\(\int \frac {1}{(b d+2 c d x)^4 (a+b x+c x^2)^3} \, dx\) [82]

Optimal result

Integrand size = 24, antiderivative size = 168 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\frac {140 c^2}{3 \left (b^2-4 a c\right )^3 d^4 (b+2 c x)^3}+\frac {140 c^2}{\left (b^2-4 a c\right )^4 d^4 (b+2 c x)}-\frac {1}{2 \left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \left (a+b x+c x^2\right )^2}+\frac {7 c}{\left (b^2-4 a c\right )^2 d^4 (b+2 c x)^3 \left (a+b x+c x^2\right )}-\frac {140 c^2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/2} d^4} \] Output:

140/3*c^2/(-4*a*c+b^2)^3/d^4/(2*c*x+b)^3+140*c^2/(-4*a*c+b^2)^4/d^4/(2*c*x 
+b)-1/2/(-4*a*c+b^2)/d^4/(2*c*x+b)^3/(c*x^2+b*x+a)^2+7*c/(-4*a*c+b^2)^2/d^ 
4/(2*c*x+b)^3/(c*x^2+b*x+a)-140*c^2*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/ 
(-4*a*c+b^2)^(9/2)/d^4
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\frac {\frac {64 c^2 \left (b^2-4 a c\right )}{(b+2 c x)^3}+\frac {576 c^2}{b+2 c x}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x)}{(a+x (b+c x))^2}+\frac {66 c (b+2 c x)}{a+x (b+c x)}+\frac {840 c^2 \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{6 \left (b^2-4 a c\right )^4 d^4} \] Input:

Integrate[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^3),x]
 

Output:

((64*c^2*(b^2 - 4*a*c))/(b + 2*c*x)^3 + (576*c^2)/(b + 2*c*x) - (3*(b^2 - 
4*a*c)*(b + 2*c*x))/(a + x*(b + c*x))^2 + (66*c*(b + 2*c*x))/(a + x*(b + c 
*x)) + (840*c^2*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] 
)/(6*(b^2 - 4*a*c)^4*d^4)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1111, 27, 1111, 1117, 1117, 1083, 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.

\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^3 (b d+2 c d x)^4} \, dx\)

\(\Big \downarrow \) 1111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1111

\(\displaystyle -\frac {7 c \left (-\frac {10 c \int \frac {1}{(b+2 c x)^4 \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )}\right )}{d^4 \left (b^2-4 a c\right )}-\frac {1}{2 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1117

\(\displaystyle -\frac {7 c \left (-\frac {10 c \left (\frac {\int \frac {1}{(b+2 c x)^2 \left (c x^2+b x+a\right )}dx}{b^2-4 a c}+\frac {2}{3 \left (b^2-4 a c\right ) (b+2 c x)^3}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )}\right )}{d^4 \left (b^2-4 a c\right )}-\frac {1}{2 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1117

\(\displaystyle -\frac {7 c \left (-\frac {10 c \left (\frac {\frac {\int \frac {1}{c x^2+b x+a}dx}{b^2-4 a c}+\frac {2}{\left (b^2-4 a c\right ) (b+2 c x)}}{b^2-4 a c}+\frac {2}{3 \left (b^2-4 a c\right ) (b+2 c x)^3}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )}\right )}{d^4 \left (b^2-4 a c\right )}-\frac {1}{2 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {7 c \left (-\frac {10 c \left (\frac {\frac {2}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {2 \int \frac {1}{b^2-(b+2 c x)^2-4 a c}d(b+2 c x)}{b^2-4 a c}}{b^2-4 a c}+\frac {2}{3 \left (b^2-4 a c\right ) (b+2 c x)^3}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )}\right )}{d^4 \left (b^2-4 a c\right )}-\frac {1}{2 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {7 c \left (-\frac {10 c \left (\frac {\frac {2}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}}{b^2-4 a c}+\frac {2}{3 \left (b^2-4 a c\right ) (b+2 c x)^3}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )}\right )}{d^4 \left (b^2-4 a c\right )}-\frac {1}{2 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^2}\)

Input:

Int[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^3),x]
 

Output:

