Integrand size = 31, antiderivative size = 214 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{3/2} (a+b x)}-\frac {2 b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b^3 B \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]
-2/7*a^3*A*((b*x+a)^2)^(1/2)/x^(7/2)/(b*x+a)-2/5*a^2*(3*A*b+B*a)*((b*x+a)^ 2)^(1/2)/x^(5/2)/(b*x+a)-2*a*b*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^(3/2)/(b*x+a) -2*b^2*(A*b+3*B*a)*((b*x+a)^2)^(1/2)/(b*x+a)/x^(1/2)+2*b^3*B*x^(1/2)*((b*x +a)^2)^(1/2)/(b*x+a)
Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (35 b^3 x^3 (A-B x)+35 a b^2 x^2 (A+3 B x)+7 a^2 b x (3 A+5 B x)+a^3 (5 A+7 B x)\right )}{35 x^{7/2} (a+b x)} \]
(-2*Sqrt[(a + b*x)^2]*(35*b^3*x^3*(A - B*x) + 35*a*b^2*x^2*(A + 3*B*x) + 7 *a^2*b*x*(3*A + 5*B*x) + a^3*(5*A + 7*B*x)))/(35*x^(7/2)*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 107, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^{9/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^{9/2}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^{9/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^{9/2}}+\frac {(3 A b+a B) a^2}{x^{7/2}}+\frac {3 b (A b+a B) a}{x^{5/2}}+\frac {b^3 B}{\sqrt {x}}+\frac {b^2 (A b+3 a B)}{x^{3/2}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 a^3 A}{7 x^{7/2}}-\frac {2 a^2 (a B+3 A b)}{5 x^{5/2}}-\frac {2 b^2 (3 a B+A b)}{\sqrt {x}}-\frac {2 a b (a B+A b)}{x^{3/2}}+2 b^3 B \sqrt {x}\right )}{a+b x}\) |
(((-2*a^3*A)/(7*x^(7/2)) - (2*a^2*(3*A*b + a*B))/(5*x^(5/2)) - (2*a*b*(A*b + a*B))/x^(3/2) - (2*b^2*(A*b + 3*a*B))/Sqrt[x] + 2*b^3*B*Sqrt[x])*Sqrt[a ^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
3.8.99.3.1 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 = 0.43
method | result | size |
gosper | \(-\frac {2 \left (-35 x^{4} B \,b^{3}+35 A \,b^{3} x^{3}+105 B a \,b^{2} x^{3}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 x^{\frac {7}{2}} \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {2 \left (-35 x^{4} B \,b^{3}+35 A \,b^{3} x^{3}+105 B a \,b^{2} x^{3}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 x^{\frac {7}{2}} \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-35 x^{4} B \,b^{3}+35 A \,b^{3} x^{3}+105 B a \,b^{2} x^{3}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right )}{35 \left (b x +a \right ) x^{\frac {7}{2}}}\) | \(92\) |
-2/35*(-35*B*b^3*x^4+35*A*b^3*x^3+105*B*a*b^2*x^3+35*A*a*b^2*x^2+35*B*a^2* b*x^2+21*A*a^2*b*x+7*B*a^3*x+5*A*a^3)*((b*x+a)^2)^(3/2)/x^(7/2)/(b*x+a)^3
Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.34 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=\frac {2 \, {\left (35 \, B b^{3} x^{4} - 5 \, A a^{3} - 35 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 7 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, x^{\frac {7}{2}}} \]
2/35*(35*B*b^3*x^4 - 5*A*a^3 - 35*(3*B*a*b^2 + A*b^3)*x^3 - 35*(B*a^2*b + A*a*b^2)*x^2 - 7*(B*a^3 + 3*A*a^2*b)*x)/x^(7/2)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{\frac {9}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=\frac {2}{15} \, B {\left (\frac {15 \, {\left (b^{3} x^{2} - a b^{2} x\right )}}{x^{\frac {3}{2}}} - \frac {10 \, {\left (3 \, a b^{2} x^{2} + a^{2} b x\right )}}{x^{\frac {5}{2}}} - \frac {5 \, a^{2} b x^{2} + 3 \, a^{3} x}{x^{\frac {7}{2}}}\right )} - \frac {2}{105} \, A {\left (\frac {35 \, {\left (3 \, b^{3} x^{2} + a b^{2} x\right )}}{x^{\frac {5}{2}}} + \frac {14 \, {\left (5 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )}}{x^{\frac {7}{2}}} + \frac {3 \, {\left (7 \, a^{2} b x^{2} + 5 \, a^{3} x\right )}}{x^{\frac {9}{2}}}\right )} \]
2/15*B*(15*(b^3*x^2 - a*b^2*x)/x^(3/2) - 10*(3*a*b^2*x^2 + a^2*b*x)/x^(5/2 ) - (5*a^2*b*x^2 + 3*a^3*x)/x^(7/2)) - 2/105*A*(35*(3*b^3*x^2 + a*b^2*x)/x ^(5/2) + 14*(5*a*b^2*x^2 + 3*a^2*b*x)/x^(7/2) + 3*(7*a^2*b*x^2 + 5*a^3*x)/ x^(9/2))
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.58 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=2 \, B b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (105 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 21 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{35 \, x^{\frac {7}{2}}} \]
2*B*b^3*sqrt(x)*sgn(b*x + a) - 2/35*(105*B*a*b^2*x^3*sgn(b*x + a) + 35*A*b ^3*x^3*sgn(b*x + a) + 35*B*a^2*b*x^2*sgn(b*x + a) + 35*A*a*b^2*x^2*sgn(b*x + a) + 7*B*a^3*x*sgn(b*x + a) + 21*A*a^2*b*x*sgn(b*x + a) + 5*A*a^3*sgn(b *x + a))/x^(7/2)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^{9/2}} \,d x \]