3.11.89 \(\int (e x)^m (A+B x) (a+b x+c x^2)^{3/2} \, dx\) [1089]

3.11.89.1 Optimal result

Integrand size = 25, antiderivative size = 281 \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {A (e x)^{1+m} \left (a+b x+c x^2\right )^{3/2} \operatorname {AppellF1}\left (1+m,-\frac {3}{2},-\frac {3}{2},2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (1+m) \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{3/2}}+\frac {B (e x)^{2+m} \left (a+b x+c x^2\right )^{3/2} \operatorname {AppellF1}\left (2+m,-\frac {3}{2},-\frac {3}{2},3+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (2+m) \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{3/2}} \]

output

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

3.11.89.2 Mathematica [A] (warning: unable to verify)

Time = 1.78 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.44 \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {x (e x)^m \sqrt {a+x (b+c x)} \left (a A \left (24+26 m+9 m^2+m^3\right ) \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x \left ((A b+a B) \left (12+7 m+m^2\right ) \operatorname {AppellF1}\left (2+m,-\frac {1}{2},-\frac {1}{2},3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+(2+m) x \left ((b B+A c) (4+m) \operatorname {AppellF1}\left (3+m,-\frac {1}{2},-\frac {1}{2},4+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+B c (3+m) x \operatorname {AppellF1}\left (4+m,-\frac {1}{2},-\frac {1}{2},5+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )\right )}{(1+m) (2+m) (3+m) (4+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}}} \]

input

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

output

(x*(e*x)^m*Sqrt[a + x*(b + c*x)]*(a*A*(24 + 26*m + 9*m^2 + m^3)*AppellF1[1 
 + m, -1/2, -1/2, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + S 
qrt[b^2 - 4*a*c])] + (1 + m)*x*((A*b + a*B)*(12 + 7*m + m^2)*AppellF1[2 + 
m, -1/2, -1/2, 3 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt 
[b^2 - 4*a*c])] + (2 + m)*x*((b*B + A*c)*(4 + m)*AppellF1[3 + m, -1/2, -1/ 
2, 4 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c 
])] + B*c*(3 + m)*x*AppellF1[4 + m, -1/2, -1/2, 5 + m, (-2*c*x)/(b + Sqrt[ 
b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])]))))/((1 + m)*(2 + m)*(3 + 
 m)*(4 + m)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c])]* 
Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c])])
 

3.11.89.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1269, 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.

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

\(\Big \downarrow \) 1269

\(\displaystyle A \int (e x)^m \left (c x^2+b x+a\right )^{3/2}dx+\frac {B \int (e x)^{m+1} \left (c x^2+b x+a\right )^{3/2}dx}{e}\)

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

input

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

output

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

3.11.89.3.1 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]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   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] &&  !IGtQ[m, 0]
 

3.11.89.4 Maple [F]

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

input

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

output

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

3.11.89.5 Fricas [F]

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

input

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

output

integral((B*c*x^3 + (B*b + A*c)*x^2 + A*a + (B*a + A*b)*x)*sqrt(c*x^2 + b* 
x + a)*(e*x)^m, x)
 

3.11.89.6 Sympy [F]

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

input

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

output

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

3.11.89.7 Maxima [F]

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

input

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

output

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

3.11.89.8 Giac [F]

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

input

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

output

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

3.11.89.9 Mupad [F(-1)]

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

input

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

output

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