\(\int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx\) [38]

Optimal result

Integrand size = 38, antiderivative size = 206 \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\frac {2 (c C-3 B d) (c-3 d x) (2 c+3 d x)}{27 d^3 \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}+\frac {2 C (c-3 d x)^2 (2 c+3 d x)}{81 d^3 \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}-\frac {2 \left (4 c^2 C-6 B c d+9 A d^2\right ) (2 c+3 d x) \sqrt {1-\frac {3 d x}{c}} \text {arctanh}\left (\frac {\sqrt {1-\frac {3 d x}{c}}}{\sqrt {3}}\right )}{27 \sqrt {3} d^3 \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \] Output:

2/27*(-3*B*d+C*c)*(-3*d*x+c)*(3*d*x+2*c)/d^3/(-27*d^3*x^3-27*c*d^2*x^2+4*c 
^3)^(1/2)+2/81*C*(-3*d*x+c)^2*(3*d*x+2*c)/d^3/(-27*d^3*x^3-27*c*d^2*x^2+4* 
c^3)^(1/2)-2/81*(9*A*d^2-6*B*c*d+4*C*c^2)*(3*d*x+2*c)*(1-3*d*x/c)^(1/2)*ar 
ctanh(1/3*(1-3*d*x/c)^(1/2)*3^(1/2))*3^(1/2)/d^3/(-27*d^3*x^3-27*c*d^2*x^2 
+4*c^3)^(1/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\frac {2 (2 c+3 d x) \left (\sqrt {c} (c-3 d x) (4 c C-3 d (3 B+C x))-\sqrt {3} \left (4 c^2 C-6 B c d+9 A d^2\right ) \sqrt {c-3 d x} \text {arctanh}\left (\frac {\sqrt {c-3 d x}}{\sqrt {3} \sqrt {c}}\right )\right )}{81 \sqrt {c} d^3 \sqrt {(c-3 d x) (2 c+3 d x)^2}} \] Input:

Integrate[(A + B*x + C*x^2)/Sqrt[4*c^3 - 27*c*d^2*x^2 - 27*d^3*x^3],x]
 

Output:

(2*(2*c + 3*d*x)*(Sqrt[c]*(c - 3*d*x)*(4*c*C - 3*d*(3*B + C*x)) - Sqrt[3]* 
(4*c^2*C - 6*B*c*d + 9*A*d^2)*Sqrt[c - 3*d*x]*ArcTanh[Sqrt[c - 3*d*x]/(Sqr 
t[3]*Sqrt[c])]))/(81*Sqrt[c]*d^3*Sqrt[(c - 3*d*x)*(2*c + 3*d*x)^2])
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2526, 27, 2490, 2483, 27, 90, 73, 221}

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.

\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx\)

\(\Big \downarrow \) 2526

\(\displaystyle -\frac {\int -\frac {27 d^2 (3 A d-(2 c C-3 B d) x)}{\sqrt {4 c^3-27 d^2 x^2 c-27 d^3 x^3}}dx}{81 d^3}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 A d-(2 c C-3 B d) x}{\sqrt {4 c^3-27 d^2 x^2 c-27 d^3 x^3}}dx}{3 d}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 2490

\(\displaystyle \frac {\int \frac {(3 B d-2 c C) \left (\frac {c}{3 d}+x\right )-\frac {27 c d^2 (3 B d-2 c C)-243 A d^4}{81 d^3}}{\sqrt {2 c^3+9 d \left (\frac {c}{3 d}+x\right ) c^2-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}d\left (\frac {c}{3 d}+x\right )}{3 d}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 2483

\(\displaystyle \frac {2 \sqrt {3} c^2 \left (3 d \left (\frac {c}{3 d}+x\right )+c\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} \int -\frac {-\frac {2 C c^2}{d}+3 B c-9 A d+3 (2 c C-3 B d) \left (\frac {c}{3 d}+x\right )}{18 \sqrt {3} c^2 \left (c+3 d \left (\frac {c}{3 d}+x\right )\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}}d\left (\frac {c}{3 d}+x\right )}{d \sqrt {2 c^3+9 c^2 d \left (\frac {c}{3 d}+x\right )-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\left (3 d \left (\frac {c}{3 d}+x\right )+c\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} \int \frac {-\frac {2 C c^2}{d}+3 B c-9 A d+3 (2 c C-3 B d) \left (\frac {c}{3 d}+x\right )}{\left (c+3 d \left (\frac {c}{3 d}+x\right )\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}}d\left (\frac {c}{3 d}+x\right )}{9 d \sqrt {2 c^3+9 c^2 d \left (\frac {c}{3 d}+x\right )-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 90

