\(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)} \, dx\) [115]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt {d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt {d}} \] Output:

-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)^(3/ 
4)/d^(1/2)-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)^(3/4)/d^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.49 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=\frac {(1-i) \sqrt {b+2 c x} \left (\arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt {d (b+2 c x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 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. \(227\) vs. \(2(83)=166\).

Time = 1.38 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.21

method	result	size
derivativedivides	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\)	\(228\)
default	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\)	\(228\)
pseudoelliptic	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}\)	\(243\)

Input:

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

Output:

1/2*d/(4*a*c*d^2-b^2*d^2)^(3/4)*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 = 0.13 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.95 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=-\left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} \log \left ({\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} + \sqrt {2 \, c d x + b d}\right ) - i \, \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} \log \left (i \, {\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} + \sqrt {2 \, c d x + b d}\right ) + i \, \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} \log \left (-i \, {\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} + \sqrt {2 \, c d x + b d}\right ) + \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} \log \left (-{\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac {1}{4}} + \sqrt {2 \, c d x + b d}\right ) \] Input:

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

Output:

-(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4)*log((b^2 
 - 4*a*c)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/ 
4) + sqrt(2*c*d*x + b*d)) - I*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64* 
a^3*c^3)*d^2))^(1/4)*log(I*(b^2 - 4*a*c)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2* 
b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4) + sqrt(2*c*d*x + b*d)) + I*(1/((b^6 - 12 
*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4)*log(-I*(b^2 - 4*a*c)*d 
*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4) + sqrt(2 
*c*d*x + b*d)) + (1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2) 
)^(1/4)*log(-(b^2 - 4*a*c)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a 
^3*c^3)*d^2))^(1/4) + sqrt(2*c*d*x + b*d))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a),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. 393 vs. \(2 (83) = 166\).

Time = 0.12 (sec) , antiderivative size = 393, normalized size of antiderivative = 3.82 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} d - 4 \, a c d} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} d - 4 \, a c d} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} d - 4 \, \sqrt {2} a c d} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} d - 4 \, \sqrt {2} a c d} \] Input:

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

Output:

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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^2*d - 4*a*c*d) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*arctan(-1/2*sqr 
t(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^2*d - 4*a*c*d) - (-b^2*d^2 + 4*a*c*d^2)^(1/4)*l 
og(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^2*d - 4*sqrt(2)*a*c*d) + (-b^2* 
d^2 + 4*a*c*d^2)^(1/4)*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^2*d - 4 
*sqrt(2)*a*c*d)
 

Mupad [B] (verification not implemented)

Time = 5.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.56 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {128\,d^{3/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {2\,\mathrm {atanh}\left (\frac {128\,d^{3/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{3/4}} \] Input:

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

Output:

- (2*atan((128*d^(3/2)*(b*d + 2*c*d*x)^(1/2))/(((128*b^2*d^2)/(b^2 - 4*a*c 
)^(3/2) - (512*a*c*d^2)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(d^(1/ 
2)*(b^2 - 4*a*c)^(3/4)) - (2*atanh((128*d^(3/2)*(b*d + 2*c*d*x)^(1/2))/((( 
128*b^2*d^2)/(b^2 - 4*a*c)^(3/2) - (512*a*c*d^2)/(b^2 - 4*a*c)^(3/2))*(b^2 
 - 4*a*c)^(3/4))))/(d^(1/2)*(b^2 - 4*a*c)^(3/4))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.89 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx=\frac {\sqrt {d}\, \sqrt {2}\, \left (-2 \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}-2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )-\mathrm {log}\left (-\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )+\mathrm {log}\left (\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )\right )}{2 \left (4 a c -b^{2}\right )^{\frac {3}{4}} d} \] Input:

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

Output:

(sqrt(d)*(4*a*c - b**2)**(1/4)*sqrt(2)*( - 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))) + 2*atan(((4* 
a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqr 
t(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) + 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)))/(2*d*(4*a*c - b**2))