\(\int \frac {\sqrt {c+d x} (A+B x+C x^2)}{\sqrt {e+f x}} \, dx\) [71]

Optimal result

Integrand size = 29, antiderivative size = 246 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{8 d^2 f^3}-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {(d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{8 d^{5/2} f^{7/2}} \] Output:

1/8*(C*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+2*d*f*(4*A*d*f-B*(c*f+3*d*e)))*(d*x+c 
)^(1/2)*(f*x+e)^(1/2)/d^2/f^3-1/12*(-6*B*d*f+7*C*c*f+5*C*d*e)*(d*x+c)^(3/2 
)*(f*x+e)^(1/2)/d^2/f^2+1/3*C*(d*x+c)^(5/2)*(f*x+e)^(1/2)/d^2/f-1/8*(-c*f+ 
d*e)*(C*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+2*d*f*(4*A*d*f-B*(c*f+3*d*e)))*arcta 
nh(f^(1/2)*(d*x+c)^(1/2)/d^(1/2)/(f*x+e)^(1/2))/d^(5/2)/f^(7/2)
 

Mathematica [A] (verified)

Time = 3.55 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\frac {\sqrt {c+d x} \sqrt {e+f x} \left (6 d f (4 A d f+B (-3 d e+c f+2 d f x))+C \left (-3 c^2 f^2+2 c d f (-2 e+f x)+d^2 \left (15 e^2-10 e f x+8 f^2 x^2\right )\right )\right )}{24 d^2 f^3}+\frac {(d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \left (\sqrt {c-\frac {d e}{f}}-\sqrt {c+d x}\right )}\right )}{4 d^{5/2} f^{7/2}} \] Input:

Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]
 

Output:

(Sqrt[c + d*x]*Sqrt[e + f*x]*(6*d*f*(4*A*d*f + B*(-3*d*e + c*f + 2*d*f*x)) 
 + C*(-3*c^2*f^2 + 2*c*d*f*(-2*e + f*x) + d^2*(15*e^2 - 10*e*f*x + 8*f^2*x 
^2))))/(24*d^2*f^3) + ((d*e - c*f)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 
2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/(Sqrt[f 
]*(Sqrt[c - (d*e)/f] - Sqrt[c + d*x]))])/(4*d^(5/2)*f^(7/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1194, 27, 90, 60, 66, 221}

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.

\(\displaystyle \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\int -\frac {\sqrt {c+d x} \left (C f c^2+5 C d e c-6 A d^2 f+d (5 C d e+7 c C f-6 B d f) x\right )}{2 \sqrt {e+f x}}dx}{3 d^2 f}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\int \frac {\sqrt {c+d x} \left (C f c^2+5 C d e c-6 A d^2 f+d (5 C d e+7 c C f-6 B d f) x\right )}{\sqrt {e+f x}}dx}{6 d^2 f}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{2 f}-\frac {3 \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {e+f x}}dx}{4 f}}{6 d^2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{2 f}-\frac {3 \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right ) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{2 f}\right )}{4 f}}{6 d^2 f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{2 f}-\frac {3 \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right ) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{f}\right )}{4 f}}{6 d^2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{2 f}-\frac {3 \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{\sqrt {d} f^{3/2}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{4 f}}{6 d^2 f}\)

Input:

Int[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]
 

Output:

(C*(c + d*x)^(5/2)*Sqrt[e + f*x])/(3*d^2*f) - (((5*C*d*e + 7*c*C*f - 6*B*d 
*f)*(c + d*x)^(3/2)*Sqrt[e + f*x])/(2*f) - (3*(C*(5*d^2*e^2 + 2*c*d*e*f + 
c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*((Sqrt[c + d*x]*Sqrt[e + f*x 
])/f - ((d*e - c*f)*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x] 
)])/(Sqrt[d]*f^(3/2))))/(4*f))/(6*d^2*f)
 

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1194

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(762\) vs. \(2(214)=428\).

Time = 0.53 (sec) , antiderivative size = 763, normalized size of antiderivative = 3.10

method	result	size
default	\(\frac {\sqrt {x d +c}\, \sqrt {f x +e}\, \left (16 C \,d^{2} f^{2} x^{2} \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+24 A \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c \,d^{2} f^{3}-24 A \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{3} e \,f^{2}-6 B \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c^{2} d \,f^{3}-12 B \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c \,d^{2} e \,f^{2}+18 B \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{3} e^{2} f +24 B \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, d^{2} f^{2} x +3 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c^{3} f^{3}+3 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c^{2} d e \,f^{2}+9 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c \,d^{2} e^{2} f -15 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{3} e^{3}+4 C \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, c d \,f^{2} x -20 C \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, d^{2} e f x +48 A \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d^{2} f^{2}+12 B \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, c d \,f^{2}-36 B \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d^{2} e f -6 C \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, c^{2} f^{2}-8 C \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, c d e f +30 C \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d^{2} e^{2}\right )}{48 f^{3} \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d^{2} \sqrt {d f}}\)	\(763\)

