Integrand size = 33, antiderivative size = 169 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(A b-a B) (d+e x)^{1+m}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (b (2 B d-A e (1-m))-a B e (1+m)) (a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{2 b (b d-a e)^3 (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/2*(A*b-B*a)*(e*x+d)^(1+m)/b/(-a*e+b*d)/(b*x+a)/((b*x+a)^2)^(1/2)+1/2*e* (b*(2*B*d-A*e*(1-m))-a*B*e*(1+m))*(b*x+a)*(e*x+d)^(1+m)*hypergeom([2, 1+m] ,[2+m],b*(e*x+d)/(-a*e+b*d))/b/(-a*e+b*d)^3/(1+m)/((b*x+a)^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) (d+e x)^{1+m} \left (-A b+a B+\frac {e (2 b B d+A b e (-1+m)-a B e (1+m)) (a+b x)^2 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2 (1+m)}\right )}{2 b (b d-a e) \left ((a+b x)^2\right )^{3/2}} \]
((a + b*x)*(d + e*x)^(1 + m)*(-(A*b) + a*B + (e*(2*b*B*d + A*b*e*(-1 + m) - a*B*e*(1 + m))*(a + b*x)^2*Hypergeometric2F1[2, 1 + m, 2 + m, (b*(d + e* x))/(b*d - a*e)])/((b*d - a*e)^2*(1 + m))))/(2*b*(b*d - a*e)*((a + b*x)^2) ^(3/2))
Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {(A+B x) (d+e x)^m}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^m}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (\frac {(-a B e (m+1)-A b e (1-m)+2 b B d) \int \frac {(d+e x)^m}{(a+b x)^2}dx}{2 b (b d-a e)}-\frac {(A b-a B) (d+e x)^{m+1}}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {(a+b x) \left (\frac {e (d+e x)^{m+1} (-a B e (m+1)-A b e (1-m)+2 b B d) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{2 b (m+1) (b d-a e)^3}-\frac {(A b-a B) (d+e x)^{m+1}}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*((A*b - a*B)*(d + e*x)^(1 + m))/(b*(b*d - a*e)*(a + b*x)^ 2) + (e*(2*b*B*d - A*b*e*(1 - m) - a*B*e*(1 + m))*(d + e*x)^(1 + m)*Hyperg eometric2F1[2, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(2*b*(b*d - a*e)^ 3*(1 + m))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.92.3.1 Defintions of rubi rules used
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[(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]
\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(b^2*x^2 + 2*a*b*x + a^2)*(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)
Exception generated. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]