\(\int \frac {(a+b x) \sqrt {c+d x} (A+B x+C x^2)}{\sqrt {e+f x}} \, dx\) [70]

Optimal result

Integrand size = 34, antiderivative size = 535 \[ \int \frac {(a+b x) \sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=-\frac {\left (8 a d f \left (2 d f (3 B d e+B c f-4 A d f)-C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )+b \left (C \left (35 d^3 e^3+15 c d^2 e^2 f+9 c^2 d e f^2+5 c^3 f^3\right )+8 d f \left (2 A d f (3 d e+c f)-B \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )\right )\right ) \sqrt {c+d x} \sqrt {e+f x}}{64 d^3 f^4}-\frac {\left (8 b d f (5 B d e+7 B c f-6 A d f)+8 a d f (5 C d e+7 c C f-6 B d f)-b C \left (35 d^2 e^2+50 c d e f+59 c^2 f^2\right )\right ) (c+d x)^{3/2} \sqrt {e+f x}}{96 d^3 f^3}+\frac {(8 a C d f-b (7 C d e+17 c C f-8 B d f)) (c+d x)^{5/2} \sqrt {e+f x}}{24 d^3 f^2}+\frac {b C (c+d x)^{7/2} \sqrt {e+f x}}{4 d^3 f}+\frac {(d e-c f) \left (8 a d f \left (2 d f (3 B d e+B c f-4 A d f)-C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )+b \left (C \left (35 d^3 e^3+15 c d^2 e^2 f+9 c^2 d e f^2+5 c^3 f^3\right )+8 d f \left (2 A d f (3 d e+c f)-B \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{64 d^{7/2} f^{9/2}} \] Output:

-1/64*(8*a*d*f*(2*d*f*(-4*A*d*f+B*c*f+3*B*d*e)-C*(c^2*f^2+2*c*d*e*f+5*d^2* 
e^2))+b*(C*(5*c^3*f^3+9*c^2*d*e*f^2+15*c*d^2*e^2*f+35*d^3*e^3)+8*d*f*(2*A* 
d*f*(c*f+3*d*e)-B*(c^2*f^2+2*c*d*e*f+5*d^2*e^2))))*(d*x+c)^(1/2)*(f*x+e)^( 
1/2)/d^3/f^4-1/96*(8*b*d*f*(-6*A*d*f+7*B*c*f+5*B*d*e)+8*a*d*f*(-6*B*d*f+7* 
C*c*f+5*C*d*e)-b*C*(59*c^2*f^2+50*c*d*e*f+35*d^2*e^2))*(d*x+c)^(3/2)*(f*x+ 
e)^(1/2)/d^3/f^3+1/24*(8*a*C*d*f-b*(-8*B*d*f+17*C*c*f+7*C*d*e))*(d*x+c)^(5 
/2)*(f*x+e)^(1/2)/d^3/f^2+1/4*b*C*(d*x+c)^(7/2)*(f*x+e)^(1/2)/d^3/f+1/64*( 
-c*f+d*e)*(8*a*d*f*(2*d*f*(-4*A*d*f+B*c*f+3*B*d*e)-C*(c^2*f^2+2*c*d*e*f+5* 
d^2*e^2))+b*(C*(5*c^3*f^3+9*c^2*d*e*f^2+15*c*d^2*e^2*f+35*d^3*e^3)+8*d*f*( 
2*A*d*f*(c*f+3*d*e)-B*(c^2*f^2+2*c*d*e*f+5*d^2*e^2))))*arctanh(f^(1/2)*(d* 
x+c)^(1/2)/d^(1/2)/(f*x+e)^(1/2))/d^(7/2)/f^(9/2)
 

Mathematica [A] (verified)

Time = 9.42 (sec) , antiderivative size = 474, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x) \sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\frac {\sqrt {d} \sqrt {f} \sqrt {c+d x} \sqrt {e+f x} \left (8 a d f \left (6 d f (4 A d f+B (-3 d e+c f+2 d f x))+C \left (-3 c^2 f^2+2 c d f (-2 e+f x)+d^2 \left (15 e^2-10 e f x+8 f^2 x^2\right )\right )\right )+b \left (C \left (15 c^3 f^3+c^2 d f^2 (17 e-10 f x)+c d^2 f \left (25 e^2-12 e f x+8 f^2 x^2\right )+d^3 \left (-105 e^3+70 e^2 f x-56 e f^2 x^2+48 f^3 x^3\right )\right )+8 d f \left (6 A d f (-3 d e+c f+2 d f x)+B \left (-3 c^2 f^2+2 c d f (-2 e+f x)+d^2 \left (15 e^2-10 e f x+8 f^2 x^2\right )\right )\right )\right )\right )-6 (d e-c f) \left (-8 a d f \left (2 d f (-3 B d e-B c f+4 A d f)+C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )+b \left (C \left (35 d^3 e^3+15 c d^2 e^2 f+9 c^2 d e f^2+5 c^3 f^3\right )+8 d f \left (2 A d f (3 d e+c f)-B \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \left (\sqrt {c-\frac {d e}{f}}-\sqrt {c+d x}\right )}\right )}{192 d^{7/2} f^{9/2}} \] Input:

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

Output:

(Sqrt[d]*Sqrt[f]*Sqrt[c + d*x]*Sqrt[e + f*x]*(8*a*d*f*(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))) + b*(C*(15*c^3*f^3 + c^2*d*f^2*(17*e - 10*f 
*x) + c*d^2*f*(25*e^2 - 12*e*f*x + 8*f^2*x^2) + d^3*(-105*e^3 + 70*e^2*f*x 
 - 56*e*f^2*x^2 + 48*f^3*x^3)) + 8*d*f*(6*A*d*f*(-3*d*e + c*f + 2*d*f*x) + 
 B*(-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 
))))) - 6*(d*e - c*f)*(-8*a*d*f*(2*d*f*(-3*B*d*e - B*c*f + 4*A*d*f) + C*(5 
*d^2*e^2 + 2*c*d*e*f + c^2*f^2)) + b*(C*(35*d^3*e^3 + 15*c*d^2*e^2*f + 9*c 
^2*d*e*f^2 + 5*c^3*f^3) + 8*d*f*(2*A*d*f*(3*d*e + c*f) - B*(5*d^2*e^2 + 2* 
c*d*e*f + c^2*f^2))))*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/(Sqrt[f]*(Sqrt[c - ( 
d*e)/f] - Sqrt[c + d*x]))])/(192*d^(7/2)*f^(9/2))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 419, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2118, 27, 164, 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.

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

\(\Big \downarrow \) 2118

\(\displaystyle \frac {\int -\frac {b (a+b x) \sqrt {c+d x} (4 b c C e+3 a C d e+a c C f-8 A b d f+(4 a C d f+b (7 C d e+5 c C f-8 B d f)) x)}{2 \sqrt {e+f x}}dx}{4 b^2 d f}+\frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}-\frac {\int \frac {(a+b x) \sqrt {c+d x} (4 b c C e+3 a C d e+a c C f-8 A b d f+(4 a C d f+b (7 C d e+5 c C f-8 B d f)) x)}{\sqrt {e+f x}}dx}{8 b d f}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}-\frac {\frac {b \left (8 a d f \left (2 d f (-4 A d f+B c f+3 B d e)-C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+b \left (8 d f \left (2 A d f (c f+3 d e)-B \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+C \left (5 c^3 f^3+9 c^2 d e f^2+15 c d^2 e^2 f+35 d^3 e^3\right )\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {e+f x}}dx}{8 d^2 f^2}+\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (24 a^2 C d^2 f^2+4 b d f x (4 a C d f+b (-8 B d f+5 c C f+7 C d e))+8 a b d f (-6 B d f+3 c C f+5 C d e)+b^2 \left (8 d f (-6 A d f+3 B c f+5 B d e)-C \left (15 c^2 f^2+22 c d e f+35 d^2 e^2\right )\right )\right )}{12 d^2 f^2}}{8 b d f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}-\frac {\frac {b \left (8 a d f \left (2 d f (-4 A d f+B c f+3 B d e)-C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+b \left (8 d f \left (2 A d f (c f+3 d e)-B \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+C \left (5 c^3 f^3+9 c^2 d e f^2+15 c d^2 e^2 f+35 d^3 e^3\right )\right )\right ) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{2 f}\right )}{8 d^2 f^2}+\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (24 a^2 C d^2 f^2+4 b d f x (4 a C d f+b (-8 B d f+5 c C f+7 C d e))+8 a b d f (-6 B d f+3 c C f+5 C d e)+b^2 \left (8 d f (-6 A d f+3 B c f+5 B d e)-C \left (15 c^2 f^2+22 c d e f+35 d^2 e^2\right )\right )\right )}{12 d^2 f^2}}{8 b d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}-\frac {\frac {b \left (8 a d f \left (2 d f (-4 A d f+B c f+3 B d e)-C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+b \left (8 d f \left (2 A d f (c f+3 d e)-B \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+C \left (5 c^3 f^3+9 c^2 d e f^2+15 c d^2 e^2 f+35 d^3 e^3\right )\right )\right ) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{f}\right )}{8 d^2 f^2}+\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (24 a^2 C d^2 f^2+4 b d f x (4 a C d f+b (-8 B d f+5 c C f+7 C d e))+8 a b d f (-6 B d f+3 c C f+5 C d e)+b^2 \left (8 d f (-6 A d f+3 B c f+5 B d e)-C \left (15 c^2 f^2+22 c d e f+35 d^2 e^2\right )\right )\right )}{12 d^2 f^2}}{8 b d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {C (a+b x)^2 (c+d x)^{3/2} \sqrt {e+f x}}{4 b d f}-\frac {\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (24 a^2 C d^2 f^2+4 b d f x (4 a C d f+b (-8 B d f+5 c C f+7 C d e))+8 a b d f (-6 B d f+3 c C f+5 C d e)+b^2 \left (8 d f (-6 A d f+3 B c f+5 B d e)-C \left (15 c^2 f^2+22 c d e f+35 d^2 e^2\right )\right )\right )}{12 d^2 f^2}+\frac {b \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{\sqrt {d} f^{3/2}}\right ) \left (8 a d f \left (2 d f (-4 A d f+B c f+3 B d e)-C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+b \left (8 d f \left (2 A d f (c f+3 d e)-B \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )+C \left (5 c^3 f^3+9 c^2 d e f^2+15 c d^2 e^2 f+35 d^3 e^3\right )\right )\right )}{8 d^2 f^2}}{8 b d f}\)

