Integrand size = 21, antiderivative size = 376 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=-\frac {A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac {4^{-1+p} \left (2 a b B+A \left (2 a c-b^2 (1-p)\right )\right ) \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{a^2}-\frac {2^p c (2 a B-A b (1-p)) (1+2 p) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} (1+p)} \]
-1/2*A*(c*x^2+b*x+a)^(p+1)/a/x^2-1/2*(2*B*a-A*b*(1-p))*(c*x^2+b*x+a)^(p+1) /a^2/x-2^p*c*(2*B*a-A*b*(1-p))*(1+2*p)*(c*x^2+b*x+a)^(p+1)*hypergeom([-p, p+1],[2+p],1/2*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))*((-b-2*c*x +(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(-1-p)/a^2/(p+1)/(-4*a*c+b^2)^(1/ 2)+4^(-1+p)*(2*a*b*B+A*(2*a*c-b^2*(1-p)))*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p ,-p,1-2*p,1/2*(-b-(-4*a*c+b^2)^(1/2))/c/x,1/2*(-b+(-4*a*c+b^2)^(1/2))/c/x) /a^2/(((b+2*c*x-(-4*a*c+b^2)^(1/2))/c/x)^p)/(((b+2*c*x+(-4*a*c+b^2)^(1/2)) /c/x)^p)
Time = 0.73 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\frac {\left (1+\frac {b-\sqrt {b^2-4 a c}}{2 c x}\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c}\right )^p \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} (a+x (b+c x))^p \left (2 B (-1+p) x \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )+A (-1+2 p) \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )\right )}{2 (-1+p) (-1+2 p) x^2} \]
(((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/c)^p*(a + x*(b + c*x))^p*(2*B*(-1 + p)*x *AppellF1[1 - 2*p, -p, -p, 2 - 2*p, -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x), (- b + Sqrt[b^2 - 4*a*c])/(2*c*x)] + A*(-1 + 2*p)*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x), (-b + Sqrt[b^2 - 4*a*c])/(2*c* x)]))/(2*(-1 + p)*(-1 + 2*p)*(1 + (b - Sqrt[b^2 - 4*a*c])/(2*c*x))^p*x^2*( (b - Sqrt[b^2 - 4*a*c])/(2*c) + x)^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x ))^p)
Time = 0.57 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1237, 25, 1237, 25, 1269, 1096, 1178, 150}
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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle -\frac {\int -\frac {(2 a B-A (b-b p)+2 A c p x) \left (c x^2+b x+a\right )^p}{x^2}dx}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(2 a B-A b (1-p)+2 A c p x) \left (c x^2+b x+a\right )^p}{x^2}dx}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {-\frac {\int -\frac {\left (\left (-A (1-p) b^2+2 a B b+2 a A c\right ) p+c (2 a B-A b (1-p)) (2 p+1) x\right ) \left (c x^2+b x+a\right )^p}{x}dx}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {\left (\left (-A (1-p) b^2+2 a B b+2 a A c\right ) p+c (2 a B-A b (1-p)) (2 p+1) x\right ) \left (c x^2+b x+a\right )^p}{x}dx}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {\frac {p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) \int \frac {\left (c x^2+b x+a\right )^p}{x}dx+c (2 p+1) (2 a B-A b (1-p)) \int \left (c x^2+b x+a\right )^pdx}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 1096 |
\(\displaystyle \frac {\frac {p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) \int \frac {\left (c x^2+b x+a\right )^p}{x}dx-\frac {c 2^{p+1} (2 p+1) (2 a B-A b (1-p)) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 1178 |
\(\displaystyle \frac {\frac {-4^p p \left (\frac {1}{x}\right )^{2 p} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) \int \left (\frac {b-\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {b+\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {1}{x}\right )^{-2 p-1}d\frac {1}{x}-\frac {c 2^{p+1} (2 p+1) (2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {\frac {2^{2 p-1} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )-\frac {c 2^{p+1} (2 p+1) (2 a B-A b (1-p)) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{a x}}{2 a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2}\) |
-1/2*(A*(a + b*x + c*x^2)^(1 + p))/(a*x^2) + (-(((2*a*B - A*b*(1 - p))*(a + b*x + c*x^2)^(1 + p))/(a*x)) + ((2^(-1 + 2*p)*(2*a*b*B + 2*a*A*c - A*b^2 *(1 - p))*(a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b - Sq rt[b^2 - 4*a*c])/(c*x), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x)])/(((b - Sqrt[b ^2 - 4*a*c] + 2*c*x)/(c*x))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p) - (2^(1 + p)*c*(2*a*B - A*b*(1 - p))*(1 + 2*p)*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]))^(-1 - p)*(a + b*x + c*x^2)^(1 + p)*Hypergeometr ic2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a* c])])/(Sqrt[b^2 - 4*a*c]*(1 + p)))/a)/(2*a)
3.12.1.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(a + b*x + c*x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x) /(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q) ], x]] /; FreeQ[{a, b, c, p}, x] && !IntegerQ[4*p] && !IntegerQ[3*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(1/(d + e*x))^(2*p))*((a + b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q + 2*c* x)/(2*c*(d + e*x))))^p)) Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b - q)/(2*c)))*x, x]^p*Simp[1 - (d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[m, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
\[\int \frac {\left (B x +A \right ) \left (c \,x^{2}+b x +a \right )^{p}}{x^{3}}d x\]
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{3}} \,d x } \]
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{p}}{x^{3}}\, dx \]
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{3}} \,d x } \]
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p}{x^3} \,d x \]