-1/2*1/((b^2 - 4*a*c)*d^4*(b + 2*c*x)^3*(a + b*x + c*x^2)^2) - (7*c*(-(1/( 
(b^2 - 4*a*c)*(b + 2*c*x)^3*(a + b*x + c*x^2))) - (10*c*(2/(3*(b^2 - 4*a*c 
)*(b + 2*c*x)^3) + (2/((b^2 - 4*a*c)*(b + 2*c*x)) - (2*ArcTanh[(b + 2*c*x) 
/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2))/(b^2 - 4*a*c)))/(b^2 - 4*a*c)))/ 
((b^2 - 4*a*c)*d^4)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1111

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> 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] - Simp[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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !G 
tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
 

rule 1117

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

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\frac {96 c^{2}}{\left (4 a c -b^{2}\right )^{4} \left (2 c x +b \right )}-\frac {32 c^{2}}{3 \left (4 a c -b^{2}\right )^{3} \left (2 c x +b \right )^{3}}+\frac {\frac {22 c^{3} x^{3}+33 b \,c^{2} x^{2}+2 c \left (13 a c +5 b^{2}\right ) x +\frac {b \left (26 a c -b^{2}\right )}{2}}{\left (c \,x^{2}+b x +a \right )^{2}}+\frac {140 c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (4 a c -b^{2}\right )^{4}}}{d^{4}}\)	\(165\)
risch	\(\frac {\frac {560 c^{6} x^{6}}{256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}}+\frac {1680 b \,c^{5} x^{5}}{256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}}+\frac {2800 c^{4} \left (a c +2 b^{2}\right ) x^{4}}{3 \left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right )}+\frac {2800 b \,c^{3} \left (2 a c +b^{2}\right ) x^{3}}{3 \left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right )}+\frac {7 c^{2} \left (128 a^{2} c^{2}+536 c a \,b^{2}+83 b^{4}\right ) x^{2}}{3 \left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right )}+\frac {7 b c \left (128 a^{2} c^{2}+136 c a \,b^{2}+3 b^{4}\right ) x}{3 \left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right )}-\frac {256 a^{3} c^{3}-640 a^{2} b^{2} c^{2}-78 a \,b^{4} c +3 b^{6}}{6 \left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right )}}{d^{4} \left (2 c x +b \right )^{3} \left (c \,x^{2}+b x +a \right )^{2}}+70 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (262144 a^{9} c^{9} d^{8}-589824 a^{8} b^{2} c^{8} d^{8}+589824 a^{7} b^{4} c^{7} d^{8}-344064 a^{6} b^{6} c^{6} d^{8}+129024 a^{5} b^{8} c^{5} d^{8}-32256 a^{4} b^{10} c^{4} d^{8}+5376 a^{3} b^{12} c^{3} d^{8}-576 a^{2} b^{14} c^{2} d^{8}+36 a \,b^{16} c \,d^{8}-b^{18} d^{8}\right ) \textit {\_Z}^{2}+c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (1572864 a^{9} c^{10} d^{8}-3538944 a^{8} b^{2} c^{9} d^{8}+3538944 a^{7} b^{4} c^{8} d^{8}-2064384 a^{6} b^{6} c^{7} d^{8}+774144 a^{5} b^{8} c^{6} d^{8}-193536 a^{4} b^{10} c^{5} d^{8}+32256 a^{3} b^{12} c^{4} d^{8}-3456 a^{2} b^{14} c^{3} d^{8}+216 a \,b^{16} c^{2} d^{8}-6 b^{18} c \,d^{8}\right ) \textit {\_R}^{2}+4 c^{5}\right ) x +\left (786432 a^{9} b \,c^{9} d^{8}-1769472 a^{8} b^{3} c^{8} d^{8}+1769472 a^{7} b^{5} c^{7} d^{8}-1032192 a^{6} b^{7} c^{6} d^{8}+387072 a^{5} b^{9} c^{5} d^{8}-96768 a^{4} b^{11} c^{4} d^{8}+16128 a^{3} b^{13} c^{3} d^{8}-1728 a^{2} b^{15} c^{2} d^{8}+108 a \,b^{17} c \,d^{8}-3 b^{19} d^{8}\right ) \textit {\_R}^{2}+\left (-1024 a^{5} c^{7} d^{4}+1280 a^{4} b^{2} c^{6} d^{4}-640 a^{3} b^{4} c^{5} d^{4}+160 a^{2} b^{6} c^{4} d^{4}-20 a \,b^{8} c^{3} d^{4}+b^{10} c^{2} d^{4}\right ) \textit {\_R} +2 b \,c^{4}\right )\right )\)	\(974\)

