\(\int \frac {1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)^3} \, dx\) [1317]

Optimal result

Integrand size = 26, antiderivative size = 223 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {45 c^2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{3/2}}-\frac {45 c^2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{3/2}} \]

[Out]

45*c^2*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(13/4)/d^(3/2)-45*c^2*arctanh((d*(2
*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(13/4)/d^(3/2)+90*c^2/(-4*a*c+b^2)^3/d/(2*c*d*x+b*d)^(
1/2)-1/2/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(1/2)+9/2*c/(-4*a*c+b^2)^2/d/(c*x^2+b*x+a)/(2*c*d*x+b*d)
^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {701, 707, 708, 335, 304, 209, 212} \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {45 c^2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}-\frac {45 c^2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}+\frac {90 c^2}{d \left (b^2-4 a c\right )^3 \sqrt {b d+2 c d x}}+\frac {9 c}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}} \]

[In]

Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x]

[Out]

(90*c^2)/((b^2 - 4*a*c)^3*d*Sqrt[b*d + 2*c*d*x]) - 1/(2*(b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^
2) + (9*c)/(2*(b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) + (45*c^2*ArcTan[Sqrt[d*(b + 2*c*x)]/((
b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(3/2)) - (45*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a
*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(3/2))

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[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 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}-\frac {(9 c) \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )} \\ & = -\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {\left (45 c^2\right ) \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^2} \\ & = \frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {\left (45 c^2\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{2 \left (b^2-4 a c\right )^3 d^2} \\ & = \frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {(45 c) \text {Subst}\left (\int \frac {\sqrt {x}}{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{4 \left (b^2-4 a c\right )^3 d^3} \\ & = \frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {(45 c) \text {Subst}\left (\int \frac {x^2}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )}{2 \left (b^2-4 a c\right )^3 d^3} \\ & = \frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {\left (45 c^2\right ) \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 d}+\frac {\left (45 c^2\right ) \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 d} \\ & = \frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {45 c^2 \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 )^{13/4} d^{3/2}}-\frac {45 c^2 \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 )^{13/4} d^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 \left (\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (b+2 c x) \left (32 b^4-256 a b^2 c+512 a^2 c^2-81 b^2 (b+2 c x)^2+324 a c (b+2 c x)^2+45 (b+2 c x)^4\right )}{c^2 \left (b^2-4 a c\right )^3 (a+x (b+c x))^2}-\frac {45 (b+2 c x)^{3/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}+\frac {45 (b+2 c x)^{3/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}-\frac {45 (b+2 c x)^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{13/4}}\right )}{(d (b+2 c x))^{3/2}} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x]

[Out]

((1/2 + I/2)*c^2*(((1/8 - I/8)*(b + 2*c*x)*(32*b^4 - 256*a*b^2*c + 512*a^2*c^2 - 81*b^2*(b + 2*c*x)^2 + 324*a*
c*(b + 2*c*x)^2 + 45*(b + 2*c*x)^4))/(c^2*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^2) - (45*(b + 2*c*x)^(3/2)*ArcTan[
1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(13/4) + (45*(b + 2*c*x)^(3/2)*ArcTan[1 + ((
1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(13/4) - (45*(b + 2*c*x)^(3/2)*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)^(13/4)))/(d*(b + 2*c*x)
)^(3/2)

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(64 c^{2} d^{5} \left (-\frac {\frac {\frac {13 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32}+16 \left (\frac {17}{128} a c \,d^{2}-\frac {17}{512} b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {45 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}-\frac {1}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}\right )\)	\(355\)
default	\(64 c^{2} d^{5} \left (-\frac {\frac {\frac {13 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32}+16 \left (\frac {17}{128} a c \,d^{2}-\frac {17}{512} b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {45 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}-\frac {1}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}\right )\)	\(355\)
pseudoelliptic	\(\frac {32 d^{5} c^{2} \left (-\frac {13 \left (d \left (2 c x +b \right )\right )^{\frac {7}{2}}}{256 c^{2} d^{10} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {17 \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}} a}{64 c \,d^{8} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {17 \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}} b^{2}}{256 c^{2} d^{8} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {45 \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{128 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}-\frac {45 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}+\frac {45 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}-\frac {2}{d^{6} \sqrt {d \left (2 c x +b \right )}}\right )}{\left (4 a c -b^{2}\right )^{3}}\)	\(417\)

