\(\int (d+e x)^m (a+b x+c x^2)^{3/2} \, dx\) [768]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {(d+e x)^{1+m} \left (a+b x+c x^2\right )^{3/2} \operatorname {AppellF1}\left (1+m,-\frac {3}{2},-\frac {3}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2}} \] Output:

(e*x+d)^(1+m)*(c*x^2+b*x+a)^(3/2)*AppellF1(1+m,-3/2,-3/2,2+m,2*c*(e*x+d)/( 
2*c*d-(b-(-4*a*c+b^2)^(1/2))*e),2*c*(e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))* 
e))/e/(1+m)/(1-2*c*(e*x+d)/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e))^(3/2)/(1-2*c* 
(e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e))^(3/2)
 

Mathematica [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx \] Input:

Integrate[(d + e*x)^m*(a + b*x + c*x^2)^(3/2),x]
 

Output:

Integrate[(d + e*x)^m*(a + b*x + c*x^2)^(3/2), x]
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 189, 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.

\(\displaystyle \int \left (a+b x+c x^2\right )^{3/2} (d+e x)^m \, dx\)

\(\Big \downarrow \) 1179

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2} \int (d+e x)^m \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2}d(d+e x)}{e \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{2},-\frac {3}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2}}\)

Input:

Int[(d + e*x)^m*(a + b*x + c*x^2)^(3/2),x]
 

Output:

((d + e*x)^(1 + m)*(a + b*x + c*x^2)^(3/2)*AppellF1[1 + m, -3/2, -3/2, 2 + 
 m, (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)])/(e*(1 + m)*(1 - (2*c*(d + e*x))/(2*c* 
d - (b - Sqrt[b^2 - 4*a*c])*e))^(3/2)*(1 - (2*c*(d + e*x))/(2*c*d - (b + S 
qrt[b^2 - 4*a*c])*e))^(3/2))
 

Defintions of rubi rules used

rule 150

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])
 

rule 1179

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]
 

Maple [F]

\[\int \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}d x\]

Input:

int((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x)
 

Output:

int((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x)
 

Fricas [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{m} \,d x } \] Input:

integrate((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

integral((c*x^2 + b*x + a)^(3/2)*(e*x + d)^m, x)
 

Sympy [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int \left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((e*x+d)**m*(c*x**2+b*x+a)**(3/2),x)
 

Output:

Integral((d + e*x)**m*(a + b*x + c*x**2)**(3/2), x)
 

Maxima [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{m} \,d x } \] Input:

integrate((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)*(e*x + d)^m, x)
 

Giac [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{m} \,d x } \] Input:

integrate((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)*(e*x + d)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int {\left (d+e\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \] Input:

int((d + e*x)^m*(a + b*x + c*x^2)^(3/2),x)
 

Output:

int((d + e*x)^m*(a + b*x + c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}d x \] Input:

int((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x)
 

Output:

int((e*x+d)^m*(c*x^2+b*x+a)^(3/2),x)