Integrand size = 31, antiderivative size = 214 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 a^2 (3 A b+a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 a b (A b+a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^2 (A b+3 a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 b^3 B x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \]
2*a*b*(A*b+B*a)*x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/5*b^2*(A*b+3*B*a)*x^(5 /2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/7*b^3*B*x^(7/2)*((b*x+a)^2)^(1/2)/(b*x+a)- 2*a^3*A*((b*x+a)^2)^(1/2)/(b*x+a)/x^(1/2)+2*a^2*(3*A*b+B*a)*x^(1/2)*((b*x+ a)^2)^(1/2)/(b*x+a)
Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \sqrt {(a+b x)^2} \left (-35 a^3 (A-B x)+35 a^2 b x (3 A+B x)+7 a b^2 x^2 (5 A+3 B x)+b^3 x^3 (7 A+5 B x)\right )}{35 \sqrt {x} (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(-35*a^3*(A - B*x) + 35*a^2*b*x*(3*A + B*x) + 7*a*b^2 *x^2*(5*A + 3*B*x) + b^3*x^3*(7*A + 5*B*x)))/(35*Sqrt[x]*(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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^{3/2}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^{3/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^{3/2}}+\frac {(3 A b+a B) a^2}{\sqrt {x}}+3 b (A b+a B) \sqrt {x} a+b^3 B x^{5/2}+b^2 (A b+3 a B) x^{3/2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 a^3 A}{\sqrt {x}}+2 a^2 \sqrt {x} (a B+3 A b)+\frac {2}{5} b^2 x^{5/2} (3 a B+A b)+2 a b x^{3/2} (a B+A b)+\frac {2}{7} b^3 B x^{7/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*a^3*A)/Sqrt[x] + 2*a^2*(3*A*b + a*B)*S qrt[x] + 2*a*b*(A*b + a*B)*x^(3/2) + (2*b^2*(A*b + 3*a*B)*x^(5/2))/5 + (2* b^3*B*x^(7/2))/7))/(a + b*x)
3.8.96.3.1 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 = 0.43
method | result | size |
gosper | \(-\frac {2 \left (-5 x^{4} B \,b^{3}-7 A \,b^{3} x^{3}-21 B a \,b^{2} x^{3}-35 A a \,b^{2} x^{2}-35 B \,a^{2} b \,x^{2}-105 A \,a^{2} b x -35 a^{3} B x +35 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \sqrt {x}\, \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {2 \left (-5 x^{4} B \,b^{3}-7 A \,b^{3} x^{3}-21 B a \,b^{2} x^{3}-35 A a \,b^{2} x^{2}-35 B \,a^{2} b \,x^{2}-105 A \,a^{2} b x -35 a^{3} B x +35 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \sqrt {x}\, \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-5 x^{4} B \,b^{3}-7 A \,b^{3} x^{3}-21 B a \,b^{2} x^{3}-35 A a \,b^{2} x^{2}-35 B \,a^{2} b \,x^{2}-105 A \,a^{2} b x -35 a^{3} B x +35 A \,a^{3}\right )}{35 \left (b x +a \right ) \sqrt {x}}\) | \(92\) |
-2/35*(-5*B*b^3*x^4-7*A*b^3*x^3-21*B*a*b^2*x^3-35*A*a*b^2*x^2-35*B*a^2*b*x ^2-105*A*a^2*b*x-35*B*a^3*x+35*A*a^3)*((b*x+a)^2)^(3/2)/x^(1/2)/(b*x+a)^3
Time = 0.41 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.34 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \, {\left (5 \, B b^{3} x^{4} - 35 \, A a^{3} + 7 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 35 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 35 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, \sqrt {x}} \]
2/35*(5*B*b^3*x^4 - 35*A*a^3 + 7*(3*B*a*b^2 + A*b^3)*x^3 + 35*(B*a^2*b + A *a*b^2)*x^2 + 35*(B*a^3 + 3*A*a^2*b)*x)/sqrt(x)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2}{15} \, {\left ({\left (3 \, b^{3} x^{2} + 5 \, a b^{2} x\right )} \sqrt {x} + \frac {10 \, {\left (a b^{2} x^{2} + 3 \, a^{2} b x\right )}}{\sqrt {x}} + \frac {15 \, {\left (a^{2} b x^{2} - a^{3} x\right )}}{x^{\frac {3}{2}}}\right )} A + \frac {2}{105} \, {\left (3 \, {\left (5 \, b^{3} x^{2} + 7 \, a b^{2} x\right )} x^{\frac {3}{2}} + 14 \, {\left (3 \, a b^{2} x^{2} + 5 \, a^{2} b x\right )} \sqrt {x} + \frac {35 \, {\left (a^{2} b x^{2} + 3 \, a^{3} x\right )}}{\sqrt {x}}\right )} B \]
2/15*((3*b^3*x^2 + 5*a*b^2*x)*sqrt(x) + 10*(a*b^2*x^2 + 3*a^2*b*x)/sqrt(x) + 15*(a^2*b*x^2 - a^3*x)/x^(3/2))*A + 2/105*(3*(5*b^3*x^2 + 7*a*b^2*x)*x^ (3/2) + 14*(3*a*b^2*x^2 + 5*a^2*b*x)*sqrt(x) + 35*(a^2*b*x^2 + 3*a^3*x)/sq rt(x))*B
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.58 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2}{7} \, B b^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{5} \, B a b^{2} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A b^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{2} b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a^{2} b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{\sqrt {x}} \]
2/7*B*b^3*x^(7/2)*sgn(b*x + a) + 6/5*B*a*b^2*x^(5/2)*sgn(b*x + a) + 2/5*A* b^3*x^(5/2)*sgn(b*x + a) + 2*B*a^2*b*x^(3/2)*sgn(b*x + a) + 2*A*a*b^2*x^(3 /2)*sgn(b*x + a) + 2*B*a^3*sqrt(x)*sgn(b*x + a) + 6*A*a^2*b*sqrt(x)*sgn(b* x + a) - 2*A*a^3*sgn(b*x + a)/sqrt(x)
Time = 10.51 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.50 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {x\,\left (70\,B\,a^3+210\,A\,b\,a^2\right )}{35\,b}-\frac {2\,A\,a^3}{b}+\frac {2\,B\,b^2\,x^4}{7}+\frac {x^3\,\left (14\,A\,b^3+42\,B\,a\,b^2\right )}{35\,b}+2\,a\,x^2\,\left (A\,b+B\,a\right )\right )}{x^{3/2}+\frac {a\,\sqrt {x}}{b}} \]