[In]

int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

64*c^2*d^5*(-1/d^6/(4*a*c-b^2)^3*(16*(13/512*(2*c*d*x+b*d)^(7/2)+(17/128*a*c*d^2-17/512*b^2*d^2)*(2*c*d*x+b*d)
^(3/2))/((2*c*d*x+b*d)^2+4*a*c*d^2-b^2*d^2)^2+45/256/(4*a*c*d^2-b^2*d^2)^(1/4)*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)))-1/d^6/(4*a*c-b^2)^3/(2*c*
d*x+b*d)^(1/2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 3252, normalized size of antiderivative = 14.58 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(45*(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^
2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^
2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c
+ 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c
^3)*d^2)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^
5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 29
9892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*lo
g(91125*(b^20 - 40*a*b^18*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5
+ 860160*a^6*b^8*c^6 - 1966080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*(c
^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5
+ 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^1
0*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(3/4)*d^5 + 91125*s
qrt(2*c*d*x + b*d)*c^6) - 45*(2*I*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*I*(b^7*c^
2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*I*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^
3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + I*(b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x
^2 + 2*I*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + I*(a^2*b^7 - 12*a^3*b^5
*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 +
183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*
c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67
108864*a^13*c^13)*d^6))^(1/4)*log(91125*I*(b^20 - 40*a*b^18*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a
^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^6 - 1966080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a
^9*b^2*c^9 + 1048576*a^10*c^10)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^
4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187
432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^
13*c^13)*d^6))^(3/4)*d^5 + 91125*sqrt(2*c*d*x + b*d)*c^6) - 45*(-2*I*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5
- 64*a^3*c^6)*d^2*x^5 - 5*I*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 - 4*I*(b^8*c - 11
*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 - I*(b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224
*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 - 2*I*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c
^4)*d^2*x - I*(a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)*(c^8/((b^26 - 52*a*b^24*c + 1248*a
^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 2811494
4*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*
c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*log(-91125*I*(b^20 - 40*a*b^18*c + 720*a^2*b^
16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^6 - 1966080*a^7*b^6*c
^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*
c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^
12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 2
18103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(3/4)*d^5 + 91125*sqrt(2*c*d*x + b*d)*c^6) - 45*(2*(b^6*c^3
- 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b
*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b
^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 -
16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)*(c^8/((b^26
- 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736
*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^1
0 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*log(-91125*(b^20 - 40*
a*b^18*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^
6 - 1966080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*(c^8/((b^26 - 52*a*b^
24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*
c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155
712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(3/4)*d^5 + 91125*sqrt(2*c*d*x + b*d)*
c^6) - (180*c^4*x^4 + 360*b*c^3*x^3 - b^4 + 17*a*b^2*c + 128*a^2*c^2 + 27*(7*b^2*c^2 + 12*a*c^3)*x^2 + 9*(b^3*
c + 36*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*
(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 -
16*a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^
2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*
c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,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. 813 vs. \(2 (193) = 386\).