\(\displaystyle -\frac {\left (3 d \left (\frac {c}{3 d}+x\right )+c\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} \left (-\frac {\left (9 A d^2-6 B c d+4 c^2 C\right ) \int \frac {1}{\left (c+3 d \left (\frac {c}{3 d}+x\right )\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}}d\left (\frac {c}{3 d}+x\right )}{d}-\frac {2 \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} (2 c C-3 B d)}{3 c^2 d^2}\right )}{9 d \sqrt {2 c^3+9 c^2 d \left (\frac {c}{3 d}+x\right )-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\left (3 d \left (\frac {c}{3 d}+x\right )+c\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} \left (\frac {2 \left (9 A d^2-6 B c d+4 c^2 C\right ) \int \frac {1}{3 c-\frac {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}{c^2}}d\sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}}{3 c^2 d^2}-\frac {2 \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} (2 c C-3 B d)}{3 c^2 d^2}\right )}{9 d \sqrt {2 c^3+9 c^2 d \left (\frac {c}{3 d}+x\right )-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\left (3 d \left (\frac {c}{3 d}+x\right )+c\right ) \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )}}{\sqrt {3} c^{3/2}}\right ) \left (9 A d^2-6 B c d+4 c^2 C\right )}{3 \sqrt {3} c^{3/2} d^2}-\frac {2 \sqrt {2 c^3-3 c^2 d \left (\frac {c}{3 d}+x\right )} (2 c C-3 B d)}{3 c^2 d^2}\right )}{9 d \sqrt {2 c^3+9 c^2 d \left (\frac {c}{3 d}+x\right )-27 d^3 \left (\frac {c}{3 d}+x\right )^3}}-\frac {2 C \sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}}{81 d^3}\)

Input:

Int[(A + B*x + C*x^2)/Sqrt[4*c^3 - 27*c*d^2*x^2 - 27*d^3*x^3],x]
 

Output:

(-2*C*Sqrt[4*c^3 - 27*c*d^2*x^2 - 27*d^3*x^3])/(81*d^3) - ((c + 3*d*(c/(3* 
d) + x))*Sqrt[2*c^3 - 3*c^2*d*(c/(3*d) + x)]*((-2*(2*c*C - 3*B*d)*Sqrt[2*c 
^3 - 3*c^2*d*(c/(3*d) + x)])/(3*c^2*d^2) + (2*(4*c^2*C - 6*B*c*d + 9*A*d^2 
)*ArcTanh[Sqrt[2*c^3 - 3*c^2*d*(c/(3*d) + x)]/(Sqrt[3]*c^(3/2))])/(3*Sqrt[ 
3]*c^(3/2)*d^2)))/(9*d*Sqrt[2*c^3 + 9*c^2*d*(c/(3*d) + x) - 27*d^3*(c/(3*d 
) + x)^3])
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 221

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

rule 2483

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_S 
ymbol] :> Simp[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*x)^(2*p))   In 
t[(e + f*x)^m*(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, 
e, f, m, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0] &&  !IntegerQ[p]
 

rule 2490

Int[(P3_)^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{a = Coeff[P3 
, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3, x, 3]}, Su 
bst[Int[((3*d*e - c*f)/(3*d) + f*x)^m*Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27 
*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x, x + c/(3*d)] /; NeQ[c 
, 0]] /; FreeQ[{e, f, m, p}, x] && PolyQ[P3, x, 3]
 

rule 2526

Int[(Pm_)*(Qn_)^(p_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x] 
}, Simp[Coeff[Pm, x, m]*(Qn^(p + 1)/(n*(p + 1)*Coeff[Qn, x, n])), x] + Simp 
[1/(n*Coeff[Qn, x, n])   Int[ExpandToSum[n*Coeff[Qn, x, n]*Pm - Coeff[Pm, x 
, m]*D[Qn, x], x]*Qn^p, x], x] /; EqQ[m, n - 1]] /; FreeQ[p, x] && PolyQ[Pm 
, x] && PolyQ[Qn, x] && NeQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {2 \left (3 d x +2 c \right ) \sqrt {-3 d x +c}\, \left (9 A \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 d x +c}\, \sqrt {3}}{3 \sqrt {c}}\right ) d^{2}-6 B \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 d x +c}\, \sqrt {3}}{3 \sqrt {c}}\right ) c d +4 C \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 d x +c}\, \sqrt {3}}{3 \sqrt {c}}\right ) c^{2}-C \left (-3 d x +c \right )^{\frac {3}{2}} \sqrt {c}+9 \sqrt {-3 d x +c}\, B d \sqrt {c}-3 \sqrt {-3 d x +c}\, C \,c^{\frac {3}{2}}\right )}{81 \sqrt {-27 d^{3} x^{3}-27 c \,d^{2} x^{2}+4 c^{3}}\, d^{3} \sqrt {c}}\)	\(171\)

