Integrand size = 35, antiderivative size = 162 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e) (B d-A e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b B (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \] Output:
2*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-2/3*(- A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/5*b*B*( e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (5 A b e (-2 d+e x)+5 a e (-2 B d+3 A e+B e x)+b B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \] Input:
Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/Sqrt[d + e*x],x]
Output:
(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(5*A*b*e*(-2*d + e*x) + 5*a*e*(-2*B*d + 3*A*e + B*e*x) + b*B*(8*d^2 - 4*d*e*x + 3*e^2*x^2)))/(15*e^3*(a + b*x))
Time = 0.44 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.67, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} (A+B x)}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b (a+b x) (A+B x)}{\sqrt {d+e x}}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b B (d+e x)^{3/2}}{e^2}+\frac {(-2 b B d+A b e+a B e) \sqrt {d+e x}}{e^2}+\frac {(a e-b d) (A e-B d)}{e^2 \sqrt {d+e x}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3}+\frac {2 \sqrt {d+e x} (b d-a e) (B d-A e)}{e^3}+\frac {2 b B (d+e x)^{5/2}}{5 e^3}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/Sqrt[d + e*x],x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)*(B*d - A*e)*Sqrt[d + e*x])/ e^3 - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(3/2))/(3*e^3) + (2*b*B*(d + e*x)^(5/2))/(5*e^3)))/(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[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 2.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right )}{15 e^{3}}\) | \(79\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(89\) |
orering | \(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(89\) |
Input:
int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/15*csgn(b*x+a)*(e*x+d)^(1/2)*(3*B*b*e^2*x^2+5*A*b*e^2*x+5*B*a*e^2*x-4*B* b*d*e*x+15*A*a*e^2-10*A*b*d*e-10*B*a*d*e+8*B*b*d^2)/e^3
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 15 \, A a e^{2} - 10 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 5 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")
Output:
2/15*(3*B*b*e^2*x^2 + 8*B*b*d^2 + 15*A*a*e^2 - 10*(B*a + A*b)*d*e - (4*B*b *d*e - 5*(B*a + A*b)*e^2)*x)*sqrt(e*x + d)/e^3
\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{\sqrt {d + e x}}\, dx \] Input:
integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**(1/2),x)
Output:
Integral((A + B*x)*sqrt((a + b*x)**2)/sqrt(d + e*x), x)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} A}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} B}{15 \, \sqrt {e x + d} e^{3}} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")
Output:
2/3*(b*e^2*x^2 - 2*b*d^2 + 3*a*d*e - (b*d*e - 3*a*e^2)*x)*A/(sqrt(e*x + d) *e^2) + 2/15*(3*b*e^3*x^3 + 8*b*d^3 - 10*a*d^2*e - (b*d*e^2 - 5*a*e^3)*x^2 + (4*b*d^2*e - 5*a*d*e^2)*x)*B/(sqrt(e*x + d)*e^3)
Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} A a \mathrm {sgn}\left (b x + a\right ) + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{15 \, e} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")
Output:
2/15*(15*sqrt(e*x + d)*A*a*sgn(b*x + a) + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*sgn(b*x + a)/e + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*s gn(b*x + a)/e + (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*sgn(b*x + a)/e^2)/e
Time = 11.47 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^3}{5}+\frac {16\,B\,b\,d^3+30\,A\,a\,d\,e^2-20\,A\,b\,d^2\,e-20\,B\,a\,d^2\,e}{15\,b\,e^3}+\frac {x\,\left (30\,A\,a\,e^3-10\,A\,b\,d\,e^2-10\,B\,a\,d\,e^2+8\,B\,b\,d^2\,e\right )}{15\,b\,e^3}+\frac {x^2\,\left (10\,A\,b\,e^3+10\,B\,a\,e^3-2\,B\,b\,d\,e^2\right )}{15\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \] Input:
int((((a + b*x)^2)^(1/2)*(A + B*x))/(d + e*x)^(1/2),x)
Output:
(((a + b*x)^2)^(1/2)*((2*B*x^3)/5 + (16*B*b*d^3 + 30*A*a*d*e^2 - 20*A*b*d^ 2*e - 20*B*a*d^2*e)/(15*b*e^3) + (x*(30*A*a*e^3 - 10*A*b*d*e^2 - 10*B*a*d* e^2 + 8*B*b*d^2*e))/(15*b*e^3) + (x^2*(10*A*b*e^3 + 10*B*a*e^3 - 2*B*b*d*e ^2))/(15*b*e^3)))/(x*(d + e*x)^(1/2) + (a*(d + e*x)^(1/2))/b)
Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right )}{15 e^{3}} \] Input:
int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x)
Output:
(2*sqrt(d + e*x)*(15*a**2*e**2 - 20*a*b*d*e + 10*a*b*e**2*x + 8*b**2*d**2 - 4*b**2*d*e*x + 3*b**2*e**2*x**2))/(15*e**3)