Integrand size = 27, antiderivative size = 237 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\frac {2 \left (a+\frac {e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c (d+e x)^{1+m} \sqrt {e+f x}}{e f^2 (3+2 m)}+\frac {2 \left (c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )-e f (3+2 m) \left (a e f (1+2 m)+b \left (d f-2 e^2 (1+m)\right )\right )\right ) (d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \sqrt {e+f x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {e (e+f x)}{e^2-d f}\right )}{e f^3 \left (e^2-d f\right ) (3+2 m)} \]
2*(a+e*(-b*f+c*e)/f^2)*(e*x+d)^(1+m)/(-d*f+e^2)/(f*x+e)^(1/2)+2*c*(e*x+d)^ (1+m)*(f*x+e)^(1/2)/e/f^2/(3+2*m)+2*(c*(d^2*f^2+4*d*e^2*f*(1+m)-4*e^4*(m^2 +3*m+2))-e*f*(3+2*m)*(a*e*f*(1+2*m)+b*(d*f-2*e^2*(1+m))))*(e*x+d)^m*hyperg eom([1/2, -m],[3/2],e*(f*x+e)/(-d*f+e^2))*(f*x+e)^(1/2)/e/f^3/(-d*f+e^2)/( 3+2*m)/((-f*(e*x+d)/(-d*f+e^2))^m)
Time = 0.50 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\frac {2 (d+e x)^m \left (\frac {f (d+e x)}{-e^2+d f}\right )^{-m} \left (-3 \left (c e^2+f (-b e+a f)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},\frac {e (e+f x)}{e^2-d f}\right )-(e+f x) \left ((6 c e-3 b f) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {e (e+f x)}{e^2-d f}\right )-c (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},\frac {e (e+f x)}{e^2-d f}\right )\right )\right )}{3 f^3 \sqrt {e+f x}} \]
(2*(d + e*x)^m*(-3*(c*e^2 + f*(-(b*e) + a*f))*Hypergeometric2F1[-1/2, -m, 1/2, (e*(e + f*x))/(e^2 - d*f)] - (e + f*x)*((6*c*e - 3*b*f)*Hypergeometri c2F1[1/2, -m, 3/2, (e*(e + f*x))/(e^2 - d*f)] - c*(e + f*x)*Hypergeometric 2F1[3/2, -m, 5/2, (e*(e + f*x))/(e^2 - d*f)])))/(3*f^3*((f*(d + e*x))/(-e^ 2 + d*f))^m*Sqrt[e + f*x])
Time = 0.48 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1193, 27, 90, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right ) (d+e x)^m}{(e+f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1193 |
| \(\displaystyle \frac {2 \int \frac {(d+e x)^m \left (-c \left (d-\frac {e^2}{f}\right ) x f^2-\left (-2 b (m+1) e^2+a f (2 m+1) e+b d f\right ) f+c \left (d e f-2 e^3 (m+1)\right )\right )}{2 f^2 \sqrt {e+f x}}dx}{e^2-d f}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d+e x)^m \left (c \left (d e f-2 e^3 (m+1)\right )-f \left (-2 b (m+1) e^2+a f (2 m+1) e+b d f\right )+c f \left (e^2-d f\right ) x\right )}{\sqrt {e+f x}}dx}{f^2 \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\frac {2 c \left (e^2-d f\right ) \sqrt {e+f x} (d+e x)^{m+1}}{e (2 m+3)}-\left (f \left (a e f (2 m+1)+b d f-2 b e^2 (m+1)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )}{e (2 m+3)}\right ) \int \frac {(d+e x)^m}{\sqrt {e+f x}}dx}{f^2 \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {\frac {2 c \left (e^2-d f\right ) \sqrt {e+f x} (d+e x)^{m+1}}{e (2 m+3)}-(d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \left (f \left (a e f (2 m+1)+b d f-2 b e^2 (m+1)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )}{e (2 m+3)}\right ) \int \frac {\left (-\frac {e x f}{e^2-d f}-\frac {d f}{e^2-d f}\right )^m}{\sqrt {e+f x}}dx}{f^2 \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\frac {2 c \left (e^2-d f\right ) \sqrt {e+f x} (d+e x)^{m+1}}{e (2 m+3)}-\frac {2 \sqrt {e+f x} (d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {e (e+f x)}{e^2-d f}\right ) \left (f \left (a e f (2 m+1)+b d f-2 b e^2 (m+1)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )}{e (2 m+3)}\right )}{f}}{f^2 \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}\) |
(2*(a + (e*(c*e - b*f))/f^2)*(d + e*x)^(1 + m))/((e^2 - d*f)*Sqrt[e + f*x] ) + ((2*c*(e^2 - d*f)*(d + e*x)^(1 + m)*Sqrt[e + f*x])/(e*(3 + 2*m)) - (2* (f*(b*d*f - 2*b*e^2*(1 + m) + a*e*f*(1 + 2*m)) - (c*(d^2*f^2 + 4*d*e^2*f*( 1 + m) - 4*e^4*(2 + 3*m + m^2)))/(e*(3 + 2*m)))*(d + e*x)^m*Sqrt[e + f*x]* Hypergeometric2F1[1/2, -m, 3/2, (e*(e + f*x))/(e^2 - d*f)])/(f*(-((f*(d + e*x))/(e^2 - d*f)))^m))/(f^2*(e^2 - d*f))
3.10.44.3.1 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[((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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) ), x] + Simp[1/((m + 1)*(e*f - d*g)) Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a , b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] && !IntegerQ[n] && !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
\[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )}{\left (f x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,\left (c\,x^2+b\,x+a\right )}{{\left (e+f\,x\right )}^{3/2}} \,d x \]