Integrand size = 31, antiderivative size = 218 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 a^3 A \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 a^2 (3 A b+a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {6 a b (A b+a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 b^3 B x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \] Output:
2*a^3*A*x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*a^2*(3*A*b+B*a)*x^(3/2)*((b*x+ a)^2)^(1/2)/(3*b*x+3*a)+6*a*b*(A*b+B*a)*x^(5/2)*((b*x+a)^2)^(1/2)/(5*b*x+5 *a)+2*b^2*(A*b+3*B*a)*x^(7/2)*((b*x+a)^2)^(1/2)/(7*b*x+7*a)+2*b^3*B*x^(9/2 )*((b*x+a)^2)^(1/2)/(9*b*x+9*a)
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \sqrt {x} \sqrt {(a+b x)^2} \left (105 a^3 (3 A+B x)+63 a^2 b x (5 A+3 B x)+27 a b^2 x^2 (7 A+5 B x)+5 b^3 x^3 (9 A+7 B x)\right )}{315 (a+b x)} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/Sqrt[x],x]
Output:
(2*Sqrt[x]*Sqrt[(a + b*x)^2]*(105*a^3*(3*A + B*x) + 63*a^2*b*x*(5*A + 3*B* x) + 27*a*b^2*x^2*(7*A + 5*B*x) + 5*b^3*x^3*(9*A + 7*B*x)))/(315*(a + b*x) )
Time = 0.42 (sec) , antiderivative size = 111, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{\sqrt {x}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{\sqrt {x}}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^{7/2}+b^2 (A b+3 a B) x^{5/2}+3 a b (A b+a B) x^{3/2}+a^2 (3 A b+a B) \sqrt {x}+\frac {a^3 A}{\sqrt {x}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (2 a^3 A \sqrt {x}+\frac {2}{3} a^2 x^{3/2} (a B+3 A b)+\frac {2}{7} b^2 x^{7/2} (3 a B+A b)+\frac {6}{5} a b x^{5/2} (a B+A b)+\frac {2}{9} b^3 B x^{9/2}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/Sqrt[x],x]
Output:
(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)*x^( 3/2))/3 + (6*a*b*(A*b + a*B)*x^(5/2))/5 + (2*b^2*(A*b + 3*a*B)*x^(7/2))/7 + (2*b^3*B*x^(9/2))/9))/(a + b*x)
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 = 0.92 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(\frac {2 \sqrt {x}\, \left (35 B \,b^{3} x^{4}+45 A \,b^{3} x^{3}+135 B a \,b^{2} x^{3}+189 A a \,b^{2} x^{2}+189 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 B \,a^{3} x +315 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {2 \sqrt {x}\, \left (35 B \,b^{3} x^{4}+45 A \,b^{3} x^{3}+135 B a \,b^{2} x^{3}+189 A a \,b^{2} x^{2}+189 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 B \,a^{3} x +315 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (35 B \,b^{3} x^{4}+45 A \,b^{3} x^{3}+135 B a \,b^{2} x^{3}+189 A a \,b^{2} x^{2}+189 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 B \,a^{3} x +315 a^{3} A \right ) \sqrt {x}}{315 \left (b x +a \right )}\) | \(92\) |
orering | \(\frac {2 \left (35 B \,b^{3} x^{4}+45 A \,b^{3} x^{3}+135 B a \,b^{2} x^{3}+189 A a \,b^{2} x^{2}+189 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 B \,a^{3} x +315 a^{3} A \right ) \sqrt {x}\, \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{315 \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)
Output:
2/315*x^(1/2)*(35*B*b^3*x^4+45*A*b^3*x^3+135*B*a*b^2*x^3+189*A*a*b^2*x^2+1 89*B*a^2*b*x^2+315*A*a^2*b*x+105*B*a^3*x+315*A*a^3)*((b*x+a)^2)^(3/2)/(b*x +a)^3
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.33 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2}{315} \, {\left (35 \, B b^{3} x^{4} + 315 \, A a^{3} + 45 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 189 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 105 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )} \sqrt {x} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^(1/2),x, algorithm="fricas ")
Output:
2/315*(35*B*b^3*x^4 + 315*A*a^3 + 45*(3*B*a*b^2 + A*b^3)*x^3 + 189*(B*a^2* b + A*a*b^2)*x^2 + 105*(B*a^3 + 3*A*a^2*b)*x)*sqrt(x)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\sqrt {x}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**(1/2),x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/sqrt(x), x)
Time = 0.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2}{105} \, {\left (3 \, {\left (5 \, b^{3} x^{2} + 7 \, a b^{2} x\right )} x^{\frac {3}{2}} + 14 \, {\left (3 \, a b^{2} x^{2} + 5 \, a^{2} b x\right )} \sqrt {x} + \frac {35 \, {\left (a^{2} b x^{2} + 3 \, a^{3} x\right )}}{\sqrt {x}}\right )} A + \frac {2}{315} \, {\left (5 \, {\left (7 \, b^{3} x^{2} + 9 \, a b^{2} x\right )} x^{\frac {5}{2}} + 18 \, {\left (5 \, a b^{2} x^{2} + 7 \, a^{2} b x\right )} x^{\frac {3}{2}} + 21 \, {\left (3 \, a^{2} b x^{2} + 5 \, a^{3} x\right )} \sqrt {x}\right )} B \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^(1/2),x, algorithm="maxima ")
Output:
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)*sq rt(x) + 35*(a^2*b*x^2 + 3*a^3*x)/sqrt(x))*A + 2/315*(5*(7*b^3*x^2 + 9*a*b^ 2*x)*x^(5/2) + 18*(5*a*b^2*x^2 + 7*a^2*b*x)*x^(3/2) + 21*(3*a^2*b*x^2 + 5* a^3*x)*sqrt(x))*B
Time = 0.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2}{9} \, B b^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, B a b^{2} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, A b^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{5} \, B a^{2} b x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{5} \, A a b^{2} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{2} b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^(1/2),x, algorithm="giac")
Output:
2/9*B*b^3*x^(9/2)*sgn(b*x + a) + 6/7*B*a*b^2*x^(7/2)*sgn(b*x + a) + 2/7*A* b^3*x^(7/2)*sgn(b*x + a) + 6/5*B*a^2*b*x^(5/2)*sgn(b*x + a) + 6/5*A*a*b^2* x^(5/2)*sgn(b*x + a) + 2/3*B*a^3*x^(3/2)*sgn(b*x + a) + 2*A*a^2*b*x^(3/2)* sgn(b*x + a) + 2*A*a^3*sqrt(x)*sgn(b*x + a)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{\sqrt {x}} \,d x \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^(1/2),x)
Output:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^(1/2), x)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.21 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \sqrt {x}\, \left (35 b^{4} x^{4}+180 a \,b^{3} x^{3}+378 a^{2} b^{2} x^{2}+420 a^{3} b x +315 a^{4}\right )}{315} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^(1/2),x)
Output:
(2*sqrt(x)*(315*a**4 + 420*a**3*b*x + 378*a**2*b**2*x**2 + 180*a*b**3*x**3 + 35*b**4*x**4))/315