Integrand size = 35, antiderivative size = 150 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {4 b (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \] Output:
2*(-a*e+b*d)^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-4/3*b*(-a*e+b*d )*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/5*b^2*(e*x+d)^(5/2)*((b*x+ a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \] 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]*(15*a^2*e^2 + 10*a*b*e*(-2*d + e*x) + b ^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2)))/(15*e^3*(a + b*x))
Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 27, 53, 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.
\(\displaystyle \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b (a+b x)^2}{\sqrt {d+e x}}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^2}{\sqrt {d+e x}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^{3/2} b^2}{e^2}-\frac {2 (b d-a e) \sqrt {d+e x} b}{e^2}+\frac {(a e-b d)^2}{e^2 \sqrt {d+e x}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {4 b (d+e x)^{3/2} (b d-a e)}{3 e^3}+\frac {2 \sqrt {d+e x} (b d-a e)^2}{e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3}\right )}{a+b x}\) |
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)^2*Sqrt[d + e*x])/e^3 - (4*b *(b*d - a*e)*(d + e*x)^(3/2))/(3*e^3) + (2*b^2*(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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
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 = 1.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right )}{15 e^{3}}\) | \(69\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(79\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(79\) |
orering | \(\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(79\) |
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^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)/e^3
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.43 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 20 \, a b d e + 15 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \] Input:
integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")
Output:
2/15*(3*b^2*e^2*x^2 + 8*b^2*d^2 - 20*a*b*d*e + 15*a^2*e^2 - 2*(2*b^2*d*e - 5*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 = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} a}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} b}{15 \, \sqrt {e x + d} e^{3}} \] 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 = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {10 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{15 \, e} \] 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^2*sgn(b*x + a) + 10*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*b*sgn(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^2*sgn(b*x + a)/e^2)/e
Time = 11.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,b\,x^3}{5}+\frac {2\,x^2\,\left (10\,a\,e-b\,d\right )}{15\,e}+\frac {30\,a^2\,d\,e^2-40\,a\,b\,d^2\,e+16\,b^2\,d^3}{15\,b\,e^3}+\frac {x\,\left (30\,a^2\,e^3-20\,a\,b\,d\,e^2+8\,b^2\,d^2\,e\right )}{15\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \] 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 + (2*x^2*(10*a*e - b*d))/(15*e) + (16*b^ 2*d^3 + 30*a^2*d*e^2 - 40*a*b*d^2*e)/(15*b*e^3) + (x*(30*a^2*e^3 + 8*b^2*d ^2*e - 20*a*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 = 0.41 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right )}{15 e^{3}} \] 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)