\(\int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx\) [194]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {3}{4} \left (b^2-4 a c\right ) d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {3 \left (b^2-4 a c\right )^2 d^4 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}} \] Output:

3/4*(-4*a*c+b^2)*d^4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)+1/2*d^4*(2*c*x+b)^3*(c* 
x^2+b*x+a)^(1/2)+3/8*(-4*a*c+b^2)^2*d^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x 
^2+b*x+a)^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} d^4 \left ((b+2 c x) \sqrt {a+x (b+c x)} \left (5 b^2+8 b c x+4 c \left (-3 a+2 c x^2\right )\right )+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}\right ) \] Input:

Integrate[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]
 

Output:

(d^4*((b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(5*b^2 + 8*b*c*x + 4*c*(-3*a + 2*c 
*x^2)) + (3*(b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b 
+ c*x)])])/Sqrt[c]))/4
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1116, 27, 1116, 1092, 219}

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.

\(\displaystyle \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{4} d^4 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+(b+2 c x) \sqrt {a+b x+c x^2}\right )+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}\)

Input:

Int[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]
 

Output:

(d^4*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/2 + (3*(b^2 - 4*a*c)*d^4*((b + 2 
*c*x)*Sqrt[a + b*x + c*x^2] + ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c 
]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c])))/4
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {\left (-16 c^{3} x^{3}-24 b \,c^{2} x^{2}+24 a \,c^{2} x -18 b^{2} c x +12 a b c -5 b^{3}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{4}}{4}+\frac {\left (\frac {3}{8} b^{4}+6 a^{2} c^{2}-3 c a \,b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{4}}{\sqrt {c}}\)	\(114\)
default	\(d^{4} \left (\frac {b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+16 c^{4} \left (\frac {x^{3} \sqrt {c \,x^{2}+b x +a}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )}{8 c}-\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+32 b \,c^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+24 b^{2} c^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+8 b^{3} c \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\right )\)	\(722\)

Input:

int((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-16*c^3*x^3-24*b*c^2*x^2+24*a*c^2*x-18*b^2*c*x+12*a*b*c-5*b^3)*(c*x^ 
2+b*x+a)^(1/2)*d^4+(3/8*b^4+6*a^2*c^2-3*c*a*b^2)*ln((1/2*b+c*x)/c^(1/2)+(c 
*x^2+b*x+a)^(1/2))/c^(1/2)*d^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.68 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} d^{4} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{16 \, c}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} d^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{8 \, c}\right ] \] Input:

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*d^4*log(-8*c^2*x^2 - 8*b*c 
*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(16*c^ 
4*d^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(3*b^2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c - 
12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a))/c, -1/8*(3*(b^4 - 8*a*b^2*c + 16*a 
^2*c^2)*sqrt(-c)*d^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c) 
/(c^2*x^2 + b*c*x + a*c)) - 2*(16*c^4*d^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(3*b^ 
2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c - 12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a) 
)/c]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (110) = 220\).

