\(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)^2} \, dx\) [127]

Optimal result

Integrand size = 26, antiderivative size = 145 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=-\frac {\sqrt {b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac {6 c \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d}}+\frac {6 c \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d}} \] Output:

-(2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)/d/(c*x^2+b*x+a)+6*c*arctan((2*c*d*x+b*d) 
^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(7/4)/d^(1/2)+6*c*arctanh( 
(2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(7/4)/d^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.75 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {i c \left (\frac {i \left (b^2-4 a c\right )^{3/4} (b+2 c x)}{c (a+x (b+c x))}+(3+3 i) \sqrt {b+2 c x} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-(3+3 i) \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-(3+3 i) \sqrt {b+2 c x} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d (b+2 c x)}} \] Input:

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

Output:

(I*c*((I*(b^2 - 4*a*c)^(3/4)*(b + 2*c*x))/(c*(a + x*(b + c*x))) + (3 + 3*I 
)*Sqrt[b + 2*c*x]*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4) 
] - (3 + 3*I)*Sqrt[b + 2*c*x]*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 
4*a*c)^(1/4)] - (3 + 3*I)*Sqrt[b + 2*c*x]*ArcTanh[((1 + I)*(b^2 - 4*a*c)^( 
1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b + 2*c*x))]))/((b^2 - 4*a*c 
)^(7/4)*Sqrt[d*(b + 2*c*x)])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1111, 1118, 27, 25, 266, 756, 216, 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.

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

\(\Big \downarrow \) 1111

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {12 c d \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{b^2-4 a c}-\frac {\sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {12 c d \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )}{b^2-4 a c}-\frac {\sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {12 c d \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{b^2-4 a c}-\frac {\sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {12 c d \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{b^2-4 a c}-\frac {\sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

Input:

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

Output:

-(Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (12*c*d*(ArcT 
an[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c)^(3/ 
4)*d^(3/2)) + ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/( 
2*(b^2 - 4*a*c)^(3/4)*d^(3/2))))/(b^2 - 4*a*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

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 1118

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(123)=246\).

Time = 1.28 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.01

method	result	size
pseudoelliptic	\(c \,d^{3} \left (\frac {\sqrt {d \left (2 c x +b \right )}}{d^{4} \left (4 a c -b^{2}\right ) c \left (c \,x^{2}+b x +a \right )}+\frac {3 \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}+\frac {3 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}+\frac {3 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}\right )\)	\(292\)
derivativedivides	\(16 c \,d^{3} \left (\frac {\sqrt {2 c d x +b d}}{4 \left (4 a \,d^{2} c -b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {7}{4}}}\right )\)	\(294\)
default	\(16 c \,d^{3} \left (\frac {\sqrt {2 c d x +b d}}{4 \left (4 a \,d^{2} c -b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {7}{4}}}\right )\)	\(294\)

Input:

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

Output:

c*d^3*((d*(2*c*x+b))^(1/2)/d^4/(4*a*c-b^2)/c/(c*x^2+b*x+a)+3/2*ln(((d^2*(4 
*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+(d^2*(4*a*c-b^2))^(1/2)+d*(2* 
c*x+b))/((d^2*(4*a*c-b^2))^(1/2)-(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/ 
2)*2^(1/2)+d*(2*c*x+b)))/(d^2*(4*a*c-b^2))^(7/4)*2^(1/2)+3*arctan(2^(1/2)/ 
(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)+1)/(d^2*(4*a*c-b^2))^(7/4)*2^( 
1/2)+3*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)-1)/(d^2* 
(4*a*c-b^2))^(7/4)*2^(1/2))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 1083, normalized size of antiderivative = 7.47 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(3*((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)*( 
c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4* 
b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/ 
4)*log(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c^4/((b^14 - 28*a*b^12*c + 336*a^ 
2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 286 
72*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/4)*d + 3*sqrt(2*c*d*x + b*d)*c) - 
 3*(-I*(b^2*c - 4*a*c^2)*d*x^2 - I*(b^3 - 4*a*b*c)*d*x - I*(a*b^2 - 4*a^2* 
c)*d)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 89 
60*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^ 
2))^(1/4)*log(3*I*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c^4/((b^14 - 28*a*b^12*c 
 + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4* 
c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/4)*d + 3*sqrt(2*c*d*x + 
b*d)*c) - 3*(I*(b^2*c - 4*a*c^2)*d*x^2 + I*(b^3 - 4*a*b*c)*d*x + I*(a*b^2 
- 4*a^2*c)*d)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8* 
c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7 
*c^7)*d^2))^(1/4)*log(-3*I*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c^4/((b^14 - 28 
*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504 
*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/4)*d + 3*sqrt(2 
*c*d*x + b*d)*c) - 3*((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b 
^2 - 4*a^2*c)*d)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^...
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(d*(b + 2*c*x))*(a + b*x + c*x**2)**2), x)
 