Input:

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

Output:

(C*(a + b*x)^2*(c + d*x)^(3/2)*Sqrt[e + f*x])/(4*b*d*f) - (((c + d*x)^(3/2 
)*Sqrt[e + f*x]*(24*a^2*C*d^2*f^2 + 8*a*b*d*f*(5*C*d*e + 3*c*C*f - 6*B*d*f 
) + b^2*(8*d*f*(5*B*d*e + 3*B*c*f - 6*A*d*f) - C*(35*d^2*e^2 + 22*c*d*e*f 
+ 15*c^2*f^2)) + 4*b*d*f*(4*a*C*d*f + b*(7*C*d*e + 5*c*C*f - 8*B*d*f))*x)) 
/(12*d^2*f^2) + (b*(8*a*d*f*(2*d*f*(3*B*d*e + B*c*f - 4*A*d*f) - C*(5*d^2* 
e^2 + 2*c*d*e*f + c^2*f^2)) + b*(C*(35*d^3*e^3 + 15*c*d^2*e^2*f + 9*c^2*d* 
e*f^2 + 5*c^3*f^3) + 8*d*f*(2*A*d*f*(3*d*e + c*f) - B*(5*d^2*e^2 + 2*c*d*e 
*f + c^2*f^2))))*((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))))/(8*d^ 
2*f^2))/(8*b*d*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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 2118

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f 
_.)*(x_))^(p_.), x_Symbol] :> With[{q = Expon[Px, x], k = Coeff[Px, x, Expo 
n[Px, x]]}, Simp[k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 
1)/(d*f*b^(q - 1)*(m + n + p + q + 1))), x] + Simp[1/(d*f*b^q*(m + n + p + 
q + 1))   Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + 
n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a + b*x)^(q 
- 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + 
 c*f*(p + 1))) + b*(a*d*f*(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m 
 + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] && PolyQ[Px, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2001\) vs. \(2(497)=994\).

Time = 0.55 (sec) , antiderivative size = 2002, normalized size of antiderivative = 3.74

method	result	size
default	\(\text {Expression too large to display}\)	\(2002\)

Input:

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

Output:

1/384*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(72*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))*a*c*d^3*e^2*f^2-60*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))*b*c*d^3*e 
^3*f-288*A*(d*f)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*b*d^3*e*f^2+192*A*((f*x+e)* 
(d*x+c))^(1/2)*(d*f)^(1/2)*b*d^3*f^3*x-12*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))*b*c^3*d*e*f^3+50*C*((f*x+e)* 
(d*x+c))^(1/2)*(d*f)^(1/2)*b*c*d^2*e^2*f+32*B*((f*x+e)*(d*x+c))^(1/2)*(d*f 
)^(1/2)*b*c*d^2*f^3*x-64*B*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*b*c*d^2*e*f 
^2-64*C*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*a*c*d^2*e*f^2-160*B*((f*x+e)*( 
d*x+c))^(1/2)*(d*f)^(1/2)*b*d^3*e*f^2*x-20*C*((f*x+e)*(d*x+c))^(1/2)*(d*f) 
^(1/2)*b*c^2*d*f^3*x+32*C*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*a*c*d^2*f^3* 
x+140*C*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*b*d^3*e^2*f*x-160*C*((f*x+e)*( 
d*x+c))^(1/2)*(d*f)^(1/2)*a*d^3*e*f^2*x+34*C*((f*x+e)*(d*x+c))^(1/2)*(d*f) 
^(1/2)*b*c^2*d*e*f^2+192*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))*a*c*d^3*f^4+96*C*b*d^3*f^3*x^3*((f*x+e)*(d*x+ 
c))^(1/2)*(d*f)^(1/2)+128*B*b*d^3*f^3*x^2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1 
/2)+128*C*a*d^3*f^3*x^2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)-24*C*((f*x+e)* 
(d*x+c))^(1/2)*(d*f)^(1/2)*b*c*d^2*e*f^2*x-18*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))*b*c^2*d^2*e^2*f^2+96*A*( 
(f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)*b*c*d^2*f^3-210*C*(d*f)^(1/2)*((f*x+...
 

Fricas [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 1114, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b x) \sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/768*(3*(35*C*b*d^4*e^4 - 20*(C*b*c*d^3 + 2*(C*a + B*b)*d^4)*e^3*f - 6*( 
C*b*c^2*d^2 - 4*(C*a + B*b)*c*d^3 - 8*(B*a + A*b)*d^4)*e^2*f^2 - 4*(C*b*c^ 
3*d + 16*A*a*d^4 - 2*(C*a + B*b)*c^2*d^2 + 8*(B*a + A*b)*c*d^3)*e*f^3 - (5 
*C*b*c^4 - 64*A*a*c*d^3 - 8*(C*a + B*b)*c^3*d + 16*(B*a + A*b)*c^2*d^2)*f^ 
4)*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*(48*C*b*d^4*f^4*x^3 - 105*C*b*d^4*e^3*f + 5*(5*C*b*c*d^3 + 24*(C 
*a + B*b)*d^4)*e^2*f^2 + (17*C*b*c^2*d^2 - 32*(C*a + B*b)*c*d^3 - 144*(B*a 
 + A*b)*d^4)*e*f^3 + 3*(5*C*b*c^3*d + 64*A*a*d^4 - 8*(C*a + B*b)*c^2*d^2 + 
 16*(B*a + A*b)*c*d^3)*f^4 - 8*(7*C*b*d^4*e*f^3 - (C*b*c*d^3 + 8*(C*a + B* 
b)*d^4)*f^4)*x^2 + 2*(35*C*b*d^4*e^2*f^2 - 2*(3*C*b*c*d^3 + 20*(C*a + B*b) 
*d^4)*e*f^3 - (5*C*b*c^2*d^2 - 8*(C*a + B*b)*c*d^3 - 48*(B*a + A*b)*d^4)*f 
^4)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^4*f^5), -1/384*(3*(35*C*b*d^4*e^4 - 
 20*(C*b*c*d^3 + 2*(C*a + B*b)*d^4)*e^3*f - 6*(C*b*c^2*d^2 - 4*(C*a + B*b) 
*c*d^3 - 8*(B*a + A*b)*d^4)*e^2*f^2 - 4*(C*b*c^3*d + 16*A*a*d^4 - 2*(C*a + 
 B*b)*c^2*d^2 + 8*(B*a + A*b)*c*d^3)*e*f^3 - (5*C*b*c^4 - 64*A*a*c*d^3 - 8 
*(C*a + B*b)*c^3*d + 16*(B*a + A*b)*c^2*d^2)*f^4)*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*(48*C*b*d^4*f^4*x^3 - 105*C*b*d^4*e^ 
3*f + 5*(5*C*b*c*d^3 + 24*(C*a + B*b)*d^4)*e^2*f^2 + (17*C*b*c^2*d^2 - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)*(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 = 0.19 (sec) , antiderivative size = 733, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b x) \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} {\left (2 \, {\left (d x + c\right )} {\left (4 \, {\left (d x + c\right )} {\left (\frac {6 \, {\left (d x + c\right )} C b}{d^{4} f} - \frac {7 \, C b d^{13} e f^{5} + 17 \, C b c d^{12} f^{6} - 8 \, C a d^{13} f^{6} - 8 \, B b d^{13} f^{6}}{d^{16} f^{7}}\right )} + \frac {35 \, C b d^{14} e^{2} f^{4} + 50 \, C b c d^{13} e f^{5} - 40 \, C a d^{14} e f^{5} - 40 \, B b d^{14} e f^{5} + 59 \, C b c^{2} d^{12} f^{6} - 56 \, C a c d^{13} f^{6} - 56 \, B b c d^{13} f^{6} + 48 \, B a d^{14} f^{6} + 48 \, A b d^{14} f^{6}}{d^{16} f^{7}}\right )} - \frac {3 \, {\left (35 \, C b d^{15} e^{3} f^{3} + 15 \, C b c d^{14} e^{2} f^{4} - 40 \, C a d^{15} e^{2} f^{4} - 40 \, B b d^{15} e^{2} f^{4} + 9 \, C b c^{2} d^{13} e f^{5} - 16 \, C a c d^{14} e f^{5} - 16 \, B b c d^{14} e f^{5} + 48 \, B a d^{15} e f^{5} + 48 \, A b d^{15} e f^{5} + 5 \, C b c^{3} d^{12} f^{6} - 8 \, C a c^{2} d^{13} f^{6} - 8 \, B b c^{2} d^{13} f^{6} + 16 \, B a c d^{14} f^{6} + 16 \, A b c d^{14} f^{6} - 64 \, A a d^{15} f^{6}\right )}}{d^{16} f^{7}}\right )} \sqrt {d x + c} - \frac {3 \, {\left (35 \, C b d^{4} e^{4} - 20 \, C b c d^{3} e^{3} f - 40 \, C a d^{4} e^{3} f - 40 \, B b d^{4} e^{3} f - 6 \, C b c^{2} d^{2} e^{2} f^{2} + 24 \, C a c d^{3} e^{2} f^{2} + 24 \, B b c d^{3} e^{2} f^{2} + 48 \, B a d^{4} e^{2} f^{2} + 48 \, A b d^{4} e^{2} f^{2} - 4 \, C b c^{3} d e f^{3} + 8 \, C a c^{2} d^{2} e f^{3} + 8 \, B b c^{2} d^{2} e f^{3} - 32 \, B a c d^{3} e f^{3} - 32 \, A b c d^{3} e f^{3} - 64 \, A a d^{4} e f^{3} - 5 \, C b c^{4} f^{4} + 8 \, C a c^{3} d f^{4} + 8 \, B b c^{3} d f^{4} - 16 \, B a c^{2} d^{2} f^{4} - 16 \, A b c^{2} d^{2} f^{4} + 64 \, A a c d^{3} f^{4}\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^{3} f^{4}}\right )} d}{192 \, {\left | d \right |}} \] Input:

