\(\int \frac {1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)^2} \, dx\) [1306]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 203 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {18 c \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{7/2}}+\frac {18 c \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{7/2}} \]

[Out]

-36/5*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(5/2)-1/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)-18*c*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^(7/2)+18*c*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^(7/2)-36*c/(-4*a*c+b^2)^3/d^3/(2*c*d*x+b*d)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 203, 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} \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=-\frac {18 c \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac {18 c \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}-\frac {36 c}{d^3 \left (b^2-4 a c\right )^3 \sqrt {b d+2 c d x}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{5/2}}-\frac {36 c}{5 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}} \]

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-36*c)/(5*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(5/2)) - (36*c)/((b^2 - 4*a*c)^3*d^3*Sqrt[b*d + 2*c*d*x]) - 1/((b
^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)) - (18*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(7/2)) + (18*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])
/((b^2 - 4*a*c)^(13/4)*d^(7/2))

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}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2 d^2} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^4} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {9 \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 )}{2 \left (b^2-4 a c\right )^3 d^5} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {9 \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 )}{\left (b^2-4 a c\right )^3 d^5} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}+\frac {(18 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 d^3}-\frac {(18 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 d^3} \\ & = -\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {18 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 )^{13/4} d^{7/2}}+\frac {18 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 )^{13/4} d^{7/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.88 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {\left (\frac {1}{5}+\frac {i}{5}\right ) c \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 c x) \left (4 b^4-32 a b^2 c+64 a^2 c^2+36 b^2 (b+2 c x)^2-144 a c (b+2 c x)^2-45 (b+2 c x)^4\right )}{c \left (b^2-4 a c\right )^3 (a+x (b+c x))}+\frac {45 (b+2 c x)^{7/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}-\frac {45 (b+2 c x)^{7/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}+\frac {45 (b+2 c x)^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{13/4}}\right )}{(d (b+2 c x))^{7/2}} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

((1/5 + I/5)*c*(((1/2 - I/2)*(b + 2*c*x)*(4*b^4 - 32*a*b^2*c + 64*a^2*c^2 + 36*b^2*(b + 2*c*x)^2 - 144*a*c*(b
+ 2*c*x)^2 - 45*(b + 2*c*x)^4))/(c*(b^2 - 4*a*c)^3*(a + x*(b + c*x))) + (45*(b + 2*c*x)^(7/2)*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)^(7/2)*ArcTan[1 + ((1 + I)*Sq
rt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(13/4) + (45*(b + 2*c*x)^(7/2)*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))^(7/2)

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.72

method result size
derivativedivides \(16 c \,d^{3} \left (-\frac {1}{5 d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {2}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}+\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 a c \,d^{2}-4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) \(350\)
default \(16 c \,d^{3} \left (-\frac {1}{5 d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {2}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}+\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 a c \,d^{2}-4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) \(350\)
pseudoelliptic \(8 d^{3} c \left (\frac {\left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{8 c \,d^{8} \left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right )^{3}}+\frac {9 \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{16 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6} \left (4 a c -b^{2}\right )^{3}}+\frac {9 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6} \left (4 a c -b^{2}\right )^{3}}-\frac {9 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6} \left (4 a c -b^{2}\right )^{3}}-\frac {2}{5 d^{4} \left (4 a c -b^{2}\right )^{2} \left (d \left (2 c x +b \right )\right )^{\frac {5}{2}}}+\frac {4}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {d \left (2 c x +b \right )}}\right )\) \(426\)

[In]

