Integrand size = 44, antiderivative size = 210 \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(b e g (1-2 m)-2 c (e f (1-m)-d g m)) (d+e x)^m \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {c d-b e-c e x}{2 c d-b e}\right )}{c e^2 (2 c d-b e) (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]
g*(e*x+d)^m/c/e^2/(1-m)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)-(b*e*g*(1-2 *m)-2*c*(e*f*(1-m)-d*g*m))*(e*x+d)^m*(c*(e*x+d)/(-b*e+2*c*d))^(1/2-m)*hype rgeom([-1/2, 3/2-m],[1/2],(-c*e*x-b*e+c*d)/(-b*e+2*c*d))/c/e^2/(-b*e+2*c*d )/(1-m)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)
Time = 0.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\frac {2 (d+e x)^m \left (e (2 c d-b e) (c e f+c d g-b e g)-e (b e g (1-2 m)+2 c (e f (-1+m)+d g m)) \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (-c d+b e+c e x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}-m,\frac {3}{2},\frac {-c d+b e+c e x}{-2 c d+b e}\right )\right )}{c e^3 (-2 c d+b e)^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \]
(2*(d + e*x)^m*(e*(2*c*d - b*e)*(c*e*f + c*d*g - b*e*g) - e*(b*e*g*(1 - 2* m) + 2*c*(e*f*(-1 + m) + d*g*m))*((c*(d + e*x))/(2*c*d - b*e))^(1/2 - m)*( -(c*d) + b*e + c*e*x)*Hypergeometric2F1[1/2, 3/2 - m, 3/2, (-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)]))/(c*e^3*(-2*c*d + b*e)^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])
Time = 0.51 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1221, 1139, 1138, 80, 79}
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.
| \(\displaystyle \int \frac {(f+g x) (d+e x)^m}{\left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(b e g (1-2 m)-2 c (e f (1-m)-d g m)) \int \frac {(d+e x)^m}{\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}dx}{2 c e (1-m)}\) |
| \(\Big \downarrow \) 1139 |
| \(\displaystyle \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} (b e g (1-2 m)-2 c (e f (1-m)-d g m)) \int \frac {\left (\frac {e x}{d}+1\right )^m}{\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}dx}{2 c e (1-m)}\) |
| \(\Big \downarrow \) 1138 |
| \(\displaystyle \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \sqrt {d (c d-b e)-c d e x} (b e g (1-2 m)-2 c (e f (1-m)-d g m)) \int \frac {\left (\frac {e x}{d}+1\right )^{m-\frac {3}{2}}}{(d (c d-b e)-c d e x)^{3/2}}dx}{2 c e (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {d (d+e x)^m \sqrt {d (c d-b e)-c d e x} \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (b e g (1-2 m)-2 c (e f (1-m)-d g m)) \int \frac {\left (\frac {c d}{2 c d-b e}+\frac {c e x}{2 c d-b e}\right )^{m-\frac {3}{2}}}{(d (c d-b e)-c d e x)^{3/2}}dx}{2 e (1-m) (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(d+e x)^m \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (b e g (1-2 m)-2 c (e f (1-m)-d g m)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {c d-b e-c e x}{2 c d-b e}\right )}{c e^2 (1-m) (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
(g*(d + e*x)^m)/(c*e^2*(1 - m)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]) - ((b*e*g*(1 - 2*m) - 2*c*(e*f*(1 - m) - d*g*m))*(d + e*x)^m*((c*(d + e*x) )/(2*c*d - b*e))^(1/2 - m)*Hypergeometric2F1[-1/2, 3/2 - m, 1/2, (c*d - b* e - c*e*x)/(2*c*d - b*e)])/(c*e^2*(2*c*d - b*e)*(1 - m)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])
3.23.89.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy mbol] :> Simp[d^m*((a + b*x + c*x^2)^FracPart[p]/((1 + e*(x/d))^FracPart[p] *(a/d + (c*x)/e)^FracPart[p])) Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^ p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[m] || GtQ[d, 0]) && !(IGtQ[m, 0] && (IntegerQ[3*p] || Integer Q[4*p]))
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy mbol] :> Simp[d^IntPart[m]*((d + e*x)^FracPart[m]/(1 + e*(x/d))^FracPart[m] ) Int[(1 + e*(x/d))^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !(IntegerQ[m] || GtQ[d, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) 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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
\[\int \frac {\left (e x +d \right )^{m} \left (g x +f \right )}{\left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(g*x + f)*(e*x + d)^m/ (c^2*e^4*x^4 + 2*b*c*e^4*x^3 + c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 - (2*c^ 2*d^2*e^2 - 2*b*c*d*e^3 - b^2*e^4)*x^2 - 2*(b*c*d^2*e^2 - b^2*d*e^3)*x), x )
\[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{m} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}} \,d x \]