Integrand size = 18, antiderivative size = 139 \[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=-\frac {d (a+b x)^{1+n}}{c (b c-a d) (c+d x)}+\frac {d (a d-b c (1-n)) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {d (a+b x)}{b c-a d}\right )}{c^2 (b c-a d)^2 (1+n)}-\frac {(a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )}{a c^2 (1+n)} \]
-d*(b*x+a)^(1+n)/c/(-a*d+b*c)/(d*x+c)+d*(a*d-b*c*(1-n))*(b*x+a)^(1+n)*hype rgeom([1, 1+n],[2+n],-d*(b*x+a)/(-a*d+b*c))/c^2/(-a*d+b*c)^2/(1+n)-(b*x+a) ^(1+n)*hypergeom([1, 1+n],[2+n],1+b*x/a)/a/c^2/(1+n)
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\frac {(a+b x)^{1+n} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b x}{a}\right )}{a+a n}+\frac {d \left (\frac {c (-b c+a d)}{c+d x}+\frac {(a d+b c (-1+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )}{1+n}\right )}{(b c-a d)^2}\right )}{c^2} \]
((a + b*x)^(1 + n)*(-(Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*x)/a]/(a + a*n)) + (d*((c*(-(b*c) + a*d))/(c + d*x) + ((a*d + b*c*(-1 + n))*Hypergeo metric2F1[1, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)])/(1 + n)))/(b*c - a*d)^2))/c^2
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {114, 25, 174, 75, 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.
\(\displaystyle \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx\) |
\(\Big \downarrow \) 114 |
\(\displaystyle -\frac {\int -\frac {(a+b x)^n (b c-a d+b d n x)}{x (c+d x)}dx}{c (b c-a d)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(a+b x)^n (b c-a d+b d n x)}{x (c+d x)}dx}{c (b c-a d)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {\frac {(b c-a d) \int \frac {(a+b x)^n}{x}dx}{c}+\frac {d (a d-b c (1-n)) \int \frac {(a+b x)^n}{c+d x}dx}{c}}{c (b c-a d)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {\frac {d (a d-b c (1-n)) \int \frac {(a+b x)^n}{c+d x}dx}{c}-\frac {(b c-a d) (a+b x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b x}{a}+1\right )}{a c (n+1)}}{c (b c-a d)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {\frac {d (a+b x)^{n+1} (a d-b c (1-n)) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {d (a+b x)}{b c-a d}\right )}{c (n+1) (b c-a d)}-\frac {(b c-a d) (a+b x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b x}{a}+1\right )}{a c (n+1)}}{c (b c-a d)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)}\) |
-((d*(a + b*x)^(1 + n))/(c*(b*c - a*d)*(c + d*x))) + ((d*(a*d - b*c*(1 - n ))*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b *c - a*d))])/(c*(b*c - a*d)*(1 + n)) - ((b*c - a*d)*(a + b*x)^(1 + n)*Hype rgeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*c*(1 + n)))/(c*(b*c - a*d) )
3.10.46.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
\[\int \frac {\left (b x +a \right )^{n}}{x \left (d x +c \right )^{2}}d x\]
\[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x} \,d x } \]
\[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\int \frac {\left (a + b x\right )^{n}}{x \left (c + d x\right )^{2}}\, dx \]
\[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x} \,d x } \]
\[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^n}{x (c+d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^n}{x\,{\left (c+d\,x\right )}^2} \,d x \]