Input:

int((d*x+c)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(16*C*d^2*f^2*x^2*((f*x+e)*(d*x+c))^(1/2) 
*(d*f)^(1/2)+24*A*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c* 
f+d*e)/(d*f)^(1/2))*c*d^2*f^3-24*A*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/ 
2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^3*e*f^2-6*B*ln(1/2*(2*d*f*x+2*((f*x 
+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^2*d*f^3-12*B*ln(1/2 
*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c*d^ 
2*e*f^2+18*B*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e 
)/(d*f)^(1/2))*d^3*e^2*f+24*B*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*d^2*f^2* 
x+3*C*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f) 
^(1/2))*c^3*f^3+3*C*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+ 
c*f+d*e)/(d*f)^(1/2))*c^2*d*e*f^2+9*C*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^ 
(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c*d^2*e^2*f-15*C*ln(1/2*(2*d*f*x+2 
*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^3*e^3+4*C*((f 
*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*c*d*f^2*x-20*C*((f*x+e)*(d*x+c))^(1/2)*(d 
*f)^(1/2)*d^2*e*f*x+48*A*(d*f)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*d^2*f^2+12*B* 
((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*c*d*f^2-36*B*(d*f)^(1/2)*((f*x+e)*(d*x 
+c))^(1/2)*d^2*e*f-6*C*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*c^2*f^2-8*C*((f 
*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*c*d*e*f+30*C*(d*f)^(1/2)*((f*x+e)*(d*x+c) 
)^(1/2)*d^2*e^2)/f^3/((f*x+e)*(d*x+c))^(1/2)/d^2/(d*f)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 576, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\left [-\frac {3 \, {\left (5 \, C d^{3} e^{3} - 3 \, {\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f - {\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} - {\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt {d f} \log \left (8 \, d^{2} f^{2} x^{2} + d^{2} e^{2} + 6 \, c d e f + c^{2} f^{2} + 4 \, {\left (2 \, d f x + d e + c f\right )} \sqrt {d f} \sqrt {d x + c} \sqrt {f x + e} + 8 \, {\left (d^{2} e f + c d f^{2}\right )} x\right ) - 4 \, {\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \, {\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \, {\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \, {\left (5 \, C d^{3} e f^{2} - {\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{96 \, d^{3} f^{4}}, \frac {3 \, {\left (5 \, C d^{3} e^{3} - 3 \, {\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f - {\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} - {\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt {-d f} \arctan \left (\frac {{\left (2 \, d f x + d e + c f\right )} \sqrt {-d f} \sqrt {d x + c} \sqrt {f x + e}}{2 \, {\left (d^{2} f^{2} x^{2} + c d e f + {\left (d^{2} e f + c d f^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \, {\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \, {\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \, {\left (5 \, C d^{3} e f^{2} - {\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{48 \, d^{3} f^{4}}\right ] \] Input:

integrate((d*x+c)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 2*B*d^3)*e^2*f - (C*c^2*d - 4*B*c*d^ 
2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^2*d + 8*A*c*d^2)*f^3)*sqrt(d*f)*log(8* 
d^2*f^2*x^2 + d^2*e^2 + 6*c*d*e*f + c^2*f^2 + 4*(2*d*f*x + d*e + c*f)*sqrt 
(d*f)*sqrt(d*x + c)*sqrt(f*x + e) + 8*(d^2*e*f + c*d*f^2)*x) - 4*(8*C*d^3* 
f^3*x^2 + 15*C*d^3*e^2*f - 2*(2*C*c*d^2 + 9*B*d^3)*e*f^2 - 3*(C*c^2*d - 2* 
B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*d^2 + 6*B*d^3)*f^3)*x)*sq 
rt(d*x + c)*sqrt(f*x + e))/(d^3*f^4), 1/48*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 
2*B*d^3)*e^2*f - (C*c^2*d - 4*B*c*d^2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^2* 
d + 8*A*c*d^2)*f^3)*sqrt(-d*f)*arctan(1/2*(2*d*f*x + d*e + c*f)*sqrt(-d*f) 
*sqrt(d*x + c)*sqrt(f*x + e)/(d^2*f^2*x^2 + c*d*e*f + (d^2*e*f + c*d*f^2)* 
x)) + 2*(8*C*d^3*f^3*x^2 + 15*C*d^3*e^2*f - 2*(2*C*c*d^2 + 9*B*d^3)*e*f^2 
- 3*(C*c^2*d - 2*B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*d^2 + 6* 
B*d^3)*f^3)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^4)]
 

Sympy [F]