Time = 0.32 (sec) , antiderivative size = 813, normalized size of antiderivative = 3.65 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=-\frac {45 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \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 )}{\sqrt {2} b^{8} d^{3} - 16 \, \sqrt {2} a b^{6} c d^{3} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{3} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{3} + 256 \, \sqrt {2} a^{4} c^{4} d^{3}} - \frac {45 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \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 )}{\sqrt {2} b^{8} d^{3} - 16 \, \sqrt {2} a b^{6} c d^{3} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{3} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{3} + 256 \, \sqrt {2} a^{4} c^{4} d^{3}} + \frac {45 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \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 )}{2 \, {\left (\sqrt {2} b^{8} d^{3} - 16 \, \sqrt {2} a b^{6} c d^{3} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{3} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{3} + 256 \, \sqrt {2} a^{4} c^{4} d^{3}\right )}} - \frac {45 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \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 )}{2 \, {\left (\sqrt {2} b^{8} d^{3} - 16 \, \sqrt {2} a b^{6} c d^{3} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{3} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{3} + 256 \, \sqrt {2} a^{4} c^{4} d^{3}\right )}} + \frac {64 \, c^{2}}{{\left (b^{6} d - 12 \, a b^{4} c d + 48 \, a^{2} b^{2} c^{2} d - 64 \, a^{3} c^{3} d\right )} \sqrt {2 \, c d x + b d}} - \frac {2 \, {\left (17 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{2} - 68 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{2} - 13 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2}\right )}}{{\left (b^{6} d - 12 \, a b^{4} c d + 48 \, a^{2} b^{2} c^{2} d - 64 \, a^{3} c^{3} d\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-45*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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))/(sqrt(2)*b^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3
- 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 45*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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))/(sqrt(2)*b^8
*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4
*d^3) + 45/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*
d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 45/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^8
*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4
*d^3) + 64*c^2/((b^6*d - 12*a*b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d)*sqrt(2*c*d*x + b*d)) - 2*(17*(2*c*d*x
+ b*d)^(3/2)*b^2*c^2*d^2 - 68*(2*c*d*x + b*d)^(3/2)*a*c^3*d^2 - 13*(2*c*d*x + b*d)^(7/2)*c^2)/((b^6*d - 12*a*
b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {45\,c^2\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}-64\,a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}+48\,a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-12\,a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}}-\frac {\frac {64\,c^2\,d^3}{4\,a\,c-b^2}-\frac {90\,c^2\,{\left (b\,d+2\,c\,d\,x\right )}^4}{-64\,d\,a^3\,c^3+48\,d\,a^2\,b^2\,c^2-12\,d\,a\,b^4\,c+d\,b^6}+\frac {162\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^2}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4+b^4\,d^4\right )-{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}+\frac {c^2\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}\,64{}\mathrm {i}+a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,48{}\mathrm {i}-a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}\,12{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )\,45{}\mathrm {i}}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}} \]

[In]

int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x)

[Out]

(45*c^2*atan((b^6*(b*d + 2*c*d*x)^(1/2) - 64*a^3*c^3*(b*d + 2*c*d*x)^(1/2) + 48*a^2*b^2*c^2*(b*d + 2*c*d*x)^(1
/2) - 12*a*b^4*c*(b*d + 2*c*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(13/4))))/(d^(3/2)*(b^2 - 4*a*c)^(13/4)) - ((64
*c^2*d^3)/(4*a*c - b^2) - (90*c^2*(b*d + 2*c*d*x)^4)/(b^6*d - 64*a^3*c^3*d + 48*a^2*b^2*c^2*d - 12*a*b^4*c*d)
+ (162*c^2*d*(b*d + 2*c*d*x)^2)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/((b*d + 2*c*d*x)^(1/2)*(b^4*d^4 + 16*a^2*c^2*d
^4 - 8*a*b^2*c*d^4) - (b*d + 2*c*d*x)^(5/2)*(2*b^2*d^2 - 8*a*c*d^2) + (b*d + 2*c*d*x)^(9/2)) + (c^2*atan((b^6*
(b*d + 2*c*d*x)^(1/2)*1i - a^3*c^3*(b*d + 2*c*d*x)^(1/2)*64i + a^2*b^2*c^2*(b*d + 2*c*d*x)^(1/2)*48i - a*b^4*c
*(b*d + 2*c*d*x)^(1/2)*12i)/(d^(1/2)*(b^2 - 4*a*c)^(13/4)))*45i)/(d^(3/2)*(b^2 - 4*a*c)^(13/4))