}{2}} + 90 \, {\left (11 \, a b^{2} x^{2} + 13 \, a^{2} b x\right )} x^{\frac {9}{2}} + 65 \, {\left (9 \, a^{2} b x^{2} + 11 \, a^{3} x\right )} x^{\frac {7}{2}}\right )} A + \frac {2}{36465} \, {\left (143 \, {\left (15 \, b^{3} x^{2} + 17 \, a b^{2} x\right )} x^{\frac {13}{2}} + 374 \, {\left (13 \, a b^{2} x^{2} + 15 \, a^{2} b x\right )} x^{\frac {11}{2}} + 255 \, {\left (11 \, a^{2} b x^{2} + 13 \, a^{3} x\right )} x^{\frac {9}{2}}\right )} B \]
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))*A + 2/36465*(143*(15*b ^3*x^2 + 17*a*b^2*x)*x^(13/2) + 374*(13*a*b^2*x^2 + 15*a^2*b*x)*x^(11/2) + 255*(11*a^2*b*x^2 + 13*a^3*x)*x^(9/2))*B
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2}{17} \, B b^{3} x^{\frac {17}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, B a b^{2} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{15} \, A b^{3} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{13} \, B a^{2} b x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{13} \, A a b^{2} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, B a^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, A a^{2} b x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, A a^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/17*B*b^3*x^(17/2)*sgn(b*x + a) + 2/5*B*a*b^2*x^(15/2)*sgn(b*x + a) + 2/1 5*A*b^3*x^(15/2)*sgn(b*x + a) + 6/13*B*a^2*b*x^(13/2)*sgn(b*x + a) + 6/13* A*a*b^2*x^(13/2)*sgn(b*x + a) + 2/11*B*a^3*x^(11/2)*sgn(b*x + a) + 6/11*A* a^2*b*x^(11/2)*sgn(b*x + a) + 2/9*A*a^3*x^(9/2)*sgn(b*x + a)
Timed out. \[ \int x^{7/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^{7/2}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]