3.11.100 \(\int \frac {(A+B x) (a+b x+c x^2)^p}{x^2} \, dx\) [1100]

3.11.100.1 Optimal result

Integrand size = 21, antiderivative size = 315 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=-\frac {A \left (a+b x+c x^2\right )^{1+p}}{a x}+\frac {2^{-1+2 p} (a B+A b p) \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{a p}-\frac {2^{1+p} A c (1+2 p) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} (1+p)} \]

output

-A*(c*x^2+b*x+a)^(p+1)/a/x-2^(p+1)*A*c*(1+2*p)*(c*x^2+b*x+a)^(p+1)*hyperge 
om([-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/(p+1)/(-4*a*c+b^ 
2)^(1/2)+2^(-1+2*p)*(A*b*p+B*a)*(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/p/(((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)
 

3.11.100.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\frac {\left (1+\frac {b-\sqrt {b^2-4 a c}}{2 c x}\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c}\right )^p \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} (a+x (b+c x))^p \left (2 A p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )+B (-1+2 p) x \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )\right )}{2 p (-1+2 p) x} \]

input

Integrate[((A + B*x)*(a + b*x + c*x^2)^p)/x^2,x]
 

output

(((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/c)^p*(a + x*(b + c*x))^p*(2*A*p*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)] + B*(-1 + 2*p)*x*AppellF1[-2*p, -p, -p, 1 - 2*p, -1 
/2*(b + Sqrt[b^2 - 4*a*c])/(c*x), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x)]))/(2*p 
*(-1 + 2*p)*(1 + (b - Sqrt[b^2 - 4*a*c])/(2*c*x))^p*x*((b - Sqrt[b^2 - 4*a 
*c])/(2*c) + x)^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p)
 

3.11.100.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {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.

\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx\)

\(\Big \downarrow \) 1237

\(\displaystyle -\frac {\int -\frac {(a B+A b p+A c (2 p+1) x) \left (c x^2+b x+a\right )^p}{x}dx}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(a B+A b p+A c (2 p+1) x) \left (c x^2+b x+a\right )^p}{x}dx}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {(a B+A b p) \int \frac {\left (c x^2+b x+a\right )^p}{x}dx+A c (2 p+1) \int \left (c x^2+b x+a\right )^pdx}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

\(\Big \downarrow \) 1096

\(\displaystyle \frac {(a B+A b p) \int \frac {\left (c x^2+b x+a\right )^p}{x}dx-\frac {A c 2^{p+1} (2 p+1) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

\(\Big \downarrow \) 1178

\(\displaystyle \frac {-4^p \left (\frac {1}{x}\right )^{2 p} (a B+A b p) \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \int \left (\frac {b-\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {b+\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {1}{x}\right )^{-2 p-1}d\frac {1}{x}-\frac {A c 2^{p+1} (2 p+1) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\frac {2^{2 p-1} (a B+A b p) \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{p}-\frac {A c 2^{p+1} (2 p+1) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}}}{a}-\frac {A \left (a+b x+c x^2\right )^{p+1}}{a x}\)

input

Int[((A + B*x)*(a + b*x + c*x^2)^p)/x^2,x]
 

output

-((A*(a + b*x + c*x^2)^(1 + p))/(a*x)) + ((2^(-1 + 2*p)*(a*B + A*b*p)*(a + 
 b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b - Sqrt[b^2 - 4*a*c 
])/(c*x), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x)])/(p*((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)* 
A*c*(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)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqr 
t[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(1 + p) 
))/a
 

3.11.100.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 150

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])
 

rule 1096

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]
 

rule 1178

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]
 

rule 1237

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])
 

rule 1269

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]
 

3.11.100.4 Maple [F]

\[\int \frac {\left (B x +A \right ) \left (c \,x^{2}+b x +a \right )^{p}}{x^{2}}d x\]

input

int((B*x+A)*(c*x^2+b*x+a)^p/x^2,x)
 

output

int((B*x+A)*(c*x^2+b*x+a)^p/x^2,x)
 

3.11.100.5 Fricas [F]

\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{2}} \,d x } \]

input

integrate((B*x+A)*(c*x^2+b*x+a)^p/x^2,x, algorithm="fricas")
 

output

integral((B*x + A)*(c*x^2 + b*x + a)^p/x^2, x)
 

3.11.100.6 Sympy [F]

\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{p}}{x^{2}}\, dx \]

input

integrate((B*x+A)*(c*x**2+b*x+a)**p/x**2,x)
 

output

Integral((A + B*x)*(a + b*x + c*x**2)**p/x**2, x)
 

3.11.100.7 Maxima [F]

\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{2}} \,d x } \]

input

integrate((B*x+A)*(c*x^2+b*x+a)^p/x^2,x, algorithm="maxima")
 

output

integrate((B*x + A)*(c*x^2 + b*x + a)^p/x^2, x)
 

3.11.100.8 Giac [F]

\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}}{x^{2}} \,d x } \]

input

integrate((B*x+A)*(c*x^2+b*x+a)^p/x^2,x, algorithm="giac")
 

output

integrate((B*x + A)*(c*x^2 + b*x + a)^p/x^2, x)
 

3.11.100.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^p}{x^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p}{x^2} \,d x \]

input

int(((A + B*x)*(a + b*x + c*x^2)^p)/x^2,x)
 

output

int(((A + B*x)*(a + b*x + c*x^2)^p)/x^2, x)