\(\int \frac {(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^2} \, dx\) [120]

Optimal result

Integrand size = 26, antiderivative size = 183 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:

44/3*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(3/2)+44/7*c*d^3*(2*c*d*x+b*d)^(7/2) 
-d*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)+22*c*(-4*a*c+b^2)^(7/4)*d^(13/2)*arc 
tan((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-22*c*(-4*a*c+b^2)^(7/4 
)*d^(13/2)*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 = 0.96 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.58 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{21}+\frac {i}{21}\right ) c (d (b+2 c x))^{13/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (77 b^4-616 a b^2 c+1232 a^2 c^2-44 b^2 (b+2 c x)^2+176 a c (b+2 c x)^2-12 (b+2 c x)^4\right )}{c (b+2 c x)^5 (a+x (b+c x))}-\frac {231 \left (b^2-4 a c\right )^{7/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{13/2}}+\frac {231 \left (b^2-4 a c\right )^{7/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{13/2}}-\frac {231 \left (b^2-4 a c\right )^{7/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{13/2}}\right ) \] Input:

Integrate[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x]
 

Output:

(1/21 + I/21)*c*(d*(b + 2*c*x))^(13/2)*(((-1/2 + I/2)*(77*b^4 - 616*a*b^2* 
c + 1232*a^2*c^2 - 44*b^2*(b + 2*c*x)^2 + 176*a*c*(b + 2*c*x)^2 - 12*(b + 
2*c*x)^4))/(c*(b + 2*c*x)^5*(a + x*(b + c*x))) - (231*(b^2 - 4*a*c)^(7/4)* 
ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(13 
/2) + (231*(b^2 - 4*a*c)^(7/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 
 4*a*c)^(1/4)])/(b + 2*c*x)^(13/2) - (231*(b^2 - 4*a*c)^(7/4)*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)^(13/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 205, 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, 827, 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.

\(\displaystyle \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle 11 c d^2 \int \frac {(b d+2 c x d)^{9/2}}{c x^2+b x+a}dx-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{5/2}}{c x^2+b x+a}dx+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1118

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {d \left (b^2-4 a c\right ) \int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c}+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (2 d^3 \left (b^2-4 a c\right ) \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-2 d^3 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 266

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 827

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle 11 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )\right )+\frac {4}{7} d (b d+2 c d x)^{7/2}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}\)

Input:

Int[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x]
 

Output:

-((d*(b*d + 2*c*d*x)^(11/2))/(a + b*x + c*x^2)) + 11*c*d^2*((4*d*(b*d + 2* 
c*d*x)^(7/2))/7 + (b^2 - 4*a*c)*d^2*((4*d*(b*d + 2*c*d*x)^(3/2))/3 - 4*(b^ 
2 - 4*a*c)*d^3*(-1/2*ArcTan[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[ 
d])]/((b^2 - 4*a*c)^(1/4)*Sqrt[d]) + ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4 
*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c)^(1/4)*Sqrt[d]))))
 

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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   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. \(343\) vs. \(2(153)=306\).

Time = 1.66 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.88

method	result	size
pseudoelliptic	\(\frac {176 d^{3} \left (-\frac {\left (a c -\frac {b^{2}}{4}\right ) d^{2} \left (\frac {8 c^{2} x^{2}}{11}+\left (\frac {8 b x}{11}+a \right ) c -\frac {3 b^{2}}{44}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{3}+\left (\frac {\left (d \left (2 c x +b \right )\right )^{\frac {7}{2}} \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{77}+\frac {\sqrt {2}\, d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )\right )}{32}\right ) \left (c \,x^{2}+b x +a \right ) c \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (c \,x^{2}+b x +a \right )}\)	\(344\)
derivativedivides	\(16 c \,d^{3} \left (-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} c a \,b^{2}-\frac {1}{16} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (44 a^{2} c^{2}-22 c a \,b^{2}+\frac {11}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(373\)
default	\(16 c \,d^{3} \left (-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} c a \,b^{2}-\frac {1}{16} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (44 a^{2} c^{2}-22 c a \,b^{2}+\frac {11}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(373\)

