Integrand size = 21, antiderivative size = 169 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac {2^{-n} (\text {b1} c-b \text {c1}) \left (-\frac {b-\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-1+n} \left (a+2 b x+c x^2\right )^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {b+\sqrt {b^2-a c}+c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c} (1-n)} \] Output:
1/2*c1*(c*x^2+2*b*x+a)^(1-n)/c/(1-n)-(-b*c1+b1*c)*(c*x^2+2*b*x+a)^(1-n)*hy pergeom([n, 1-n],[2-n],1/2*(b+c*x+(-a*c+b^2)^(1/2))/(-a*c+b^2)^(1/2))*((-b -c*x+(-a*c+b^2)^(1/2))/(-a*c+b^2)^(1/2))^(-1+n)/(2^n)/c/(1-n)/(-a*c+b^2)^( 1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.51 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.56 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\frac {1}{2} (a+x (2 b+c x))^{-n} \left (\text {c1} x^2 \left (\frac {b-\sqrt {b^2-a c}+c x}{b-\sqrt {b^2-a c}}\right )^n \left (\frac {b+\sqrt {b^2-a c}+c x}{b+\sqrt {b^2-a c}}\right )^n \operatorname {AppellF1}\left (2,n,n,3,-\frac {c x}{b+\sqrt {b^2-a c}},\frac {c x}{-b+\sqrt {b^2-a c}}\right )-\frac {2^{1-n} \text {b1} \left (b-\sqrt {b^2-a c}+c x\right ) \left (\frac {b+\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^n \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {-b+\sqrt {b^2-a c}-c x}{2 \sqrt {b^2-a c}}\right )}{c (-1+n)}\right ) \] Input:
Integrate[(b1 + c1*x)/(a + 2*b*x + c*x^2)^n,x]
Output:
(c1*x^2*((b - Sqrt[b^2 - a*c] + c*x)/(b - Sqrt[b^2 - a*c]))^n*((b + Sqrt[b ^2 - a*c] + c*x)/(b + Sqrt[b^2 - a*c]))^n*AppellF1[2, n, n, 3, -((c*x)/(b + Sqrt[b^2 - a*c])), (c*x)/(-b + Sqrt[b^2 - a*c])] - (2^(1 - n)*b1*(b - Sq rt[b^2 - a*c] + c*x)*((b + Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c])^n*Hyper geometric2F1[1 - n, n, 2 - n, (-b + Sqrt[b^2 - a*c] - c*x)/(2*Sqrt[b^2 - a *c])])/(c*(-1 + n)))/(2*(a + x*(2*b + c*x))^n)
Time = 0.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1160, 1096}
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.
\(\displaystyle \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {(\text {b1} c-b \text {c1}) \int \left (c x^2+2 b x+a\right )^{-n}dx}{c}+\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}\) |
\(\Big \downarrow \) 1096 |
\(\displaystyle \frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac {2^{-n} (\text {b1} c-b \text {c1}) \left (-\frac {-\sqrt {b^2-a c}+b+c x}{\sqrt {b^2-a c}}\right )^{n-1} \left (a+2 b x+c x^2\right )^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {b+c x+\sqrt {b^2-a c}}{2 \sqrt {b^2-a c}}\right )}{c (1-n) \sqrt {b^2-a c}}\) |
Input:
Int[(b1 + c1*x)/(a + 2*b*x + c*x^2)^n,x]
Output:
(c1*(a + 2*b*x + c*x^2)^(1 - n))/(2*c*(1 - n)) - ((b1*c - b*c1)*(-((b - Sq rt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c]))^(-1 + n)*(a + 2*b*x + c*x^2)^(1 - n )*Hypergeometric2F1[1 - n, n, 2 - n, (b + Sqrt[b^2 - a*c] + c*x)/(2*Sqrt[b ^2 - a*c])])/(2^n*c*Sqrt[b^2 - a*c]*(1 - n))
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_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
\[\int \left (\operatorname {c1} x +\operatorname {b1} \right ) \left (c \,x^{2}+2 b x +a \right )^{-n}d x\]
Input:
int((c1*x+b1)/((c*x^2+2*b*x+a)^n),x)
Output:
int((c1*x+b1)/((c*x^2+2*b*x+a)^n),x)
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \] Input:
integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="fricas")
Output:
integral((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)
Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\text {Timed out} \] Input:
integrate((c1*x+b1)/((c*x**2+2*b*x+a)**n),x)
Output:
Timed out
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \] Input:
integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="maxima")
Output:
integrate((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \] Input:
integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="giac")
Output:
integrate((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)
Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int \frac {b_{1}+c_{1}\,x}{{\left (c\,x^2+2\,b\,x+a\right )}^n} \,d x \] Input:
int((b1 + c1*x)/(a + 2*b*x + c*x^2)^n,x)
Output:
int((b1 + c1*x)/(a + 2*b*x + c*x^2)^n, x)
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\left (\int \frac {x}{\left (c \,x^{2}+2 b x +a \right )^{n}}d x \right ) \mathit {c1} +\left (\int \frac {1}{\left (c \,x^{2}+2 b x +a \right )^{n}}d x \right ) \mathit {b1} \] Input:
int((c1*x+b1)/((c*x^2+2*b*x+a)^n),x)
Output:
int(x/(a + 2*b*x + c*x**2)**n,x)*c1 + int(1/(a + 2*b*x + c*x**2)**n,x)*b1