int(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

16*c*d^3*(-1/5/d^4/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(5/2)+2/d^6/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(1/2)+1/d^6/(4*a*c-b^2)
^3*(1/16*(2*c*d*x+b*d)^(3/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b*d)^2)+9/32/(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))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 3285, normalized size of antiderivative = 16.18 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*(45*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^
2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4
*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 -
 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^
2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((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 - 67108864*a^13*c^13
)*d^14))^(1/4)*log(729*(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 - 25804
8*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 + 104857
6*a^10*c^10)*d^11*(c^4/((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 - 1
317888*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 - 67108864*a^13*c^13)*d^14)
)^(3/4) + 729*sqrt(2*c*d*x + b*d)*c^3) + 45*(-8*I*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x
^5 - 20*I*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 - 2*I*(9*b^8*c^2 - 104*a*b^6*c^3 +
384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 - I*(7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a
^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 - I*(b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4
)*d^4*x - I*(a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((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 - 67108864*a^13*c^13)*d^14))^(1/4)*log(729*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)*d^11*(c^4/((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 - 67108864*a^13*c^13)*d^14))^(3/4) + 729*sqrt(2*c*d*x + b*d)*c^3) + 45*(8*I*(b^6*c^4 -
 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*I*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*
b*c^6)*d^4*x^4 + 2*I*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + I
*(7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + I*(b^10 - 6*a*b^8*c -
24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + I*(a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*
b^3*c^3)*d^4)*(c^4/((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 - 13178
88*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 - 67108864*a^13*c^13)*d^14))^(1
/4)*log(-729*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)*d^11*(c^4/((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 + 2
99892736*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^14))^(3/4)
+ 729*sqrt(2*c*d*x + b*d)*c^3) - 45*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^
7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4
 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*
a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 -
12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((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 + 843448
32*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^14))^(1/4)*log(-729*(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)*d^11*(c^4/((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^1
2 - 67108864*a^13*c^13)*d^14))^(3/4) + 729*sqrt(2*c*d*x + b*d)*c^3) - (720*c^4*x^4 + 1440*b*c^3*x^3 + 5*b^4 +
176*a*b^2*c - 64*a^2*c^2 + 72*(13*b^2*c^2 + 8*a*c^3)*x^2 + 72*(3*b^3*c + 8*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(8
*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5
- 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*
x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c
 - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4
*b^3*c^3)*d^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(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. 779 vs. \(2 (175) = 350\).

Time = 0.31 (sec) , antiderivative size = 779, normalized size of antiderivative = 3.84 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {9 \, \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^{8} d^{5} - 16 \, a b^{6} c d^{5} + 96 \, a^{2} b^{4} c^{2} d^{5} - 256 \, a^{3} b^{2} c^{3} d^{5} + 256 \, a^{4} c^{4} d^{5}} + \frac {9 \, \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^{8} d^{5} - 16 \, a b^{6} c d^{5} + 96 \, a^{2} b^{4} c^{2} d^{5} - 256 \, a^{3} b^{2} c^{3} d^{5} + 256 \, a^{4} c^{4} d^{5}} - \frac {9 \, {\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^{8} d^{5} - 16 \, \sqrt {2} a b^{6} c d^{5} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{5} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{5} + 256 \, \sqrt {2} a^{4} c^{4} d^{5}} + \frac {9 \, {\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^{8} d^{5} - 16 \, \sqrt {2} a b^{6} c d^{5} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{5} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{5} + 256 \, \sqrt {2} a^{4} c^{4} d^{5}} + \frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c}{{\left (b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}} - \frac {16 \, {\left (b^{2} c d^{2} - 4 \, a c^{2} d^{2} + 10 \, {\left (2 \, c d x + b d\right )}^{2} c\right )}}{5 \, {\left (b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*d^5 - 256*a^3*b^2*c^3*d^
5 + 256*a^4*c^4*d^5) + 9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*
d^5 - 256*a^3*b^2*c^3*d^5 + 256*a^4*c^4*d^5) - 9*(-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^8*d^5 - 16*sqrt(2)*a*b
^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 9*(-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^8*d^5 - 16*sqrt(2)*a*b^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*
c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 4*(2*c*d*x + b*d)^(3/2)*c/((b^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3
 - 64*a^3*c^3*d^3)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/5*(b^2*c*d^2 - 4*a*c^2*d^2 + 10*(2*c*d*x +
b*d)^2*c)/((b^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3)*(2*c*d*x + b*d)^(5/2))

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {\frac {36\,c\,{\left (b\,d+2\,c\,d\,x\right )}^4}{{\left (b^2\,d-4\,a\,c\,d\right )}^3}+\frac {16\,c\,d}{5\,\left (4\,a\,c-b^2\right )}-\frac {144\,c\,{\left (b\,d+2\,c\,d\,x\right )}^2}{5\,d\,{\left (4\,a\,c-b^2\right )}^2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (b^2\,d^2-4\,a\,c\,d^2\right )-{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}-\frac {18\,c\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}-64\,a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}+48\,a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-12\,a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}}-\frac {c\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}\,64{}\mathrm {i}+a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,48{}\mathrm {i}-a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}\,12{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )\,18{}\mathrm {i}}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}} \]

[In]

int(1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x)

[Out]

((36*c*(b*d + 2*c*d*x)^4)/(b^2*d - 4*a*c*d)^3 + (16*c*d)/(5*(4*a*c - b^2)) - (144*c*(b*d + 2*c*d*x)^2)/(5*d*(4
*a*c - b^2)^2))/((b*d + 2*c*d*x)^(5/2)*(b^2*d^2 - 4*a*c*d^2) - (b*d + 2*c*d*x)^(9/2)) - (18*c*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^(7/2)*(b^2 - 4*a*c)^(13/4)) - (c*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)))*18i)/(d^(7/2)*(b^2 - 4*a*c)^(13/4))