Maxima [F(-2)]

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

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^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. 507 vs. \(2 (123) = 246\).

Time = 0.14 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.50 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {4 \, \sqrt {2 \, c d x + b d} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )} {\left (b^{2} - 4 \, a c\right )}} \] Input:

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

Output:

3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2 
*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1 
/4))/(b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) + 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^ 
2)^(1/4)*c*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*s 
qrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^4*d - 8*a*b^2*c*d + 1 
6*a^2*c^2*d) + 3*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*log(2*c*d*x + b*d + sqrt(2 
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c 
*d^2))/(sqrt(2)*b^4*d - 8*sqrt(2)*a*b^2*c*d + 16*sqrt(2)*a^2*c^2*d) - 3*(- 
b^2*d^2 + 4*a*c*d^2)^(1/4)*c*log(2*c*d*x + b*d - sqrt(2)*(-b^2*d^2 + 4*a*c 
*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^4 
*d - 8*sqrt(2)*a*b^2*c*d + 16*sqrt(2)*a^2*c^2*d) + 4*sqrt(2*c*d*x + b*d)*c 
*d/((b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)*(b^2 - 4*a*c))
 

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {6\,c\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {6\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {4\,c\,d\,\sqrt {b\,d+2\,c\,d\,x}}{\left (4\,a\,c-b^2\right )\,\left ({\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2\right )} \] Input:

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

Output:

(6*c*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d^(1/2)*(b^4 + 16*a 
^2*c^2 - 8*a*b^2*c))))/(d^(1/2)*(b^2 - 4*a*c)^(7/4)) + (6*c*atanh(((b*d + 
2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c 
))))/(d^(1/2)*(b^2 - 4*a*c)^(7/4)) + (4*c*d*(b*d + 2*c*d*x)^(1/2))/((4*a*c 
 - b^2)*((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 847, normalized size of antiderivative = 5.84 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*( - 6*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*s 
qrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*c - 6*(4*a* 
c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 
2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b*c*x - 6*(4*a*c - b**2)**(1/4)*s 
qrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - 
b**2)**(1/4)*sqrt(2)))*c**2*x**2 + 6*(4*a*c - b**2)**(1/4)*sqrt(2)*atan((( 
4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*s 
qrt(2)))*a*c + 6*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4) 
*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b*c*x + 6*( 
4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt( 
b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*c**2*x**2 - 3*(4*a*c - b**2)* 
*(1/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt 
(4*a*c - b**2) + b + 2*c*x)*a*c - 3*(4*a*c - b**2)**(1/4)*sqrt(2)*log( - s 
qrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2* 
c*x)*b*c*x - 3*(4*a*c - b**2)**(1/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c 
 - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*c**2*x**2 + 3*(4 
*a*c - b**2)**(1/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt 
(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*a*c + 3*(4*a*c - b**2)**(1/4)*sqrt(2 
)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + 
 b + 2*c*x)*b*c*x + 3*(4*a*c - b**2)**(1/4)*sqrt(2)*log(sqrt(b + 2*c*x)...