Integrand size = 22, antiderivative size = 185 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{5/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{5 e} \]
2/5*(e*x+d)^(5/2)*(c*x^2+b*x+a)^p*AppellF1(5/2,-p,-p,7/2,2*c*(e*x+d)/(2*c* d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))/ e/((1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^p)/((1-2*c*(e*x+d)/(2* c*d-e*(b+(-4*a*c+b^2)^(1/2))))^p)
Time = 1.70 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.29 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\frac {2 \left (\frac {e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e-2 c x\right )}{2 c d+e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )^{-p} \left (\frac {e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e+2 c x\right )}{-2 c d+e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )^{-p} (d+e x)^{5/2} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )},\frac {2 c (d+e x)}{2 c d+e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )}{5 e} \]
(2*(d + e*x)^(5/2)*(a + x*(b + c*x))^p*AppellF1[5/2, -p, -p, 7/2, (2*c*(d + e*x))/(2*c*d - e*(b + Sqrt[(b^2 - 4*a*c)/e^2]*e)), (2*c*(d + e*x))/(2*c* d + e*(-b + Sqrt[(b^2 - 4*a*c)/e^2]*e))])/(5*e*((e*(-b + Sqrt[(b^2 - 4*a*c )/e^2]*e - 2*c*x))/(2*c*d + e*(-b + Sqrt[(b^2 - 4*a*c)/e^2]*e)))^p*((e*(b + Sqrt[(b^2 - 4*a*c)/e^2]*e + 2*c*x))/(-2*c*d + e*(b + Sqrt[(b^2 - 4*a*c)/ e^2]*e)))^p)
Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1179, 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.
\(\displaystyle \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {\left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \int (d+e x)^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^pd(d+e x)}{e}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 (d+e x)^{5/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{5 e}\) |
(2*(d + e*x)^(5/2)*(a + b*x + c*x^2)^p*AppellF1[5/2, -p, -p, 7/2, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(5*e*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^ p)
3.26.70.3.1 Defintions of rubi rules used
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])
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[(a + b*x + c*x^2)^p/(e*(1 - ( d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) ^p) Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m , p}, x]
\[\int \left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{p}d x\]
\[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
Timed out. \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
\[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
\[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
Timed out. \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\int {\left (d+e\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]