Integrand size = 31, antiderivative size = 135 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b B d+A b e-2 a B e}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B e}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-1/4*(A*b-B*a)*(-a*e+b*d)/b^3/(b*x+a)^3/((b*x+a)^2)^(1/2)-1/3*(A*b*e-2*B*a *e+B*b*d)/b^3/(b*x+a)^2/((b*x+a)^2)^(1/2)-1/2*B*e/b^3/(b*x+a)/((b*x+a)^2)^ (1/2)
Time = 1.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-A b (3 b d+a e+4 b e x)-B \left (a^2 e+2 b^2 x (2 d+3 e x)+a b (d+4 e x)\right )}{12 b^3 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:
Integrate[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
(-(A*b*(3*b*d + a*e + 4*b*e*x)) - B*(a^2*e + 2*b^2*x*(2*d + 3*e*x) + a*b*( d + 4*e*x)))/(12*b^3*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {(A+B x) (d+e x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {B e}{b^2 (a+b x)^3}+\frac {b B d+A b e-2 a B e}{b^2 (a+b x)^4}+\frac {(A b-a B) (b d-a e)}{b^2 (a+b x)^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {-2 a B e+A b e+b B d}{3 b^3 (a+b x)^3}-\frac {(A b-a B) (b d-a e)}{4 b^3 (a+b x)^4}-\frac {B e}{2 b^3 (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
((a + b*x)*(-1/4*((A*b - a*B)*(b*d - a*e))/(b^3*(a + b*x)^4) - (b*B*d + A* b*e - 2*a*B*e)/(3*b^3*(a + b*x)^3) - (B*e)/(2*b^3*(a + b*x)^2)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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]
Time = 1.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {\left (b x +a \right ) \left (6 B e \,x^{2} b^{2}+4 A \,b^{2} e x +4 B a b e x +4 B \,b^{2} d x +A a b e +3 A \,b^{2} d +B e \,a^{2}+B a b d \right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(77\) |
default | \(-\frac {\left (b x +a \right ) \left (6 B e \,x^{2} b^{2}+4 A \,b^{2} e x +4 B a b e x +4 B \,b^{2} d x +A a b e +3 A \,b^{2} d +B e \,a^{2}+B a b d \right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(77\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B e \,x^{2}}{2 b}-\frac {\left (A b e +B a e +B b d \right ) x}{3 b^{2}}-\frac {A a b e +3 A \,b^{2} d +B e \,a^{2}+B a b d}{12 b^{3}}\right )}{\left (b x +a \right )^{5}}\) | \(77\) |
orering | \(-\frac {\left (6 B e \,x^{2} b^{2}+4 A \,b^{2} e x +4 B a b e x +4 B \,b^{2} d x +A a b e +3 A \,b^{2} d +B e \,a^{2}+B a b d \right ) \left (b x +a \right )}{12 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}\) | \(86\) |
Input:
int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/12*(b*x+a)/b^3*(6*B*b^2*e*x^2+4*A*b^2*e*x+4*B*a*b*e*x+4*B*b^2*d*x+A*a*b *e+3*A*b^2*d+B*a^2*e+B*a*b*d)/((b*x+a)^2)^(5/2)
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} e x^{2} + {\left (B a b + 3 \, A b^{2}\right )} d + {\left (B a^{2} + A a b\right )} e + 4 \, {\left (B b^{2} d + {\left (B a b + A b^{2}\right )} e\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas ")
Output:
-1/12*(6*B*b^2*e*x^2 + (B*a*b + 3*A*b^2)*d + (B*a^2 + A*a*b)*e + 4*(B*b^2* d + (B*a*b + A*b^2)*e)*x)/(b^7*x^4 + 4*a*b^6*x^3 + 6*a^2*b^5*x^2 + 4*a^3*b ^4*x + a^4*b^3)
\[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Integral((A + B*x)*(d + e*x)/((a + b*x)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {B d + A e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {B e}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a e}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {A d}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {B a^{2} e}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {{\left (B d + A e\right )} a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima ")
Output:
-1/3*(B*d + A*e)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 1/2*B*e/(b^5*(x + a/b)^2) + 2/3*B*a*e/(b^6*(x + a/b)^3) - 1/4*A*d/(b^5*(x + a/b)^4) - 1/4*B *a^2*e/(b^7*(x + a/b)^4) + 1/4*(B*d + A*e)*a/(b^6*(x + a/b)^4)
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} e x^{2} + 4 \, B b^{2} d x + 4 \, B a b e x + 4 \, A b^{2} e x + B a b d + 3 \, A b^{2} d + B a^{2} e + A a b e}{12 \, {\left (b x + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
Output:
-1/12*(6*B*b^2*e*x^2 + 4*B*b^2*d*x + 4*B*a*b*e*x + 4*A*b^2*e*x + B*a*b*d + 3*A*b^2*d + B*a^2*e + A*a*b*e)/((b*x + a)^4*b^3*sgn(b*x + a))
Time = 11.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (3\,A\,b^2\,d+B\,a^2\,e+4\,A\,b^2\,e\,x+4\,B\,b^2\,d\,x+6\,B\,b^2\,e\,x^2+A\,a\,b\,e+B\,a\,b\,d+4\,B\,a\,b\,e\,x\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \] Input:
int(((A + B*x)*(d + e*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
Output:
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(3*A*b^2*d + B*a^2*e + 4*A*b^2*e*x + 4*B *b^2*d*x + 6*B*b^2*e*x^2 + A*a*b*e + B*a*b*d + 4*B*a*b*e*x))/(12*b^3*(a + b*x)^5)
Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.36 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-3 b e x -a e -2 b d}{6 b^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:
int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
( - a*e - 2*b*d - 3*b*e*x)/(6*b**2*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b* *3*x**3))