\(\int \frac {(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^2} \, dx\) [121]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:

36*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(1/2)+36/5*c*d^3*(2*c*d*x+b*d)^(5/2)-d 
*(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)-18*c*(-4*a*c+b^2)^(5/4)*d^(11/2)*arctan 
((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-18*c*(-4*a*c+b^2)^(5/4)*d 
^(11/2)*arctanh((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.95 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.64 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{5}+\frac {i}{5}\right ) c (d (b+2 c x))^{11/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (45 b^4-360 a b^2 c+720 a^2 c^2-36 b^2 (b+2 c x)^2+144 a c (b+2 c x)^2-4 (b+2 c x)^4\right )}{c (b+2 c x)^5 (a+x (b+c x))}-\frac {45 i \left (b^2-4 a c\right )^{5/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{11/2}}+\frac {45 i \left (b^2-4 a c\right )^{5/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{11/2}}+\frac {45 i \left (b^2-4 a c\right )^{5/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{11/2}}\right ) \] Input:

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]
 

Output:

(1/5 + I/5)*c*(d*(b + 2*c*x))^(11/2)*(((-1/2 + I/2)*(45*b^4 - 360*a*b^2*c 
+ 720*a^2*c^2 - 36*b^2*(b + 2*c*x)^2 + 144*a*c*(b + 2*c*x)^2 - 4*(b + 2*c* 
x)^4))/(c*(b + 2*c*x)^5*(a + x*(b + c*x))) - ((45*I)*(b^2 - 4*a*c)^(5/4)*A 
rcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(11/ 
2) + ((45*I)*(b^2 - 4*a*c)^(5/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 
 - 4*a*c)^(1/4)])/(b + 2*c*x)^(11/2) + ((45*I)*(b^2 - 4*a*c)^(5/4)*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*c*x)^(11/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1110, 1116, 1116, 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.

\(\displaystyle \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle 9 c d^2 \int \frac {(b d+2 c x d)^{7/2}}{c x^2+b x+a}dx-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{3/2}}{c x^2+b x+a}dx+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1118

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {d \left (b^2-4 a c\right ) \int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c}+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (2 d^3 \left (b^2-4 a c\right ) \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-2 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 266

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 756

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle 9 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}\)

Input:

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]
 

Output:

-((d*(b*d + 2*c*d*x)^(9/2))/(a + b*x + c*x^2)) + 9*c*d^2*((4*d*(b*d + 2*c* 
d*x)^(5/2))/5 + (b^2 - 4*a*c)*d^2*(4*d*Sqrt[b*d + 2*c*d*x] - 4*(b^2 - 4*a* 
c)*d^3*(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) 
*Sqrt[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 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[d*e*((m - 1)/(b*(p + 1)))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^ 
2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1116

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

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. \(352\) vs. \(2(153)=306\).

Time = 1.67 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.95

method	result	size
pseudoelliptic	\(\frac {72 d^{3} \left (\frac {2 \left (d \left (2 c x +b \right )\right )^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right ) c \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}{45}+\frac {d^{2} \left (4 a c -b^{2}\right ) \left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (\frac {8 c^{2} x^{2}}{9}+\left (\frac {8 b x}{9}+a \right ) c -\frac {b^{2}}{36}\right ) \sqrt {d \left (2 c x +b \right )}+c \,d^{2} \sqrt {2}\, \left (a c -\frac {b^{2}}{4}\right ) \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )\right )\right )}{4}\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (c \,x^{2}+b x +a \right )}\)	\(353\)
derivativedivides	\(16 c \,d^{3} \left (-8 a c \,d^{2} \sqrt {2 c d x +b d}+2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} c a \,b^{2}-\frac {1}{16} b^{4}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {9 \left (4 a^{2} c^{2}-2 c a \,b^{2}+\frac {1}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(373\)
default	\(16 c \,d^{3} \left (-8 a c \,d^{2} \sqrt {2 c d x +b d}+2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} c a \,b^{2}-\frac {1}{16} b^{4}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {9 \left (4 a^{2} c^{2}-2 c a \,b^{2}+\frac {1}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(373\)

Input:

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

72*d^3*(2/45*(d*(2*c*x+b))^(5/2)*(c*x^2+b*x+a)*c*(d^2*(4*a*c-b^2))^(3/4)+1 
/4*d^2*(4*a*c-b^2)*(-2*(d^2*(4*a*c-b^2))^(3/4)*(8/9*c^2*x^2+(8/9*b*x+a)*c- 
1/36*b^2)*(d*(2*c*x+b))^(1/2)+c*d^2*2^(1/2)*(a*c-1/4*b^2)*(c*x^2+b*x+a)*(l 
n(((d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+(d^2*(4*a*c-b^2))^( 
1/2)+d*(2*c*x+b))/((d^2*(4*a*c-b^2))^(1/2)-(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c 
*x+b))^(1/2)*2^(1/2)+d*(2*c*x+b)))+2*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4 
)*(d*(2*c*x+b))^(1/2)-1)+2*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c* 
x+b))^(1/2)+1))))/(d^2*(4*a*c-b^2))^(3/4)/(c*x^2+b*x+a)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 816, normalized size of antiderivative = 4.51 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
 