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

Output:

1/192*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*(2*(d*x + c)*(4*(d*x + c)*(6*(d 
*x + c)*C*b/(d^4*f) - (7*C*b*d^13*e*f^5 + 17*C*b*c*d^12*f^6 - 8*C*a*d^13*f 
^6 - 8*B*b*d^13*f^6)/(d^16*f^7)) + (35*C*b*d^14*e^2*f^4 + 50*C*b*c*d^13*e* 
f^5 - 40*C*a*d^14*e*f^5 - 40*B*b*d^14*e*f^5 + 59*C*b*c^2*d^12*f^6 - 56*C*a 
*c*d^13*f^6 - 56*B*b*c*d^13*f^6 + 48*B*a*d^14*f^6 + 48*A*b*d^14*f^6)/(d^16 
*f^7)) - 3*(35*C*b*d^15*e^3*f^3 + 15*C*b*c*d^14*e^2*f^4 - 40*C*a*d^15*e^2* 
f^4 - 40*B*b*d^15*e^2*f^4 + 9*C*b*c^2*d^13*e*f^5 - 16*C*a*c*d^14*e*f^5 - 1 
6*B*b*c*d^14*e*f^5 + 48*B*a*d^15*e*f^5 + 48*A*b*d^15*e*f^5 + 5*C*b*c^3*d^1 
2*f^6 - 8*C*a*c^2*d^13*f^6 - 8*B*b*c^2*d^13*f^6 + 16*B*a*c*d^14*f^6 + 16*A 
*b*c*d^14*f^6 - 64*A*a*d^15*f^6)/(d^16*f^7))*sqrt(d*x + c) - 3*(35*C*b*d^4 
*e^4 - 20*C*b*c*d^3*e^3*f - 40*C*a*d^4*e^3*f - 40*B*b*d^4*e^3*f - 6*C*b*c^ 
2*d^2*e^2*f^2 + 24*C*a*c*d^3*e^2*f^2 + 24*B*b*c*d^3*e^2*f^2 + 48*B*a*d^4*e 
^2*f^2 + 48*A*b*d^4*e^2*f^2 - 4*C*b*c^3*d*e*f^3 + 8*C*a*c^2*d^2*e*f^3 + 8* 
B*b*c^2*d^2*e*f^3 - 32*B*a*c*d^3*e*f^3 - 32*A*b*c*d^3*e*f^3 - 64*A*a*d^4*e 
*f^3 - 5*C*b*c^4*f^4 + 8*C*a*c^3*d*f^4 + 8*B*b*c^3*d*f^4 - 16*B*a*c^2*d^2* 
f^4 - 16*A*b*c^2*d^2*f^4 + 64*A*a*c*d^3*f^4)*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^3*f^4))*d/abs(d)
 

