\(\int \frac {(e x)^m (A+B x)}{\sqrt {a+b x+c x^2}} \, dx\) [236]

Optimal result

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

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

Mathematica [A] (warning: unable to verify)

Time = 1.72 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=\frac {x (e x)^m \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}} \left (A (2+m) \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+B (1+m) x \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )\right )}{(1+m) (2+m) \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*S 
qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c])]*(A*(2 + m)*Ap 
pellF1[1 + m, 1/2, 1/2, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/( 
-b + Sqrt[b^2 - 4*a*c])] + B*(1 + m)*x*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])]))/((1 + 
m)*(2 + m)*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.45 (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.

\(\displaystyle \int \frac {(A+B x) (e x)^m}{\sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

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]
 

Maple [F]

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

Input:

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)
 

Output:

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)
 

Fricas [F]

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

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((e*x)**m*(B*x+A)/(c*x**2+b*x+a)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(1/2),x)
 

Output:

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(1/2), x)
 

Reduce [F]

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

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)
 

Output:

e**m*(int(x**m/sqrt(a + b*x + c*x**2),x)*a + int((x**m*x)/sqrt(a + b*x + c 
*x**2),x)*b)