3.10.50 \(\int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx\) [950]

3.10.50.1 Optimal result

Integrand size = 27, antiderivative size = 388 \[ \int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx=\frac {(e f-d g) (d+e x)^{1+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e^2 (1+m) \sqrt {a+b x+c x^2}}+\frac {g (d+e x)^{2+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e^2 (2+m) \sqrt {a+b x+c x^2}} \]

output

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

3.10.50.2 Mathematica [F]

\[ \int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx \]

input

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

output

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

3.10.50.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 1269

\(\displaystyle \frac {(e f-d g) \int \frac {(d+e x)^m}{\sqrt {c x^2+b x+a}}dx}{e}+\frac {g \int \frac {(d+e x)^{m+1}}{\sqrt {c x^2+b x+a}}dx}{e}\)

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

input

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

output

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

3.10.50.3.1 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]
 

3.10.50.4 Maple [F]

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

input

int((e*x+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x)
 

output

int((e*x+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x)
 

3.10.50.5 Fricas [F]

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

input

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

output

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

3.10.50.6 Sympy [F]

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

input

integrate((e*x+d)**m*(g*x+f)/(c*x**2+b*x+a)**(1/2),x)
 

output

Integral((d + e*x)**m*(f + g*x)/sqrt(a + b*x + c*x**2), x)
 

3.10.50.7 Maxima [F]

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

input

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

output

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

3.10.50.8 Giac [F]

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

input

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

output

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

3.10.50.9 Mupad [F(-1)]

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

input

int(((f + g*x)*(d + e*x)^m)/(a + b*x + c*x^2)^(1/2),x)
 

output

int(((f + g*x)*(d + e*x)^m)/(a + b*x + c*x^2)^(1/2), x)