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [B] (verification not implemented)

Time = 47.98 (sec) , antiderivative size = 1525, normalized size of antiderivative = 2.85 \[ \int \frac {(a+b x) \sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx =\text {Too large to display} \] Input:

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

Output:

(192*sqrt(e + f*x)*sqrt(c + d*x)*a**2*d**4*f**4 + 96*sqrt(e + f*x)*sqrt(c 
+ d*x)*a*b*c*d**3*f**4 - 288*sqrt(e + f*x)*sqrt(c + d*x)*a*b*d**4*e*f**3 + 
 192*sqrt(e + f*x)*sqrt(c + d*x)*a*b*d**4*f**4*x - 24*sqrt(e + f*x)*sqrt(c 
 + d*x)*a*c**3*d**2*f**4 - 32*sqrt(e + f*x)*sqrt(c + d*x)*a*c**2*d**3*e*f* 
*3 + 16*sqrt(e + f*x)*sqrt(c + d*x)*a*c**2*d**3*f**4*x + 120*sqrt(e + f*x) 
*sqrt(c + d*x)*a*c*d**4*e**2*f**2 - 80*sqrt(e + f*x)*sqrt(c + d*x)*a*c*d** 
4*e*f**3*x + 64*sqrt(e + f*x)*sqrt(c + d*x)*a*c*d**4*f**4*x**2 - 24*sqrt(e 
 + f*x)*sqrt(c + d*x)*b**2*c**2*d**2*f**4 - 32*sqrt(e + f*x)*sqrt(c + d*x) 
*b**2*c*d**3*e*f**3 + 16*sqrt(e + f*x)*sqrt(c + d*x)*b**2*c*d**3*f**4*x + 
120*sqrt(e + f*x)*sqrt(c + d*x)*b**2*d**4*e**2*f**2 - 80*sqrt(e + f*x)*sqr 
t(c + d*x)*b**2*d**4*e*f**3*x + 64*sqrt(e + f*x)*sqrt(c + d*x)*b**2*d**4*f 
**4*x**2 + 15*sqrt(e + f*x)*sqrt(c + d*x)*b*c**4*d*f**4 + 17*sqrt(e + f*x) 
*sqrt(c + d*x)*b*c**3*d**2*e*f**3 - 10*sqrt(e + f*x)*sqrt(c + d*x)*b*c**3* 
d**2*f**4*x + 25*sqrt(e + f*x)*sqrt(c + d*x)*b*c**2*d**3*e**2*f**2 - 12*sq 
rt(e + f*x)*sqrt(c + d*x)*b*c**2*d**3*e*f**3*x + 8*sqrt(e + f*x)*sqrt(c + 
d*x)*b*c**2*d**3*f**4*x**2 - 105*sqrt(e + f*x)*sqrt(c + d*x)*b*c*d**4*e**3 
*f + 70*sqrt(e + f*x)*sqrt(c + d*x)*b*c*d**4*e**2*f**2*x - 56*sqrt(e + f*x 
)*sqrt(c + d*x)*b*c*d**4*e*f**3*x**2 + 48*sqrt(e + f*x)*sqrt(c + d*x)*b*c* 
d**4*f**4*x**3 + 192*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)* 
sqrt(e + f*x))/sqrt(c*f - d*e))*a**2*c*d**3*f**4 - 192*sqrt(f)*sqrt(d)*...