\(\int \frac {1}{(b d+2 c d x) (a+b x+c x^2)^{5/2}} \, dx\) [216]

Optimal result

Integrand size = 26, antiderivative size = 118 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {8 c}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}+\frac {16 c^{3/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d} \] Output:

-2/3/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^(3/2)+8*c/(-4*a*c+b^2)^2/d/(c*x^2+b*x+a) 
^(1/2)+16*c^(3/2)*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2)) 
/(-4*a*c+b^2)^(5/2)/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1624\) vs. \(2(118)=236\).

Time = 6.71 (sec) , antiderivative size = 1624, normalized size of antiderivative = 13.76 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(5/2)),x]
 

Output:

(-2*(-8192*a^(15/2)*(b^2 + 12*b*c*x + 12*c^2*x^2) - Sqrt[a]*b^2*x^7*(b + c 
*x)^4*(b^3 + 212*b^2*c*x + 308*b*c^2*x^2 + 96*c^3*x^3) - 2048*a^(13/2)*x*( 
14*b^3 + 203*b^2*c*x + 348*b*c^2*x^2 + 156*c^3*x^3) - 512*a^(11/2)*x^2*(77 
*b^4 + 1362*b^3*c*x + 3272*b^2*c^2*x^2 + 2760*b*c^3*x^3 + 776*c^4*x^4) - 2 
*a^(3/2)*x^6*(b + c*x)^3*(49*b^5 + 3251*b^4*c*x + 6878*b^3*c^2*x^2 + 4740* 
b^2*c^3*x^3 + 1144*b*c^4*x^4 + 64*c^5*x^5) - 8*a^(5/2)*x^5*(b + c*x)^2*(19 
6*b^5 + 7609*b^4*c*x + 19480*b^3*c^2*x^2 + 18023*b^2*c^3*x^3 + 6836*b*c^4* 
x^4 + 860*c^5*x^5) - 128*a^(9/2)*x^3*(210*b^5 + 4636*b^4*c*x + 14248*b^3*c 
^2*x^2 + 17237*b^2*c^3*x^3 + 9244*b*c^4*x^4 + 1828*c^5*x^5) - 32*a^(7/2)*x 
^4*(294*b^6 + 8464*b^5*c*x + 31565*b^4*c^2*x^2 + 48878*b^3*c^3*x^3 + 37475 
*b^2*c^4*x^4 + 14012*b*c^5*x^5 + 2020*c^6*x^6) + 16*b^3*c*x^8*(b + c*x)^5* 
Sqrt[a + x*(b + c*x)] + 8192*a^7*(b^2 + 12*b*c*x + 12*c^2*x^2)*Sqrt[a + x* 
(b + c*x)] + 6144*a^6*x*(4*b^3 + 59*b^2*c*x + 100*b*c^2*x^2 + 44*c^3*x^3)* 
Sqrt[a + x*(b + c*x)] + 2*a*b*x^6*(b + c*x)^3*(7*b^4 + 711*b^3*c*x + 1296* 
b^2*c^2*x^2 + 684*b*c^3*x^3 + 96*c^4*x^4)*Sqrt[a + x*(b + c*x)] + 512*a^5* 
x^2*(55*b^4 + 1012*b^3*c*x + 2392*b^2*c^2*x^2 + 1968*b*c^3*x^3 + 536*c^4*x 
^4)*Sqrt[a + x*(b + c*x)] + 8*a^2*x^5*(b + c*x)^2*(56*b^5 + 2695*b^4*c*x + 
 6406*b^3*c^2*x^2 + 5293*b^2*c^3*x^3 + 1700*b*c^4*x^4 + 164*c^5*x^5)*Sqrt[ 
a + x*(b + c*x)] + 128*a^4*x^3*(120*b^5 + 2844*b^4*c*x + 8568*b^3*c^2*x^2 
+ 10043*b^2*c^3*x^3 + 5172*b*c^4*x^4 + 972*c^5*x^5)*Sqrt[a + x*(b + c*x...
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1111, 27, 1111, 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.