Input:

int((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

176*d^3/(d^2*(4*a*c-b^2))^(1/4)*(-1/3*(a*c-1/4*b^2)*d^2*(8/11*c^2*x^2+(8/1 
1*b*x+a)*c-3/44*b^2)*(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(3/2)+(1/77*(d* 
(2*c*x+b))^(7/2)*(d^2*(4*a*c-b^2))^(1/4)+1/32*2^(1/2)*d^4*(4*a*c-b^2)^2*(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)+ln(((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))/(( 
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)))))*(c*x^2+b*x+a)*c)/(c*x^2+b*x+a)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 1211, normalized size of antiderivative = 6.62 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
 

Output:

1/21*(231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 
 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c 
^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*log(-1331*(b^10*c^3 - 20*a*b^8*c^4 + 16 
0*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2* 
c*d*x + b*d)*d^19 + 1331*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2 
240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^1 
0 - 16384*a^7*c^11)*d^26)^(3/4)) - 231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^ 
2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 286 
72*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(I*c*x^2 + I*b*x + I*a)*log( 
-1331*(b^10*c^3 - 20*a*b^8*c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280* 
a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c*d*x + b*d)*d^19 + 1331*I*((b^14*c^4 - 
 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 
21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) - 23 
1*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960* 
a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^2 
6)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(-1331*(b^10*c^3 - 20*a*b^8*c^4 + 160 
*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c 
*d*x + b*d)*d^19 - 1331*I*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 
2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^ 
10 - 16384*a^7*c^11)*d^26)^(3/4)) - 231*((b^14*c^4 - 28*a*b^12*c^5 + 33...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(13/2)/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^(13/2)/(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 = 0.15 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.53 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
 

Output:

32/3*(2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 128/3*(2*c*d*x + b*d)^(3/2)*a*c^2*d 
^5 + 16/7*(2*c*d*x + b*d)^(7/2)*c*d^3 - 11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2) 
^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*c^2*d^5)*arcta 
n(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)) - 11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4 
)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) + 11/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)* 
b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) - 11/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^ 
5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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*((2*c*d*x + b*d)^(3/2)*b^4*c*d^7 - 8*(2*c*d*x + b*d)^(3/2 
)*a*b^2*c^2*d^7 + 16*(2*c*d*x + b*d)^(3/2)*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 = 0.09 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.36 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (64\,a^2\,c^3\,d^7-32\,a\,b^2\,c^2\,d^7+4\,b^4\,c\,d^7\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+22\,c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}-\frac {32\,c\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )}{3}+c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}\,1{}\mathrm {i}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}\,22{}\mathrm {i} \] Input:

int((b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x)
 

Output:

(16*c*d^3*(b*d + 2*c*d*x)^(7/2))/7 - ((b*d + 2*c*d*x)^(3/2)*(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) + 22*c*d^(13/2)*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d^( 
1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^2 - 4*a*c)^(7/4) + c*d^(13/2)*ata 
n(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4)*1i)/(d^(1/2)*(b^4 + 16*a^2*c^ 
2 - 8*a*b^2*c)))*(b^2 - 4*a*c)^(7/4)*22i - (32*c*d^5*(b*d + 2*c*d*x)^(3/2) 
*(4*a*c - b^2))/3
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1750, normalized size of antiderivative = 9.56 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x)
 

Output:

(sqrt(d)*d**6*( - 1848*(4*a*c - b**2)**(3/4)*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 + 462*(4*a*c - b**2)**(3/4)*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 - 1848*( 
4*a*c - b**2)**(3/4)*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 - 1848*(4*a*c - b* 
*2)**(3/4)*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 + 462*(4*a*c - b**2)**(3/4) 
*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 + 462*(4*a*c - b**2)**(3/4)*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 + 1848*(4*a*c - b**2)**(3/4)*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 - 462*(4*a*c - b**2)**(3/4)*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 + 1848*(4*a*c - b**2)**(3/4)*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 + 1848* 
(4*a*c - b**2)**(3/4)*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 - 462*(4*a*c - b 
**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*...