\(\int \frac {(a+b x+c x^2)^{5/2}}{b d+2 c d x} \, dx\) [182]

Optimal result

Integrand size = 26, antiderivative size = 149 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d} \] Output:

1/32*(-4*a*c+b^2)^2*(c*x^2+b*x+a)^(1/2)/c^3/d-1/24*(-4*a*c+b^2)*(c*x^2+b*x 
+a)^(3/2)/c^2/d+1/10*(c*x^2+b*x+a)^(5/2)/c/d-1/64*(-4*a*c+b^2)^(5/2)*arcta 
n(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))/c^(7/2)/d
 

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^4-20 b^3 c x+28 b^2 c \left (-5 a+c x^2\right )+16 b c^2 x \left (11 a+6 c x^2\right )+16 c^2 \left (23 a^2+11 a c x^2+3 c^2 x^4\right )\right )-15 \left (-b^2+4 a c\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{480 c^{7/2} d} \] Input:

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x),x]
 

Output:

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^4 - 20*b^3*c*x + 28*b^2*c*(-5*a + c*x 
^2) + 16*b*c^2*x*(11*a + 6*c*x^2) + 16*c^2*(23*a^2 + 11*a*c*x^2 + 3*c^2*x^ 
4)) - 15*(-b^2 + 4*a*c)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[-b^2 + 4*a*c]*x)/(Sqrt 
[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/(480*c^(7/2)*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1109, 27, 1109, 1109, 1112, 218}

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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx\)

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\left (c x^2+b x+a\right )^{3/2}}{d (b+2 c x)}dx}{4 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\left (c x^2+b x+a\right )^{3/2}}{b+2 c x}dx}{4 c d}\)

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \int \frac {\sqrt {c x^2+b x+a}}{b+2 c x}dx}{4 c}\right )}{4 c d}\)

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{(b+2 c x) \sqrt {c x^2+b x+a}}dx}{4 c}\right )}{4 c}\right )}{4 c d}\)

\(\Big \downarrow \) 1112

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\left (b^2-4 a c\right ) \int \frac {1}{8 \left (c x^2+b x+a\right ) c^2+2 \left (b^2-4 a c\right ) c}d\sqrt {c x^2+b x+a}\right )}{4 c}\right )}{4 c d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\frac {\sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{4 c^{3/2}}\right )}{4 c}\right )}{4 c d}\)

Input:

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x),x]
 

Output:

(a + b*x + c*x^2)^(5/2)/(10*c*d) - ((b^2 - 4*a*c)*((a + b*x + c*x^2)^(3/2) 
/(6*c) - ((b^2 - 4*a*c)*(Sqrt[a + b*x + c*x^2]/(2*c) - (Sqrt[b^2 - 4*a*c]* 
ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(4*c^(3/2)))) 
/(4*c)))/(4*c*d)
 

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 218

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

rule 1109

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

rule 1112

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb 
ol] :> Simp[4*c   Subst[Int[1/(b^2*e - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a 
+ b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
 

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(-\frac {\left (a c -\frac {b^{2}}{4}\right )^{3} \operatorname {arctanh}\left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, c}{\sqrt {4 a \,c^{2}-b^{2} c}}\right )-\frac {23 \sqrt {c \,x^{2}+b x +a}\, \sqrt {4 a \,c^{2}-b^{2} c}\, \left (\frac {3 c^{4} x^{4}}{23}+\frac {11 x^{2} \left (\frac {6 b x}{11}+a \right ) c^{3}}{23}+\left (\frac {7}{92} b^{2} x^{2}+\frac {11}{23} a b x +a^{2}\right ) c^{2}-\frac {35 b^{2} \left (\frac {b x}{7}+a \right ) c}{92}+\frac {15 b^{4}}{368}\right )}{30}}{\sqrt {4 a \,c^{2}-b^{2} c}\, c^{3} d}\)	\(158\)
risch	\(\frac {\left (48 c^{4} x^{4}+96 b \,c^{3} x^{3}+176 a \,c^{3} x^{2}+28 b^{2} c^{2} x^{2}+176 a b \,c^{2} x -20 b^{3} c x +368 a^{2} c^{2}-140 c a \,b^{2}+15 b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}{480 c^{3} d}-\frac {\left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{64 c^{4} \sqrt {\frac {4 a c -b^{2}}{c}}\, d}\)	\(227\)
default	\(\frac {\frac {\left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{5}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 c}}{2 d c}\)	\(245\)

Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d),x,method=_RETURNVERBOSE)
 

Output:

-1/(4*a*c^2-b^2*c)^(1/2)*((a*c-1/4*b^2)^3*arctanh(2*(c*x^2+b*x+a)^(1/2)*c/ 
(4*a*c^2-b^2*c)^(1/2))-23/30*(c*x^2+b*x+a)^(1/2)*(4*a*c^2-b^2*c)^(1/2)*(3/ 
23*c^4*x^4+11/23*x^2*(6/11*b*x+a)*c^3+(7/92*b^2*x^2+11/23*a*b*x+a^2)*c^2-3 
5/92*b^2*(1/7*b*x+a)*c+15/368*b^4))/c^3/d
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.57 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\left [\frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1920 \, c^{3} d}, \frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (-\frac {2 \, \sqrt {c x^{2} + b x + a} c \sqrt {\frac {b^{2} - 4 \, a c}{c}}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{960 \, c^{3} d}\right ] \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d),x, algorithm="fricas")
 

