3.10.49 \(\int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx\) [949]

3.10.49.1 Optimal result

Integrand size = 29, antiderivative size = 502 \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) (d+e x)^{1+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (2+m) \sqrt {a+b x+c x^2}}-\frac {g (b e g (3+2 m)+c (2 d g-4 e f (2+m))) (d+e x)^{2+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m)^2 \sqrt {a+b x+c x^2}} \]

output

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

3.10.49.2 Mathematica [F]

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

input

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

output

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

3.10.49.3 Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1291, 27, 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.

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

\(\Big \downarrow \) 1291

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

input

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

output

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

3.10.49.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 1291

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b 
*x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m 
 + n + 2*p + 1))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m 
 + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d 
+ e*x)^(n - 2)*(b*d*e*(p + 1) + a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) 
 - e*(2*c*d - b*e)*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, p}, x] && IGtQ[n, 1] && NeQ[m + n + 2*p + 1, 0]
 

3.10.49.4 Maple [F]

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

input

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

output

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

3.10.49.5 Fricas [F]

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

input

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

output

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

3.10.49.6 Sympy [F]

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

input

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

output

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

3.10.49.7 Maxima [F]

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

input

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

output

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

3.10.49.8 Giac [F]

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

input

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

output

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

3.10.49.9 Mupad [F(-1)]

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

input

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

output

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