\(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^2} \, dx\) [126]

Optimal result

Integrand size = 26, antiderivative size = 145 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac {2 c \sqrt {d} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {2 c \sqrt {d} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {\sqrt {d (b+2 c x)} \left (-\sqrt [4]{b^2-4 a c} (b+2 c x)^{3/2}+(1+i) c (a+x (b+c x)) \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-(1+i) c (a+x (b+c x)) \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+(1+i) c (a+x (b+c x)) \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{5/4} \sqrt {b+2 c x} (a+x (b+c x))} \] Input:

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

Output:

(Sqrt[d*(b + 2*c*x)]*(-((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)^(3/2)) + (1 + I)*c 
*(a + x*(b + c*x))*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4 
)] - (1 + I)*c*(a + x*(b + c*x))*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 
 - 4*a*c)^(1/4)] + (1 + I)*c*(a + x*(b + 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)^(5/4)*Sqrt[b + 2*c*x]*(a + x*(b + c*x)))
 

Rubi [A] (verified)

Time = 0.31 (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, 827, 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.

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

\(\Big \downarrow \) 1111

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-((b*d + 2*c*d*x)^(3/2)/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (4*c*d*(-1/ 
2*ArcTan[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/((b^2 - 4*a*c) 
^(1/4)*Sqrt[d]) + ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d] 
)]/(2*(b^2 - 4*a*c)^(1/4)*Sqrt[d])))/(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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   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. \(293\) vs. \(2(123)=246\).

Time = 1.32 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.03

method	result	size
derivativedivides	\(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{4 \left (4 a \,d^{2} c -b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {5}{4}}}\right )\)	\(294\)
default	\(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{4 \left (4 a \,d^{2} c -b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {5}{4}}}\right )\)	\(294\)
pseudoelliptic	\(\frac {2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \left (2 c x +b \right )+\sqrt {2}\, c d \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}\)	\(314\)

Input:

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

Output:

16*c*d^3*(1/4*(2*c*d*x+b*d)^(3/2)/(4*a*c*d^2-b^2*d^2)/((2*c*d*x+b*d)^2+4*a 
*d^2*c-b^2*d^2)+1/32/(4*a*c*d^2-b^2*d^2)^(5/4)*2^(1/2)*(ln((2*c*d*x+b*d-(4 
*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1 
/2))/(2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4 
*a*c*d^2-b^2*d^2)^(1/2)))+2*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c* 
d*x+b*d)^(1/2)+1)-2*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d 
)^(1/2)+1)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 980, normalized size of antiderivative = 6.76 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^ 
4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 
+ (b^3 - 4*a*b*c)*x)*log(sqrt(2*c*d*x + b*d)*c^3*d + (b^8 - 16*a*b^6*c + 9 
6*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*(c^4*d^2/(b^10 - 20*a*b^8*c 
 + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^( 
3/4)) - (c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 
1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^ 
2)*x^2 + (b^3 - 4*a*b*c)*x)*log(sqrt(2*c*d*x + b*d)*c^3*d - (b^8 - 16*a*b^ 
6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*(c^4*d^2/(b^10 - 20* 
a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5* 
c^5))^(3/4)) - (c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4 
*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(I*a*b^2 - 4*I*a^2*c + I*(b 
^2*c - 4*a*c^2)*x^2 + I*(b^3 - 4*a*b*c)*x)*log(sqrt(2*c*d*x + b*d)*c^3*d + 
 (I*b^8 - 16*I*a*b^6*c + 96*I*a^2*b^4*c^2 - 256*I*a^3*b^2*c^3 + 256*I*a^4* 
c^4)*(c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 128 
0*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)) - (c^4*d^2/(b^10 - 20*a*b^8*c + 160* 
a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(- 
I*a*b^2 + 4*I*a^2*c - I*(b^2*c - 4*a*c^2)*x^2 - I*(b^3 - 4*a*b*c)*x)*log(s 
qrt(2*c*d*x + b*d)*c^3*d + (-I*b^8 + 16*I*a*b^6*c - 96*I*a^2*b^4*c^2 + 256 
*I*a^3*b^2*c^3 - 256*I*a^4*c^4)*(c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((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. 504 vs. \(2 (123) = 246\).

Time = 0.19 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.48 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )} {\left (b^{2} - 4 \, a c\right )}} \] Input:

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

Output:

sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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) + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^( 
3/4)*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) - (-b^2*d^2 + 4*a*c*d^2)^(3/4)*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) + (-b^2*d^2 
+ 4*a*c*d^2)^(3/4)*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*s 
qrt(2)*a*b^2*c*d + 16*sqrt(2)*a^2*c^2*d) + 4*(2*c*d*x + b*d)^(3/2)*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 = 5.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {2\,c\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {2\,c\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}+\frac {4\,c\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{\left (4\,a\,c-b^2\right )\,\left ({\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2\right )} \] Input:

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

Output:

(2*c*d^(1/2)*atanh(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/ 
(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b^2 - 4*a*c)^(5/4) - (2*c*d^(1/2)*atan((( 
b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 - 4*a*c 
)^(9/4))))/(b^2 - 4*a*c)^(5/4) + (4*c*d*(b*d + 2*c*d*x)^(3/2))/((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 = 0.21 (sec) , antiderivative size = 872, normalized size of antiderivative = 6.01 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*( - 2*(4*a*c - b**2)**(3/4)*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 - 2*(4*a* 
c - b**2)**(3/4)*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 - 2*(4*a*c - b**2)**(3/4)*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 + 2*(4*a*c - b**2)**(3/4)*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 + 2*(4*a*c - b**2)**(3/4)*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 + 2*( 
4*a*c - b**2)**(3/4)*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 + (4*a*c - b**2)**( 
3/4)*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 + (4*a*c - b**2)**(3/4)*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 + (4*a*c - b**2)**(3/4)*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 - (4*a*c - b 
**2)**(3/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sq 
rt(4*a*c - b**2) + b + 2*c*x)*a*c - (4*a*c - b**2)**(3/4)*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 - (4*a*c - b**2)**(3/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b*...