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

integrate((d*x+c)**(1/2)*(C*x**2+B*x+A)/(f*x+e)**(1/2),x)
 

Output:

Integral(sqrt(c + d*x)*(A + B*x + C*x**2)/sqrt(e + f*x), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\frac {{\left (\sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \sqrt {d x + c} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} C}{d^{3} f} - \frac {5 \, C d^{7} e f^{3} + 7 \, C c d^{6} f^{4} - 6 \, B d^{7} f^{4}}{d^{9} f^{5}}\right )} + \frac {3 \, {\left (5 \, C d^{8} e^{2} f^{2} + 2 \, C c d^{7} e f^{3} - 6 \, B d^{8} e f^{3} + C c^{2} d^{6} f^{4} - 2 \, B c d^{7} f^{4} + 8 \, A d^{8} f^{4}\right )}}{d^{9} f^{5}}\right )} + \frac {3 \, {\left (5 \, C d^{3} e^{3} - 3 \, C c d^{2} e^{2} f - 6 \, B d^{3} e^{2} f - C c^{2} d e f^{2} + 4 \, B c d^{2} e f^{2} + 8 \, A d^{3} e f^{2} - C c^{3} f^{3} + 2 \, B c^{2} d f^{3} - 8 \, A c d^{2} f^{3}\right )} \log \left ({\left | -\sqrt {d f} \sqrt {d x + c} + \sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \right |}\right )}{\sqrt {d f} d^{2} f^{3}}\right )} d}{24 \, {\left | d \right |}} \] Input:

integrate((d*x+c)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^(1/2),x, algorithm="giac")
 

Output:

1/24*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d 
*x + c)*C/(d^3*f) - (5*C*d^7*e*f^3 + 7*C*c*d^6*f^4 - 6*B*d^7*f^4)/(d^9*f^5 
)) + 3*(5*C*d^8*e^2*f^2 + 2*C*c*d^7*e*f^3 - 6*B*d^8*e*f^3 + C*c^2*d^6*f^4 
- 2*B*c*d^7*f^4 + 8*A*d^8*f^4)/(d^9*f^5)) + 3*(5*C*d^3*e^3 - 3*C*c*d^2*e^2 
*f - 6*B*d^3*e^2*f - C*c^2*d*e*f^2 + 4*B*c*d^2*e*f^2 + 8*A*d^3*e*f^2 - C*c 
^3*f^3 + 2*B*c^2*d*f^3 - 8*A*c*d^2*f^3)*log(abs(-sqrt(d*f)*sqrt(d*x + c) + 
 sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/(sqrt(d*f)*d^2*f^3))*d/abs(d)
 

Mupad [B] (verification not implemented)

Time = 61.14 (sec) , antiderivative size = 1832, normalized size of antiderivative = 7.45 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\text {Too large to display} \] Input:

int(((c + d*x)^(1/2)*(A + B*x + C*x^2))/(e + f*x)^(1/2),x)
 

Output:

