Integrand size = 26, antiderivative size = 440 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\frac {e (d+e x)^{1+m} \left (a+b x+c x^2\right )^{-1-\frac {m}{2}}}{\left (c d^2-b d e+a e^2\right ) (1+m)}+\frac {e (2 c d-b e) m (d+e x)^{2+m} \left (a+b x+c x^2\right )^{-1-\frac {m}{2}}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+m) (2+m)}-\frac {\left (b^2 e^2 m+4 c^2 d^2 (1+m)+4 c e (a e-b d (1+m))\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{\frac {4+m}{2}} (d+e x)^{3+m} \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \operatorname {Hypergeometric2F1}\left (3+m,\frac {4+m}{2},4+m,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{4 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^2 (1+m) (3+m)} \]
e*(e*x+d)^(1+m)*(c*x^2+b*x+a)^(-1-1/2*m)/(a*e^2-b*d*e+c*d^2)/(1+m)+1/2*e*( -b*e+2*c*d)*m*(e*x+d)^(2+m)*(c*x^2+b*x+a)^(-1-1/2*m)/(a*e^2-b*d*e+c*d^2)^2 /(1+m)/(2+m)-1/4*(b^2*e^2*m+4*c^2*d^2*(1+m)+4*c*e*(a*e-b*d*(1+m)))*(e*x+d) ^(3+m)*(c*x^2+b*x+a)^(-2-1/2*m)*hypergeom([3+m, 2+1/2*m],[4+m],-4*c*(e*x+d )*(-4*a*c+b^2)^(1/2)/(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4*a*c+b^2) ^(1/2))))*(b+2*c*x-(-4*a*c+b^2)^(1/2))*((2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))*( b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4*a *c+b^2)^(1/2))))^(2+1/2*m)/(a*e^2-b*d*e+c*d^2)^2/(1+m)/(3+m)/(2*c*d-e*(b-( -4*a*c+b^2)^(1/2)))
Time = 4.55 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.86 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\frac {(d+e x)^{1+m} (a+x (b+c x))^{-2-\frac {m}{2}} \left (2 e \left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))-\frac {e (-2 c d+b e) m (d+e x) (a+x (b+c x))}{2+m}+\frac {\left (b^2 e^2 m+4 c^2 d^2 (1+m)-4 c e (-a e+b d (1+m))\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{\frac {2+m}{2}} (d+e x)^2 \operatorname {Hypergeometric2F1}\left (3+m,\frac {4+m}{2},4+m,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{2 \left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (3+m)}\right )}{2 \left (c d^2+e (-b d+a e)\right )^2 (1+m)} \]
((d + e*x)^(1 + m)*(a + x*(b + c*x))^(-2 - m/2)*(2*e*(c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x)) - (e*(-2*c*d + b*e)*m*(d + e*x)*(a + x*(b + c*x))) /(2 + m) + ((b^2*e^2*m + 4*c^2*d^2*(1 + m) - 4*c*e*(-(a*e) + b*d*(1 + m))) *(b + Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))^((2 + m)/2)*(d + e*x)^2*Hypergeometric2F1[3 + m, (4 + m)/2, 4 + m, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))])/(2*(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(3 + m))))/(2*(c*d^2 + e*(-(b*d) + a*e))^2*(1 + m) )
Time = 0.59 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1167, 27, 1228, 1155}
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.
| \(\displaystyle \int (d+e x)^m \left (a+b x+c x^2\right )^{-\frac {m}{2}-2} \, dx\) |
| \(\Big \downarrow \) 1167 |
| \(\displaystyle \frac {\int -\frac {1}{2} (d+e x)^{m+1} (b e m-2 c d (m+1)-2 c e x) \left (c x^2+b x+a\right )^{-\frac {m}{2}-2}dx}{(m+1) \left (a e^2-b d e+c d^2\right )}+\frac {e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}-\frac {\int (d+e x)^{m+1} (b e m-2 c d (m+1)-2 c e x) \left (c x^2+b x+a\right )^{-\frac {m}{2}-2}dx}{2 (m+1) \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}-\frac {-\frac {\left (4 c e (a e-b d (m+1))+b^2 e^2 m+4 c^2 d^2 (m+1)\right ) \int (d+e x)^{m+2} \left (c x^2+b x+a\right )^{-\frac {m}{2}-2}dx}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e m (2 c d-b e) (d+e x)^{m+2} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+2) \left (a e^2-b d e+c d^2\right )}}{2 (m+1) \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 1155 |
| \(\displaystyle \frac {e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}-\frac {\frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{m+3} \left (a+b x+c x^2\right )^{-\frac {m}{2}-2} \left (4 c e (a e-b d (m+1))+b^2 e^2 m+4 c^2 d^2 (m+1)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{\frac {m+4}{2}} \operatorname {Hypergeometric2F1}\left (m+3,\frac {m+4}{2},m+4,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (m+3) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac {e m (2 c d-b e) (d+e x)^{m+2} \left (a+b x+c x^2\right )^{-\frac {m}{2}-1}}{(m+2) \left (a e^2-b d e+c d^2\right )}}{2 (m+1) \left (a e^2-b d e+c d^2\right )}\) |
(e*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^(-1 - m/2))/((c*d^2 - b*d*e + a*e^2 )*(1 + m)) - (-((e*(2*c*d - b*e)*m*(d + e*x)^(2 + m)*(a + b*x + c*x^2)^(-1 - m/2))/((c*d^2 - b*d*e + a*e^2)*(2 + m))) + ((b^2*e^2*m + 4*c^2*d^2*(1 + m) + 4*c*e*(a*e - b*d*(1 + m)))*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)))^((4 + m)/2)*(d + e*x)^(3 + m)*(a + b*x + c*x^2)^(-2 - m/2)*Hypergeometric2F1[3 + m, (4 + m)/2, 4 + m, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/(2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)*(3 + m)))/(2*(c*d^2 - b*d*e + a*e^2)*(1 + m))
3.26.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(-(b - q + 2*c*x))*(d + e*x)^ (m + 1)*((a + b*x + c*x^2)^p/((m + 1)*(2*c*d - b*e + e*q)*((2*c*d - b*e + e *q)*((b + q + 2*c*x)/((2*c*d - b*e - e*q)*(b - q + 2*c*x))))^p))*Hypergeome tric2F1[m + 1, -p, m + 2, -4*c*q*((d + e*x)/((2*c*d - b*e - e*q)*(b - q + 2 *c*x)))], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(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)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m , -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 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)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
\[\int \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{-2-\frac {m}{2}}d x\]
\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{-\frac {1}{2} \, m - 2} {\left (e x + d\right )}^{m} \,d x } \]
Timed out. \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\text {Timed out} \]
\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{-\frac {1}{2} \, m - 2} {\left (e x + d\right )}^{m} \,d x } \]
\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{-\frac {1}{2} \, m - 2} {\left (e x + d\right )}^{m} \,d x } \]
Timed out. \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac {m}{2}} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^{\frac {m}{2}+2}} \,d x \]