Output:

[1/1920*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4* 
c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*c*sqrt(-(b^2 - 4 
*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(48*c^4*x^4 + 96*b*c^3*x^3 + 15 
*b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2 + 44*a*c^3)*x^2 - 4*(5*b^3 
*c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d), 1/960*(15*(b^4 - 8*a*b 
^2*c + 16*a^2*c^2)*sqrt((b^2 - 4*a*c)/c)*arctan(-2*sqrt(c*x^2 + b*x + a)*c 
*sqrt((b^2 - 4*a*c)/c)/(b^2 - 4*a*c)) + 2*(48*c^4*x^4 + 96*b*c^3*x^3 + 15* 
b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2 + 44*a*c^3)*x^2 - 4*(5*b^3* 
c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d)]
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \] Input:

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d),x)
 

Output:

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(b**2*x**2 
*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(c**2*x**4*sqrt(a + b*x 
+ c*x**2)/(b + 2*c*x), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b + 2 
*c*x), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + I 
ntegral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x))/d
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d),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 = 0.41 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {1}{480} \, \sqrt {c x^{2} + b x + a} {\left (4 \, {\left ({\left (12 \, {\left (\frac {c x}{d} + \frac {2 \, b}{d}\right )} x + \frac {7 \, b^{2} c^{9} d^{5} + 44 \, a c^{10} d^{5}}{c^{10} d^{6}}\right )} x - \frac {5 \, b^{3} c^{8} d^{5} - 44 \, a b c^{9} d^{5}}{c^{10} d^{6}}\right )} x + \frac {15 \, b^{4} c^{7} d^{5} - 140 \, a b^{2} c^{8} d^{5} + 368 \, a^{2} c^{9} d^{5}}{c^{10} d^{6}}\right )} - \frac {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{32 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d} \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d),x, algorithm="giac")
 

Output:

1/480*sqrt(c*x^2 + b*x + a)*(4*((12*(c*x/d + 2*b/d)*x + (7*b^2*c^9*d^5 + 4 
4*a*c^10*d^5)/(c^10*d^6))*x - (5*b^3*c^8*d^5 - 44*a*b*c^9*d^5)/(c^10*d^6)) 
*x + (15*b^4*c^7*d^5 - 140*a*b^2*c^8*d^5 + 368*a^2*c^9*d^5)/(c^10*d^6)) - 
1/32*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*arctan(-(2*(sqrt(c)* 
x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(b^2 
*c - 4*a*c^2)*c^3*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{b\,d+2\,c\,d\,x} \,d x \] Input:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x),x)
 

Output:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 591, normalized size of antiderivative = 3.97 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {240 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {-\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}-120 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {-\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c +15 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {-\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}-240 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}+120 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c -15 \sqrt {c}\, \sqrt {4 a c -b^{2}}\, \mathrm {log}\left (\frac {\sqrt {4 a c -b^{2}}+2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}+736 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{3}-280 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{2}+352 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{3} x +352 \sqrt {c \,x^{2}+b x +a}\, a \,c^{4} x^{2}+30 \sqrt {c \,x^{2}+b x +a}\, b^{4} c -40 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{2} x +56 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{3} x^{2}+192 \sqrt {c \,x^{2}+b x +a}\, b \,c^{4} x^{3}+96 \sqrt {c \,x^{2}+b x +a}\, c^{5} x^{4}}{960 c^{4} d} \] Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d),x)
 

Output:

(240*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqr 
t(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*c**2 - 120*sqrt( 
c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x 
+ c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c + 15*sqrt(c)*sqrt(4*a* 
c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 
b + 2*c*x)/sqrt(4*a*c - b**2))*b**4 - 240*sqrt(c)*sqrt(4*a*c - b**2)*log(( 
sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4* 
a*c - b**2))*a**2*c**2 + 120*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - 
b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))* 
a*b**2*c - 15*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt( 
c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4 + 736*sqrt 
(a + b*x + c*x**2)*a**2*c**3 - 280*sqrt(a + b*x + c*x**2)*a*b**2*c**2 + 35 
2*sqrt(a + b*x + c*x**2)*a*b*c**3*x + 352*sqrt(a + b*x + c*x**2)*a*c**4*x* 
*2 + 30*sqrt(a + b*x + c*x**2)*b**4*c - 40*sqrt(a + b*x + c*x**2)*b**3*c** 
2*x + 56*sqrt(a + b*x + c*x**2)*b**2*c**3*x**2 + 192*sqrt(a + b*x + c*x**2 
)*b*c**4*x**3 + 96*sqrt(a + b*x + c*x**2)*c**5*x**4)/(960*c**4*d)