Time = 0.71 (sec) , antiderivative size = 549, normalized size of antiderivative = 4.69 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \cdot \left (6 b c^{2} d^{4} x^{2} + 4 c^{3} d^{4} x^{3} + \frac {x \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{2 c} + \frac {- 12 a b c^{2} d^{4} + 8 b^{3} c d^{4} - \frac {3 b \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{4 c}}{c}\right ) + \left (- \frac {a \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{2 c} + b^{4} d^{4} - \frac {b \left (- 12 a b c^{2} d^{4} + 8 b^{3} c d^{4} - \frac {3 b \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \cdot \left (\frac {16 c^{4} d^{4} \left (a + b x\right )^{\frac {9}{2}}}{9 b^{4}} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (- 64 a c^{4} d^{4} + 32 b^{2} c^{3} d^{4}\right )}{7 b^{4}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \cdot \left (96 a^{2} c^{4} d^{4} - 96 a b^{2} c^{3} d^{4} + 24 b^{4} c^{2} d^{4}\right )}{5 b^{4}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 64 a^{3} c^{4} d^{4} + 96 a^{2} b^{2} c^{3} d^{4} - 48 a b^{4} c^{2} d^{4} + 8 b^{6} c d^{4}\right )}{3 b^{4}} + \frac {\sqrt {a + b x} \left (16 a^{4} c^{4} d^{4} - 32 a^{3} b^{2} c^{3} d^{4} + 24 a^{2} b^{4} c^{2} d^{4} - 8 a b^{6} c d^{4} + b^{8} d^{4}\right )}{b^{4}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {16 c^{4} d^{4} x^{5}}{5 \sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((2*c*d*x+b*d)**4/(c*x**2+b*x+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x + c*x**2)*(6*b*c**2*d**4*x**2 + 4*c**3*d**4*x**3 + 
 x*(-12*a*c**3*d**4 + 9*b**2*c**2*d**4)/(2*c) + (-12*a*b*c**2*d**4 + 8*b** 
3*c*d**4 - 3*b*(-12*a*c**3*d**4 + 9*b**2*c**2*d**4)/(4*c))/c) + (-a*(-12*a 
*c**3*d**4 + 9*b**2*c**2*d**4)/(2*c) + b**4*d**4 - b*(-12*a*b*c**2*d**4 + 
8*b**3*c*d**4 - 3*b*(-12*a*c**3*d**4 + 9*b**2*c**2*d**4)/(4*c))/(2*c))*Pie 
cewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - 
b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), 
 True)), Ne(c, 0)), (2*(16*c**4*d**4*(a + b*x)**(9/2)/(9*b**4) + (a + b*x) 
**(7/2)*(-64*a*c**4*d**4 + 32*b**2*c**3*d**4)/(7*b**4) + (a + b*x)**(5/2)* 
(96*a**2*c**4*d**4 - 96*a*b**2*c**3*d**4 + 24*b**4*c**2*d**4)/(5*b**4) + ( 
a + b*x)**(3/2)*(-64*a**3*c**4*d**4 + 96*a**2*b**2*c**3*d**4 - 48*a*b**4*c 
**2*d**4 + 8*b**6*c*d**4)/(3*b**4) + sqrt(a + b*x)*(16*a**4*c**4*d**4 - 32 
*a**3*b**2*c**3*d**4 + 24*a**2*b**4*c**2*d**4 - 8*a*b**6*c*d**4 + b**8*d** 
4)/b**4)/b, Ne(b, 0)), (16*c**4*d**4*x**5/(5*sqrt(a)), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(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(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c^{3} d^{4} x + 3 \, b c^{2} d^{4}\right )} x + \frac {3 \, {\left (3 \, b^{2} c^{4} d^{4} - 4 \, a c^{5} d^{4}\right )}}{c^{3}}\right )} x + \frac {5 \, b^{3} c^{3} d^{4} - 12 \, a b c^{4} d^{4}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, \sqrt {c}} \] Input:

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/4*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c^3*d^4*x + 3*b*c^2*d^4)*x + 3*(3*b^2*c 
^4*d^4 - 4*a*c^5*d^4)/c^3)*x + (5*b^3*c^3*d^4 - 12*a*b*c^4*d^4)/c^3) - 3/8 
*(b^4*d^4 - 8*a*b^2*c*d^4 + 16*a^2*c^2*d^4)*log(abs(2*(sqrt(c)*x - sqrt(c* 
x^2 + b*x + a))*sqrt(c) + b))/sqrt(c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^4}{\sqrt {c\,x^2+b\,x+a}} \,d x \] Input:

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2),x)
 

Output:

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.18 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {d^{4} \left (-24 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2}-48 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x +10 \sqrt {c \,x^{2}+b x +a}\, b^{3} c +36 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x +48 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{2}+32 \sqrt {c \,x^{2}+b x +a}\, c^{4} x^{3}+48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}-24 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c +3 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}\right )}{8 c} \] Input:

int((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x)
 

Output:

(d**4*( - 24*sqrt(a + b*x + c*x**2)*a*b*c**2 - 48*sqrt(a + b*x + c*x**2)*a 
*c**3*x + 10*sqrt(a + b*x + c*x**2)*b**3*c + 36*sqrt(a + b*x + c*x**2)*b** 
2*c**2*x + 48*sqrt(a + b*x + c*x**2)*b*c**3*x**2 + 32*sqrt(a + b*x + c*x** 
2)*c**4*x**3 + 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c* 
x)/sqrt(4*a*c - b**2))*a**2*c**2 - 24*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x 
+ c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c + 3*sqrt(c)*log((2*sqr 
t(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4))/(8*c)