Output:

1/5*(45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 12 
80*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(c*x^2 + b*x + a)*log(-9*(b^2*c 
 - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 + 9*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^ 
2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4) 
) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 128 
0*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(-9* 
(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 + 9*I*((b^10*c^4 - 20*a*b^8*c^5 
+ 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^2 
2)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c 
^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(I*c*x^2 + I*b*x + I*a)* 
log(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 - 9*I*((b^10*c^4 - 20*a*b 
^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c 
^9)*d^22)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^ 
3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(c*x^2 + b*x + a) 
*log(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 - 9*((b^10*c^4 - 20*a*b^ 
8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^ 
9)*d^22)^(1/4)) + (64*c^4*d^5*x^4 + 128*b*c^3*d^5*x^3 + 48*(5*b^2*c^2 - 12 
*a*c^3)*d^5*x^2 + 16*(11*b^3*c - 36*a*b*c^2)*d^5*x - (5*b^4 - 216*a*b^2*c 
+ 720*a^2*c^2)*d^5)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(11/2)/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,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. 646 vs. \(2 (153) = 306\).

Time = 0.14 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.57 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=32 \, \sqrt {2 \, c d x + b d} b^{2} c d^{5} - 128 \, \sqrt {2 \, c d x + b d} a c^{2} d^{5} + \frac {16}{5} \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c d^{3} - 9 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) - 9 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) - \frac {9}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) + \frac {9}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) + \frac {4 \, {\left (\sqrt {2 \, c d x + b d} b^{4} c d^{7} - 8 \, \sqrt {2 \, c d x + b d} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2 \, c d x + b d} a^{2} c^{3} d^{7}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
 

Output:

32*sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 128*sqrt(2*c*d*x + b*d)*a*c^2*d^5 + 16/ 
5*(2*c*d*x + b*d)^(5/2)*c*d^3 - 9*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^ 
2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) - 9*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^ 
5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) - 9/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 
 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*log(2*c*d*x + b*d + sqr 
t(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)) + 9/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 4*sqrt(2 
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) 
 + 4*(sqrt(2*c*d*x + b*d)*b^4*c*d^7 - 8*sqrt(2*c*d*x + b*d)*a*b^2*c^2*d^7 
+ 16*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b* 
d)^2)
 

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.61 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x)
 

Output:

(16*c*d^3*(b*d + 2*c*d*x)^(5/2))/5 - ((b*d + 2*c*d*x)^(1/2)*(4*b^4*c*d^7 + 
 64*a^2*c^3*d^7 - 32*a*b^2*c^2*d^7))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c* 
d^2) - 18*c*d^(11/2)*atan((9*c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d* 
x)^(1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 
+ 497664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) - c*d^(11/2)*(b^2 - 
4*a*c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 276 
48*a^2*b^2*c^3*d^9)*9i) + 9*c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d*x 
)^(1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 
 497664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) + c*d^(11/2)*(b^2 - 4 
*a*c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 2764 
8*a^2*b^2*c^3*d^9)*9i))/(c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d*x)^( 
1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 49 
7664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) - c*d^(11/2)*(b^2 - 4*a* 
c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 27648*a 
^2*b^2*c^3*d^9)*9i)*9i - c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d*x)^( 
1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 49 
7664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) + c*d^(11/2)*(b^2 - 4*a* 
c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 27648*a 
^2*b^2*c^3*d^9)*9i)*9i))*(b^2 - 4*a*c)^(5/4) - 32*c*d^5*(b*d + 2*c*d*x)^(1 
/2)*(4*a*c - b^2) + c*d^(11/2)*atan((b^2*(b*d + 2*c*d*x)^(1/2)*1i - a*c...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 1696, normalized size of antiderivative = 9.37 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x)
 

Output:

(sqrt(d)*d**5*( - 360*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)** 
(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a**2*c 
**2 + 90*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) 
 - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*b**2*c - 360*(4*a 
*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 
 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*b*c**2*x - 360*(4*a*c - b**2)* 
*(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/(( 
4*a*c - b**2)**(1/4)*sqrt(2)))*a*c**3*x**2 + 90*(4*a*c - b**2)**(1/4)*sqrt 
(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b** 
2)**(1/4)*sqrt(2)))*b**3*c*x + 90*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a 
*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt 
(2)))*b**2*c**2*x**2 + 360*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b* 
*2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a 
**2*c**2 - 90*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sq 
rt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*b**2*c + 360 
*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqr 
t(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*b*c**2*x + 360*(4*a*c - b 
**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x 
))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*c**3*x**2 - 90*(4*a*c - b**2)**(1/4) 
*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a...