\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)} \, dx\)

\(\Big \downarrow \) 1111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1111

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

\(\Big \downarrow \) 1112

\(\displaystyle -\frac {4 c \left (-\frac {16 c^2 \int \frac {1}{8 \left (c x^2+b x+a\right ) c^2+2 \left (b^2-4 a c\right ) c}d\sqrt {c x^2+b x+a}}{b^2-4 a c}-\frac {2}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{d \left (b^2-4 a c\right )}-\frac {2}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 218

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

Input:

Int[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(5/2)),x]
 

Output:

-2/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) - (4*c*(-2/((b^2 - 4*a*c)*S 
qrt[a + b*x + c*x^2]) - (4*Sqrt[c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2] 
)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)))/((b^2 - 4*a*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 1111

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* 
(b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) 
  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] && LtQ[p, -1] &&  !G 
tQ[m, 1] && 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 = 1.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(-\frac {16 \left (c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, c}{\sqrt {4 a \,c^{2}-b^{2} c}}\right )-\frac {2 \left (\frac {3 c^{2} x^{2}}{4}+\left (\frac {3 b x}{4}+a \right ) c -\frac {b^{2}}{16}\right ) \sqrt {4 a \,c^{2}-b^{2} c}}{3}\right )}{\sqrt {4 a \,c^{2}-b^{2} c}\, \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )^{2} d}\)	\(132\)
default	\(\frac {\frac {4 c}{3 \left (4 a c -b^{2}\right ) \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}+\frac {4 c \left (\frac {4 c}{\left (4 a c -b^{2}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}-\frac {8 c \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}}{2 d c}\)	\(223\)

Input:

int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-16*(c^2*(c*x^2+b*x+a)^(3/2)*arctanh(2*(c*x^2+b*x+a)^(1/2)*c/(4*a*c^2-b^2* 
c)^(1/2))-2/3*(3/4*c^2*x^2+(3/4*b*x+a)*c-1/16*b^2)*(4*a*c^2-b^2*c)^(1/2))/ 
(4*a*c^2-b^2*c)^(1/2)/(c*x^2+b*x+a)^(3/2)/(4*a*c-b^2)^2/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (102) = 204\).

Time = 0.44 (sec) , antiderivative size = 620, normalized size of antiderivative = 5.25 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\left [\frac {2 \, {\left (12 \, {\left (c^{3} x^{4} + 2 \, b c^{2} x^{3} + 2 \, a b c x + a^{2} c + {\left (b^{2} c + 2 \, a c^{2}\right )} x^{2}\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c + 4 \, \sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + {\left (12 \, c^{2} x^{2} + 12 \, b c x - b^{2} + 16 \, a c\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x + {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}}, \frac {2 \, {\left (24 \, {\left (c^{3} x^{4} + 2 \, b c^{2} x^{3} + 2 \, a b c x + a^{2} c + {\left (b^{2} c + 2 \, a c^{2}\right )} x^{2}\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}} \arctan \left (-\frac {\sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (12 \, c^{2} x^{2} + 12 \, b c x - b^{2} + 16 \, a c\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x + {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}}\right ] \] Input:

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

[2/3*(12*(c^3*x^4 + 2*b*c^2*x^3 + 2*a*b*c*x + a^2*c + (b^2*c + 2*a*c^2)*x^ 
2)*sqrt(-c/(b^2 - 4*a*c))*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c + 4*sqrt 
(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^2 + 4*b*c 
*x + b^2)) + (12*c^2*x^2 + 12*b*c*x - b^2 + 16*a*c)*sqrt(c*x^2 + b*x + a)) 
/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16 
*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^ 
2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d), 2/3 
*(24*(c^3*x^4 + 2*b*c^2*x^3 + 2*a*b*c*x + a^2*c + (b^2*c + 2*a*c^2)*x^2)*s 
qrt(c/(b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt( 
c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a*c)) + (12*c^2*x^2 + 12*b*c*x - b^2 + 
 16*a*c)*sqrt(c*x^2 + b*x + a))/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^ 
4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a 
^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8* 
a^3*b^2*c + 16*a^4*c^2)*d)]
 

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {\int \frac {1}{a^{2} b \sqrt {a + b x + c x^{2}} + 2 a^{2} c x \sqrt {a + b x + c x^{2}} + 2 a b^{2} x \sqrt {a + b x + c x^{2}} + 6 a b c x^{2} \sqrt {a + b x + c x^{2}} + 4 a c^{2} x^{3} \sqrt {a + b x + c x^{2}} + b^{3} x^{2} \sqrt {a + b x + c x^{2}} + 4 b^{2} c x^{3} \sqrt {a + b x + c x^{2}} + 5 b c^{2} x^{4} \sqrt {a + b x + c x^{2}} + 2 c^{3} x^{5} \sqrt {a + b x + c x^{2}}}\, dx}{d} \] Input:

integrate(1/(2*c*d*x+b*d)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Integral(1/(a**2*b*sqrt(a + b*x + c*x**2) + 2*a**2*c*x*sqrt(a + b*x + c*x* 
*2) + 2*a*b**2*x*sqrt(a + b*x + c*x**2) + 6*a*b*c*x**2*sqrt(a + b*x + c*x* 
*2) + 4*a*c**2*x**3*sqrt(a + b*x + c*x**2) + b**3*x**2*sqrt(a + b*x + c*x* 
*2) + 4*b**2*c*x**3*sqrt(a + b*x + c*x**2) + 5*b*c**2*x**4*sqrt(a + b*x + 
c*x**2) + 2*c**3*x**5*sqrt(a + b*x + c*x**2)), x)/d
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 906 vs. \(2 (102) = 204\).