Input:

int(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d^4*(96/(4*a*c-b^2)^4*c^2/(2*c*x+b)-32/3*c^2/(4*a*c-b^2)^3/(2*c*x+b)^3+1 
/(4*a*c-b^2)^4*((22*c^3*x^3+33*b*c^2*x^2+2*c*(13*a*c+5*b^2)*x+1/2*b*(26*a* 
c-b^2))/(c*x^2+b*x+a)^2+140*c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c- 
b^2)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1028 vs. \(2 (160) = 320\).

Time = 0.13 (sec) , antiderivative size = 2077, normalized size of antiderivative = 12.36 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^3,x, algorithm="fricas")
 

Output:

[-1/6*(3*b^8 - 90*a*b^6*c - 328*a^2*b^4*c^2 + 2816*a^3*b^2*c^3 - 1024*a^4* 
c^4 - 3360*(b^2*c^6 - 4*a*c^7)*x^6 - 10080*(b^3*c^5 - 4*a*b*c^6)*x^5 - 560 
0*(2*b^4*c^4 - 7*a*b^2*c^5 - 4*a^2*c^6)*x^4 - 5600*(b^5*c^3 - 2*a*b^3*c^4 
- 8*a^2*b*c^5)*x^3 - 14*(83*b^6*c^2 + 204*a*b^4*c^3 - 2016*a^2*b^2*c^4 - 5 
12*a^3*c^5)*x^2 - 420*(8*c^7*x^7 + 28*b*c^6*x^6 + a^2*b^3*c^2 + 2*(19*b^2* 
c^5 + 8*a*c^6)*x^5 + 5*(5*b^3*c^4 + 8*a*b*c^5)*x^4 + 4*(2*b^4*c^3 + 9*a*b^ 
2*c^4 + 2*a^2*c^5)*x^3 + (b^5*c^2 + 14*a*b^3*c^3 + 12*a^2*b*c^4)*x^2 + 2*( 
a*b^4*c^2 + 3*a^2*b^2*c^3)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + 
 b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - 14*(3*b 
^7*c + 124*a*b^5*c^2 - 416*a^2*b^3*c^3 - 512*a^3*b*c^4)*x)/(8*(b^10*c^5 - 
20*a*b^8*c^6 + 160*a^2*b^6*c^7 - 640*a^3*b^4*c^8 + 1280*a^4*b^2*c^9 - 1024 
*a^5*c^10)*d^4*x^7 + 28*(b^11*c^4 - 20*a*b^9*c^5 + 160*a^2*b^7*c^6 - 640*a 
^3*b^5*c^7 + 1280*a^4*b^3*c^8 - 1024*a^5*b*c^9)*d^4*x^6 + 2*(19*b^12*c^3 - 
 372*a*b^10*c^4 + 2880*a^2*b^8*c^5 - 10880*a^3*b^6*c^6 + 19200*a^4*b^4*c^7 
 - 9216*a^5*b^2*c^8 - 8192*a^6*c^9)*d^4*x^5 + 5*(5*b^13*c^2 - 92*a*b^11*c^ 
3 + 640*a^2*b^9*c^4 - 1920*a^3*b^7*c^5 + 1280*a^4*b^5*c^6 + 5120*a^5*b^3*c 
^7 - 8192*a^6*b*c^8)*d^4*x^4 + 4*(2*b^14*c - 31*a*b^12*c^2 + 142*a^2*b^10* 
c^3 + 120*a^3*b^8*c^4 - 2880*a^4*b^6*c^5 + 8192*a^5*b^4*c^6 - 6656*a^6*b^2 
*c^7 - 2048*a^7*c^8)*d^4*x^3 + (b^15 - 6*a*b^13*c - 108*a^2*b^11*c^2 + 136 
0*a^3*b^9*c^3 - 5760*a^4*b^7*c^4 + 9216*a^5*b^5*c^5 + 1024*a^6*b^3*c^6 ...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1238 vs. \(2 (165) = 330\).

