Integrand size = 17, antiderivative size = 185 \[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m}}{3 c^2 (a c-b d) \left (d+\frac {c}{x}\right )^3}-\frac {d (6 a c-b d (4+m)) \left (a+\frac {b}{x}\right )^{1+m}}{6 c^2 (a c-b d)^2 \left (d+\frac {c}{x}\right )^2}-\frac {b \left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \left (a+\frac {b}{x}\right )^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{6 c^2 (a c-b d)^4 (1+m)} \]
1/3*d^2*(a+b/x)^(1+m)/c^2/(a*c-b*d)/(d+c/x)^3-1/6*d*(6*a*c-b*d*(4+m))*(a+b /x)^(1+m)/c^2/(a*c-b*d)^2/(d+c/x)^2-1/6*b*(6*a^2*c^2-6*a*b*c*d*(1+m)+b^2*d ^2*(m^2+3*m+2))*(a+b/x)^(1+m)*hypergeom([2, 1+m],[2+m],c*(a+b/x)/(a*c-b*d) )/c^2/(a*c-b*d)^4/(1+m)
Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\frac {\left (a+\frac {b}{x}\right )^{1+m} \left (\frac {2 d^2 (a c-b d) x^3}{(c+d x)^3}+\frac {d (-6 a c+b d (4+m)) x^2}{(c+d x)^2}-\frac {b \left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b c+a c x}{a c x-b d x}\right )}{(a c-b d)^2 (1+m)}\right )}{6 c^2 (a c-b d)^2} \]
((a + b/x)^(1 + m)*((2*d^2*(a*c - b*d)*x^3)/(c + d*x)^3 + (d*(-6*a*c + b*d *(4 + m))*x^2)/(c + d*x)^2 - (b*(6*a^2*c^2 - 6*a*b*c*d*(1 + m) + b^2*d^2*( 2 + 3*m + m^2))*Hypergeometric2F1[2, 1 + m, 2 + m, (b*c + a*c*x)/(a*c*x - b*d*x)])/((a*c - b*d)^2*(1 + m))))/(6*c^2*(a*c - b*d)^2)
Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {941, 948, 100, 25, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx\) |
\(\Big \downarrow \) 941 |
\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^m}{x^4 \left (\frac {c}{x}+d\right )^4}dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\int \frac {\left (a+\frac {b}{x}\right )^m}{\left (\frac {c}{x}+d\right )^4 x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}-\frac {\int -\frac {\left (a+\frac {b}{x}\right )^m \left (d (3 a c-b d (m+1))-\frac {3 c (a c-b d)}{x}\right )}{\left (\frac {c}{x}+d\right )^3}d\frac {1}{x}}{3 c^2 (a c-b d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\left (a+\frac {b}{x}\right )^m \left (d (3 a c-b d (m+1))-\frac {3 c (a c-b d)}{x}\right )}{\left (\frac {c}{x}+d\right )^3}d\frac {1}{x}}{3 c^2 (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {-\frac {\left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \int \frac {\left (a+\frac {b}{x}\right )^m}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}}{2 (a c-b d)}-\frac {d \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (m+4))}{2 \left (\frac {c}{x}+d\right )^2 (a c-b d)}}{3 c^2 (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {-\frac {b \left (a+\frac {b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{2 (m+1) (a c-b d)^3}-\frac {d \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (m+4))}{2 \left (\frac {c}{x}+d\right )^2 (a c-b d)}}{3 c^2 (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}\) |
(d^2*(a + b/x)^(1 + m))/(3*c^2*(a*c - b*d)*(d + c/x)^3) + (-1/2*(d*(6*a*c - b*d*(4 + m))*(a + b/x)^(1 + m))/((a*c - b*d)*(d + c/x)^2) - (b*(6*a^2*c^ 2 - 6*a*b*c*d*(1 + m) + b^2*d^2*(2 + 3*m + m^2))*(a + b/x)^(1 + m)*Hyperge ometric2F1[2, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)])/(2*(a*c - b*d)^3*( 1 + m)))/(3*c^2*(a*c - b*d))
3.6.94.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Sym bol] :> Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] || !IntegerQ[p])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \frac {\left (a +\frac {b}{x}\right )^{m}}{\left (d x +c \right )^{4}}d x\]
\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int \frac {\left (a + \frac {b}{x}\right )^{m}}{\left (c + d x\right )^{4}}\, dx \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int \frac {{\left (a+\frac {b}{x}\right )}^m}{{\left (c+d\,x\right )}^4} \,d x \]