Time = 0.35 (sec) , antiderivative size = 906, normalized size of antiderivative = 7.68 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

32*c^2*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt( 
b^2*c - 4*a*c^2))/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*sqrt(b^2*c - 4*a*c 
^2)) + 2/3*(12*((b^16*c^2*d^3 - 32*a*b^14*c^3*d^3 + 448*a^2*b^12*c^4*d^3 - 
 3584*a^3*b^10*c^5*d^3 + 17920*a^4*b^8*c^6*d^3 - 57344*a^5*b^6*c^7*d^3 + 1 
14688*a^6*b^4*c^8*d^3 - 131072*a^7*b^2*c^9*d^3 + 65536*a^8*c^10*d^3)*x/(b^ 
20*d^4 - 40*a*b^18*c*d^4 + 720*a^2*b^16*c^2*d^4 - 7680*a^3*b^14*c^3*d^4 + 
53760*a^4*b^12*c^4*d^4 - 258048*a^5*b^10*c^5*d^4 + 860160*a^6*b^8*c^6*d^4 
- 1966080*a^7*b^6*c^7*d^4 + 2949120*a^8*b^4*c^8*d^4 - 2621440*a^9*b^2*c^9* 
d^4 + 1048576*a^10*c^10*d^4) + (b^17*c*d^3 - 32*a*b^15*c^2*d^3 + 448*a^2*b 
^13*c^3*d^3 - 3584*a^3*b^11*c^4*d^3 + 17920*a^4*b^9*c^5*d^3 - 57344*a^5*b^ 
7*c^6*d^3 + 114688*a^6*b^5*c^7*d^3 - 131072*a^7*b^3*c^8*d^3 + 65536*a^8*b* 
c^9*d^3)/(b^20*d^4 - 40*a*b^18*c*d^4 + 720*a^2*b^16*c^2*d^4 - 7680*a^3*b^1 
4*c^3*d^4 + 53760*a^4*b^12*c^4*d^4 - 258048*a^5*b^10*c^5*d^4 + 860160*a^6* 
b^8*c^6*d^4 - 1966080*a^7*b^6*c^7*d^4 + 2949120*a^8*b^4*c^8*d^4 - 2621440* 
a^9*b^2*c^9*d^4 + 1048576*a^10*c^10*d^4))*x - (b^18*d^3 - 48*a*b^16*c*d^3 
+ 960*a^2*b^14*c^2*d^3 - 10752*a^3*b^12*c^3*d^3 + 75264*a^4*b^10*c^4*d^3 - 
 344064*a^5*b^8*c^5*d^3 + 1032192*a^6*b^6*c^6*d^3 - 1966080*a^7*b^4*c^7*d^ 
3 + 2162688*a^8*b^2*c^8*d^3 - 1048576*a^9*c^9*d^3)/(b^20*d^4 - 40*a*b^18*c 
*d^4 + 720*a^2*b^16*c^2*d^4 - 7680*a^3*b^14*c^3*d^4 + 53760*a^4*b^12*c^4*d 
^4 - 258048*a^5*b^10*c^5*d^4 + 860160*a^6*b^8*c^6*d^4 - 1966080*a^7*b^6...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (b\,d+2\,c\,d\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(5/2)),x)
 

Output:

int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 1193, normalized size of antiderivative = 10.11 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(2*(24*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*s 
qrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*c + 48*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*c*x + 48*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*c**2*x**2 + 24*sqrt(c)*sqrt(4*a*c - b**2)*lo 
g(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/s 
qrt(4*a*c - b**2))*b**2*c*x**2 + 48*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqr 
t(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*c**2*x**3 + 24*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)) 
*c**3*x**4 - 24*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqr 
t(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*c - 48*s 
qrt(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*c*x - 48*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*c**2*x**2 - 24*sqrt(c)*sqrt(4*a*c - b**2)*lo 
g((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**2*c*x**2 - 48*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 - ...