((((c + d*x)^(1/2) - c^(1/2))*(2*A*d^2*e + 2*A*c*d*f))/(f^3*((e + f*x)^(1/ 
2) - e^(1/2))) + ((2*A*c*f + 2*A*d*e)*((c + d*x)^(1/2) - c^(1/2))^3)/(f^2* 
((e + f*x)^(1/2) - e^(1/2))^3) - (8*A*c^(1/2)*d*e^(1/2)*((c + d*x)^(1/2) - 
 c^(1/2))^2)/(f^2*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d*x)^(1/2) - c^(1 
/2))^4/((e + f*x)^(1/2) - e^(1/2))^4 + d^2/f^2 - (2*d*((c + d*x)^(1/2) - c 
^(1/2))^2)/(f*((e + f*x)^(1/2) - e^(1/2))^2)) - ((((c + d*x)^(1/2) - c^(1/ 
2))*((C*c^3*d^3*f^3)/4 - (5*C*d^6*e^3)/4 + (C*c^2*d^4*e*f^2)/4 + (3*C*c*d^ 
5*e^2*f)/4))/(f^9*((e + f*x)^(1/2) - e^(1/2))) - (((c + d*x)^(1/2) - c^(1/ 
2))^5*((33*C*d^4*e^3)/2 + (19*C*c^3*d*f^3)/2 + (275*C*c^2*d^2*e*f^2)/2 + ( 
313*C*c*d^3*e^2*f)/2))/(f^7*((e + f*x)^(1/2) - e^(1/2))^5) - (((c + d*x)^( 
1/2) - c^(1/2))^7*((19*C*c^3*f^3)/2 + (33*C*d^3*e^3)/2 + (313*C*c*d^2*e^2* 
f)/2 + (275*C*c^2*d*e*f^2)/2))/(f^6*((e + f*x)^(1/2) - e^(1/2))^7) - (((c 
+ d*x)^(1/2) - c^(1/2))^3*((17*C*c^3*d^2*f^3)/12 - (85*C*d^5*e^3)/12 + (91 
*C*c^2*d^3*e*f^2)/4 + (17*C*c*d^4*e^2*f)/4))/(f^8*((e + f*x)^(1/2) - e^(1/ 
2))^3) + (((c + d*x)^(1/2) - c^(1/2))^11*((C*c^3*f^3)/4 - (5*C*d^3*e^3)/4 
+ (3*C*c*d^2*e^2*f)/4 + (C*c^2*d*e*f^2)/4))/(d^2*f^4*((e + f*x)^(1/2) - e^ 
(1/2))^11) - (((c + d*x)^(1/2) - c^(1/2))^9*((17*C*c^3*f^3)/12 - (85*C*d^3 
*e^3)/12 + (17*C*c*d^2*e^2*f)/4 + (91*C*c^2*d*e*f^2)/4))/(d*f^5*((e + f*x) 
^(1/2) - e^(1/2))^9) + (c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^8*(32* 
C*c^2*f + 96*C*c*d*e))/(f^4*((e + f*x)^(1/2) - e^(1/2))^8) + (c^(1/2)*e...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 648, normalized size of antiderivative = 2.63 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\frac {24 \sqrt {f x +e}\, \sqrt {d x +c}\, a \,d^{3} f^{3}+6 \sqrt {f x +e}\, \sqrt {d x +c}\, b c \,d^{2} f^{3}-18 \sqrt {f x +e}\, \sqrt {d x +c}\, b \,d^{3} e \,f^{2}+12 \sqrt {f x +e}\, \sqrt {d x +c}\, b \,d^{3} f^{3} x -3 \sqrt {f x +e}\, \sqrt {d x +c}\, c^{3} d \,f^{3}-4 \sqrt {f x +e}\, \sqrt {d x +c}\, c^{2} d^{2} e \,f^{2}+2 \sqrt {f x +e}\, \sqrt {d x +c}\, c^{2} d^{2} f^{3} x +15 \sqrt {f x +e}\, \sqrt {d x +c}\, c \,d^{3} e^{2} f -10 \sqrt {f x +e}\, \sqrt {d x +c}\, c \,d^{3} e \,f^{2} x +8 \sqrt {f x +e}\, \sqrt {d x +c}\, c \,d^{3} f^{3} x^{2}+24 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) a c \,d^{2} f^{3}-24 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) a \,d^{3} e \,f^{2}-6 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) b \,c^{2} d \,f^{3}-12 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) b c \,d^{2} e \,f^{2}+18 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) b \,d^{3} e^{2} f +3 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c^{4} f^{3}+3 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c^{3} d e \,f^{2}+9 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c^{2} d^{2} e^{2} f -15 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c \,d^{3} e^{3}}{24 d^{3} f^{4}} \] Input:

int((d*x+c)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^(1/2),x)
 

Output:

(24*sqrt(e + f*x)*sqrt(c + d*x)*a*d**3*f**3 + 6*sqrt(e + f*x)*sqrt(c + d*x 
)*b*c*d**2*f**3 - 18*sqrt(e + f*x)*sqrt(c + d*x)*b*d**3*e*f**2 + 12*sqrt(e 
 + f*x)*sqrt(c + d*x)*b*d**3*f**3*x - 3*sqrt(e + f*x)*sqrt(c + d*x)*c**3*d 
*f**3 - 4*sqrt(e + f*x)*sqrt(c + d*x)*c**2*d**2*e*f**2 + 2*sqrt(e + f*x)*s 
qrt(c + d*x)*c**2*d**2*f**3*x + 15*sqrt(e + f*x)*sqrt(c + d*x)*c*d**3*e**2 
*f - 10*sqrt(e + f*x)*sqrt(c + d*x)*c*d**3*e*f**2*x + 8*sqrt(e + f*x)*sqrt 
(c + d*x)*c*d**3*f**3*x**2 + 24*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) 
 + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*a*c*d**2*f**3 - 24*sqrt(f)*sqrt 
(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*a 
*d**3*e*f**2 - 6*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt 
(e + f*x))/sqrt(c*f - d*e))*b*c**2*d*f**3 - 12*sqrt(f)*sqrt(d)*log((sqrt(f 
)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*b*c*d**2*e*f**2 
+ 18*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/s 
qrt(c*f - d*e))*b*d**3*e**2*f + 3*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d* 
x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*c**4*f**3 + 3*sqrt(f)*sqrt(d) 
*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*c**3 
*d*e*f**2 + 9*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e 
+ f*x))/sqrt(c*f - d*e))*c**2*d**2*e**2*f - 15*sqrt(f)*sqrt(d)*log((sqrt(f 
)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*c*d**3*e**3)/(24 
*d**3*f**4)