\(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)^2} \, dx\) [1303]

Optimal result

Integrand size = 26, antiderivative size = 143 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=-\frac {\sqrt {b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac {6 c \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d}}+\frac {6 c \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {701, 708, 335, 218, 212, 209} \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {6 c \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{7/4}}+\frac {6 c \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{7/4}}-\frac {\sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[In]

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

[Out]

-(Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (6*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1
/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7/4)*Sqrt[d]) + (6*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
)/((b^2 - 4*a*c)^(7/4)*Sqrt[d])

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 701

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> 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] - Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = -\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 {(3 c) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c} \\ & = -\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 {3 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )}{2 \left (b^2-4 a c\right ) d} \\ & = -\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 {3 \text {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right ) d} \\ & = -\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) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac {(6 c) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/2}} \\ & = -\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 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c 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 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.61 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {i c \left (\frac {i \left (b^2-4 a c\right )^{3/4} (b+2 c x)}{c (a+x (b+c x))}+(3+3 i) \sqrt {b+2 c x} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-(3+3 i) \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-(3+3 i) \sqrt {b+2 c x} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{7/4} \sqrt {d (b+2 c x)}} \]

[In]

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

[Out]

(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)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(293\) vs. \(2(121)=242\).

Time = 2.88 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.06

method	result	size
derivativedivides	\(16 c \,d^{3} \left (\frac {\sqrt {2 c d x +b d}}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {7}{4}}}\right )\)	\(294\)
default	\(16 c \,d^{3} \left (\frac {\sqrt {2 c d x +b d}}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {7}{4}}}\right )\)	\(294\)
pseudoelliptic	\(d^{3} c \left (\frac {\sqrt {d \left (2 c x +b \right )}}{d^{4} \left (4 a c -b^{2}\right ) c \left (c \,x^{2}+b x +a \right )}+\frac {3 \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}+\frac {3 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}-\frac {3 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {7}{4}}}\right )\)	\(325\)

[In]

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

[Out]

16*c*d^3*(1/4*(2*c*d*x+b*d)^(1/2)/(4*a*c*d^2-b^2*d^2)/((2*c*d*x+b*d)^2+4*a*c*d^2-b^2*d^2)+3/32/(4*a*c*d^2-b^2*
d^2)^(7/4)*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 = 0.28 (sec) , antiderivative size = 1083, normalized size of antiderivative = 7.57 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(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 + 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)*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*(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^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 + 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) - sqrt(2*c*d*x + b*d))/((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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (121) = 242\).

Time = 0.29 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.55 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {4 \, \sqrt {2 \, c d x + b d} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )} {\left (b^{2} - 4 \, a c\right )}} \]

[In]

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

[Out]

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*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*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*log(2*c*d*x + b*d + sq
rt(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 = 9.22 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx=\frac {6\,c\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {6\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {4\,c\,d\,\sqrt {b\,d+2\,c\,d\,x}}{\left (4\,a\,c-b^2\right )\,\left ({\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2\right )} \]

[In]

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

[Out]

(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))