Integrand size = 31, antiderivative size = 121 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {B (d+e x)^{1+m}}{b e (2-m) (a+b x)^3}-\frac {e^2 (3 b B d-A b e (2-m)-a B e (1+m)) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{b (b d-a e)^4 (2-m) (1+m)} \] Output:
-B*(e*x+d)^(1+m)/b/e/(2-m)/(b*x+a)^3-e^2*(3*B*b*d-A*b*e*(2-m)-a*B*e*(1+m)) *(e*x+d)^(1+m)*hypergeom([4, 1+m],[2+m],b*(e*x+d)/(-a*e+b*d))/b/(-a*e+b*d) ^4/(2-m)/(1+m)
Time = 0.17 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {-A b+a B}{(a+b x)^3}-\frac {e^2 (3 b B d+A b e (-2+m)-a B e (1+m)) \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^3 (1+m)}\right )}{3 b (b d-a e)} \] Input:
Integrate[((A + B*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
((d + e*x)^(1 + m)*((-(A*b) + a*B)/(a + b*x)^3 - (e^2*(3*b*B*d + A*b*e*(-2 + m) - a*B*e*(1 + m))*Hypergeometric2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b *d - a*e)])/((b*d - a*e)^3*(1 + m))))/(3*b*(b*d - a*e))
Time = 0.44 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 87, 78}
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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle b^4 \int \frac {(A+B x) (d+e x)^m}{b^4 (a+b x)^4}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(A+B x) (d+e x)^m}{(a+b x)^4}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(-a B e (m+1)-A b e (2-m)+3 b B d) \int \frac {(d+e x)^m}{(a+b x)^3}dx}{3 b (b d-a e)}-\frac {(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle -\frac {e^2 (d+e x)^{m+1} (-a B e (m+1)-A b e (2-m)+3 b B d) \operatorname {Hypergeometric2F1}\left (3,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{3 b (m+1) (b d-a e)^4}-\frac {(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)}\) |
Input:
Int[((A + B*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
-1/3*((A*b - a*B)*(d + e*x)^(1 + m))/(b*(b*d - a*e)*(a + b*x)^3) - (e^2*(3 *b*B*d - A*b*e*(2 - m) - a*B*e*(1 + m))*(d + e*x)^(1 + m)*Hypergeometric2F 1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(3*b*(b*d - a*e)^4*(1 + m))
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] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^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}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}d x\]
Input:
int((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
int((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x)
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:
integrate((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
integral((B*x + A)*(e*x + d)^m/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4* a^3*b*x + a^4), x)
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{m}}{\left (a + b x\right )^{4}}\, dx \] Input:
integrate((B*x+A)*(e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
Integral((A + B*x)*(d + e*x)**m/(a + b*x)**4, x)
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:
integrate((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
integrate((B*x + A)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:
integrate((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
integrate((B*x + A)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)
Timed out. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^2} \,d x \] Input:
int(((A + B*x)*(d + e*x)^m)/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
int(((A + B*x)*(d + e*x)^m)/(a^2 + b^2*x^2 + 2*a*b*x)^2, x)
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {too large to display} \] Input:
int((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
((d + e*x)**m*d + int(((d + e*x)**m*x)/(a**4*d*e*m + a**4*e**2*m*x - 2*a** 3*b*d**2 + 3*a**3*b*d*e*m*x - 2*a**3*b*d*e*x + 3*a**3*b*e**2*m*x**2 - 6*a* *2*b**2*d**2*x + 3*a**2*b**2*d*e*m*x**2 - 6*a**2*b**2*d*e*x**2 + 3*a**2*b* *2*e**2*m*x**3 - 6*a*b**3*d**2*x**2 + a*b**3*d*e*m*x**3 - 6*a*b**3*d*e*x** 3 + a*b**3*e**2*m*x**4 - 2*b**4*d**2*x**3 - 2*b**4*d*e*x**4),x)*a**4*e**3* m**2 - int(((d + e*x)**m*x)/(a**4*d*e*m + a**4*e**2*m*x - 2*a**3*b*d**2 + 3*a**3*b*d*e*m*x - 2*a**3*b*d*e*x + 3*a**3*b*e**2*m*x**2 - 6*a**2*b**2*d** 2*x + 3*a**2*b**2*d*e*m*x**2 - 6*a**2*b**2*d*e*x**2 + 3*a**2*b**2*e**2*m*x **3 - 6*a*b**3*d**2*x**2 + a*b**3*d*e*m*x**3 - 6*a*b**3*d*e*x**3 + a*b**3* e**2*m*x**4 - 2*b**4*d**2*x**3 - 2*b**4*d*e*x**4),x)*a**3*b*d*e**2*m**2 - 2*int(((d + e*x)**m*x)/(a**4*d*e*m + a**4*e**2*m*x - 2*a**3*b*d**2 + 3*a** 3*b*d*e*m*x - 2*a**3*b*d*e*x + 3*a**3*b*e**2*m*x**2 - 6*a**2*b**2*d**2*x + 3*a**2*b**2*d*e*m*x**2 - 6*a**2*b**2*d*e*x**2 + 3*a**2*b**2*e**2*m*x**3 - 6*a*b**3*d**2*x**2 + a*b**3*d*e*m*x**3 - 6*a*b**3*d*e*x**3 + a*b**3*e**2* m*x**4 - 2*b**4*d**2*x**3 - 2*b**4*d*e*x**4),x)*a**3*b*d*e**2*m + 2*int((( d + e*x)**m*x)/(a**4*d*e*m + a**4*e**2*m*x - 2*a**3*b*d**2 + 3*a**3*b*d*e* m*x - 2*a**3*b*d*e*x + 3*a**3*b*e**2*m*x**2 - 6*a**2*b**2*d**2*x + 3*a**2* b**2*d*e*m*x**2 - 6*a**2*b**2*d*e*x**2 + 3*a**2*b**2*e**2*m*x**3 - 6*a*b** 3*d**2*x**2 + a*b**3*d*e*m*x**3 - 6*a*b**3*d*e*x**3 + a*b**3*e**2*m*x**4 - 2*b**4*d**2*x**3 - 2*b**4*d*e*x**4),x)*a**3*b*e**3*m**2*x + 2*int(((d ...