Input:

int((C*x^2+B*x+A)/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/81*(3*d*x+2*c)*(-3*d*x+c)^(1/2)*(9*A*3^(1/2)*arctanh(1/3*(-3*d*x+c)^(1/ 
2)*3^(1/2)/c^(1/2))*d^2-6*B*3^(1/2)*arctanh(1/3*(-3*d*x+c)^(1/2)*3^(1/2)/c 
^(1/2))*c*d+4*C*3^(1/2)*arctanh(1/3*(-3*d*x+c)^(1/2)*3^(1/2)/c^(1/2))*c^2- 
C*(-3*d*x+c)^(3/2)*c^(1/2)+9*(-3*d*x+c)^(1/2)*B*d*c^(1/2)-3*(-3*d*x+c)^(1/ 
2)*C*c^(3/2))/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2)/d^3/c^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.79 \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\left [\frac {\sqrt {3} {\left (8 \, C c^{3} - 12 \, B c^{2} d + 18 \, A c d^{2} + 3 \, {\left (4 \, C c^{2} d - 6 \, B c d^{2} + 9 \, A d^{3}\right )} x\right )} \sqrt {c} \log \left (\frac {9 \, d^{2} x^{2} - 6 \, c d x - 8 \, c^{2} + 2 \, \sqrt {3} \sqrt {-27 \, d^{3} x^{3} - 27 \, c d^{2} x^{2} + 4 \, c^{3}} \sqrt {c}}{9 \, d^{2} x^{2} + 12 \, c d x + 4 \, c^{2}}\right ) - 2 \, \sqrt {-27 \, d^{3} x^{3} - 27 \, c d^{2} x^{2} + 4 \, c^{3}} {\left (3 \, C c d x - 4 \, C c^{2} + 9 \, B c d\right )}}{81 \, {\left (3 \, c d^{4} x + 2 \, c^{2} d^{3}\right )}}, -\frac {2 \, {\left (\sqrt {3} {\left (8 \, C c^{3} - 12 \, B c^{2} d + 18 \, A c d^{2} + 3 \, {\left (4 \, C c^{2} d - 6 \, B c d^{2} + 9 \, A d^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {3} \sqrt {-27 \, d^{3} x^{3} - 27 \, c d^{2} x^{2} + 4 \, c^{3}} \sqrt {-c}}{9 \, d^{2} x^{2} + 3 \, c d x - 2 \, c^{2}}\right ) + \sqrt {-27 \, d^{3} x^{3} - 27 \, c d^{2} x^{2} + 4 \, c^{3}} {\left (3 \, C c d x - 4 \, C c^{2} + 9 \, B c d\right )}\right )}}{81 \, {\left (3 \, c d^{4} x + 2 \, c^{2} d^{3}\right )}}\right ] \] Input:

integrate((C*x^2+B*x+A)/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2),x, algorith 
m="fricas")
 

Output:

[1/81*(sqrt(3)*(8*C*c^3 - 12*B*c^2*d + 18*A*c*d^2 + 3*(4*C*c^2*d - 6*B*c*d 
^2 + 9*A*d^3)*x)*sqrt(c)*log((9*d^2*x^2 - 6*c*d*x - 8*c^2 + 2*sqrt(3)*sqrt 
(-27*d^3*x^3 - 27*c*d^2*x^2 + 4*c^3)*sqrt(c))/(9*d^2*x^2 + 12*c*d*x + 4*c^ 
2)) - 2*sqrt(-27*d^3*x^3 - 27*c*d^2*x^2 + 4*c^3)*(3*C*c*d*x - 4*C*c^2 + 9* 
B*c*d))/(3*c*d^4*x + 2*c^2*d^3), -2/81*(sqrt(3)*(8*C*c^3 - 12*B*c^2*d + 18 
*A*c*d^2 + 3*(4*C*c^2*d - 6*B*c*d^2 + 9*A*d^3)*x)*sqrt(-c)*arctan(sqrt(3)* 
sqrt(-27*d^3*x^3 - 27*c*d^2*x^2 + 4*c^3)*sqrt(-c)/(9*d^2*x^2 + 3*c*d*x - 2 
*c^2)) + sqrt(-27*d^3*x^3 - 27*c*d^2*x^2 + 4*c^3)*(3*C*c*d*x - 4*C*c^2 + 9 
*B*c*d))/(3*c*d^4*x + 2*c^2*d^3)]
 

Sympy [F]

\[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\int \frac {A + B x + C x^{2}}{\sqrt {- \left (- c + 3 d x\right ) \left (2 c + 3 d x\right )^{2}}}\, dx \] Input:

integrate((C*x**2+B*x+A)/(-27*d**3*x**3-27*c*d**2*x**2+4*c**3)**(1/2),x)
 

Output:

Integral((A + B*x + C*x**2)/sqrt(-(-c + 3*d*x)*(2*c + 3*d*x)**2), x)
 

Maxima [F]

\[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\int { \frac {C x^{2} + B x + A}{\sqrt {-27 \, d^{3} x^{3} - 27 \, c d^{2} x^{2} + 4 \, c^{3}}} \,d x } \] Input:

integrate((C*x^2+B*x+A)/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2),x, algorith 
m="maxima")
 

Output:

integrate((C*x^2 + B*x + A)/sqrt(-27*d^3*x^3 - 27*c*d^2*x^2 + 4*c^3), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.59 \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=-\frac {2 \, \sqrt {3} {\left (4 \, C c^{2} - 6 \, B c d + 9 \, A d^{2}\right )} \arctan \left (\frac {\sqrt {3} \sqrt {-3 \, d x + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{3} \mathrm {sgn}\left (-3 \, d x - 2 \, c\right )} - \frac {2 \, {\left ({\left (-3 \, d x + c\right )}^{\frac {3}{2}} C d^{6} + 3 \, \sqrt {-3 \, d x + c} C c d^{6} - 9 \, \sqrt {-3 \, d x + c} B d^{7}\right )}}{81 \, d^{9} \mathrm {sgn}\left (-3 \, d x - 2 \, c\right )} \] Input:

integrate((C*x^2+B*x+A)/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2),x, algorith 
m="giac")
 

Output:

-2/81*sqrt(3)*(4*C*c^2 - 6*B*c*d + 9*A*d^2)*arctan(1/3*sqrt(3)*sqrt(-3*d*x 
 + c)/sqrt(-c))/(sqrt(-c)*d^3*sgn(-3*d*x - 2*c)) - 2/81*((-3*d*x + c)^(3/2 
)*C*d^6 + 3*sqrt(-3*d*x + c)*C*c*d^6 - 9*sqrt(-3*d*x + c)*B*d^7)/(d^9*sgn( 
-3*d*x - 2*c))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\int \frac {C\,x^2+B\,x+A}{\sqrt {4\,c^3-27\,c\,d^2\,x^2-27\,d^3\,x^3}} \,d x \] Input:

int((A + B*x + C*x^2)/(4*c^3 - 27*d^3*x^3 - 27*c*d^2*x^2)^(1/2),x)
 

Output:

int((A + B*x + C*x^2)/(4*c^3 - 27*d^3*x^3 - 27*c*d^2*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2}{\sqrt {4 c^3-27 c d^2 x^2-27 d^3 x^3}} \, dx=\frac {-18 \sqrt {-3 d x +c}\, b c d +8 \sqrt {-3 d x +c}\, c^{3}-6 \sqrt {-3 d x +c}\, c^{2} d x +9 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}-\sqrt {c}\, \sqrt {3}\right ) a \,d^{2}-6 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}-\sqrt {c}\, \sqrt {3}\right ) b c d +4 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}-\sqrt {c}\, \sqrt {3}\right ) c^{3}-9 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}+\sqrt {c}\, \sqrt {3}\right ) a \,d^{2}+6 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}+\sqrt {c}\, \sqrt {3}\right ) b c d -4 \sqrt {c}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-3 d x +c}+\sqrt {c}\, \sqrt {3}\right ) c^{3}}{81 c \,d^{3}} \] Input:

int((C*x^2+B*x+A)/(-27*d^3*x^3-27*c*d^2*x^2+4*c^3)^(1/2),x)
 

Output:

( - 18*sqrt(c - 3*d*x)*b*c*d + 8*sqrt(c - 3*d*x)*c**3 - 6*sqrt(c - 3*d*x)* 
c**2*d*x + 9*sqrt(c)*sqrt(3)*log(sqrt(c - 3*d*x) - sqrt(c)*sqrt(3))*a*d**2 
 - 6*sqrt(c)*sqrt(3)*log(sqrt(c - 3*d*x) - sqrt(c)*sqrt(3))*b*c*d + 4*sqrt 
(c)*sqrt(3)*log(sqrt(c - 3*d*x) - sqrt(c)*sqrt(3))*c**3 - 9*sqrt(c)*sqrt(3 
)*log(sqrt(c - 3*d*x) + sqrt(c)*sqrt(3))*a*d**2 + 6*sqrt(c)*sqrt(3)*log(sq 
rt(c - 3*d*x) + sqrt(c)*sqrt(3))*b*c*d - 4*sqrt(c)*sqrt(3)*log(sqrt(c - 3* 
d*x) + sqrt(c)*sqrt(3))*c**3)/(81*c*d**3)