Time = 2.49 (sec) , antiderivative size = 1238, normalized size of antiderivative = 7.37 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(2*c*d*x+b*d)**4/(c*x**2+b*x+a)**3,x)
 

Output:

-70*c**2*sqrt(-1/(4*a*c - b**2)**9)*log(x + (-71680*a**5*c**7*sqrt(-1/(4*a 
*c - b**2)**9) + 89600*a**4*b**2*c**6*sqrt(-1/(4*a*c - b**2)**9) - 44800*a 
**3*b**4*c**5*sqrt(-1/(4*a*c - b**2)**9) + 11200*a**2*b**6*c**4*sqrt(-1/(4 
*a*c - b**2)**9) - 1400*a*b**8*c**3*sqrt(-1/(4*a*c - b**2)**9) + 70*b**10* 
c**2*sqrt(-1/(4*a*c - b**2)**9) + 70*b*c**2)/(140*c**3))/d**4 + 70*c**2*sq 
rt(-1/(4*a*c - b**2)**9)*log(x + (71680*a**5*c**7*sqrt(-1/(4*a*c - b**2)** 
9) - 89600*a**4*b**2*c**6*sqrt(-1/(4*a*c - b**2)**9) + 44800*a**3*b**4*c** 
5*sqrt(-1/(4*a*c - b**2)**9) - 11200*a**2*b**6*c**4*sqrt(-1/(4*a*c - b**2) 
**9) + 1400*a*b**8*c**3*sqrt(-1/(4*a*c - b**2)**9) - 70*b**10*c**2*sqrt(-1 
/(4*a*c - b**2)**9) + 70*b*c**2)/(140*c**3))/d**4 + (-256*a**3*c**3 + 640* 
a**2*b**2*c**2 + 78*a*b**4*c - 3*b**6 + 10080*b*c**5*x**5 + 3360*c**6*x**6 
 + x**4*(5600*a*c**5 + 11200*b**2*c**4) + x**3*(11200*a*b*c**4 + 5600*b**3 
*c**3) + x**2*(1792*a**2*c**4 + 7504*a*b**2*c**3 + 1162*b**4*c**2) + x*(17 
92*a**2*b*c**3 + 1904*a*b**3*c**2 + 42*b**5*c))/(1536*a**6*b**3*c**4*d**4 
- 1536*a**5*b**5*c**3*d**4 + 576*a**4*b**7*c**2*d**4 - 96*a**3*b**9*c*d**4 
 + 6*a**2*b**11*d**4 + x**7*(12288*a**4*c**9*d**4 - 12288*a**3*b**2*c**8*d 
**4 + 4608*a**2*b**4*c**7*d**4 - 768*a*b**6*c**6*d**4 + 48*b**8*c**5*d**4) 
 + x**6*(43008*a**4*b*c**8*d**4 - 43008*a**3*b**3*c**7*d**4 + 16128*a**2*b 
**5*c**6*d**4 - 2688*a*b**7*c**5*d**4 + 168*b**9*c**4*d**4) + x**5*(24576* 
a**5*c**8*d**4 + 33792*a**4*b**2*c**7*d**4 - 49152*a**3*b**4*c**6*d**4 ...
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^3,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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.85 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\frac {140 \, c^{2} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{8} d^{4} - 16 \, a b^{6} c d^{4} + 96 \, a^{2} b^{4} c^{2} d^{4} - 256 \, a^{3} b^{2} c^{3} d^{4} + 256 \, a^{4} c^{4} d^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {44 \, c^{3} x^{3} + 66 \, b c^{2} x^{2} + 20 \, b^{2} c x + 52 \, a c^{2} x - b^{3} + 26 \, a b c}{2 \, {\left (b^{8} d^{4} - 16 \, a b^{6} c d^{4} + 96 \, a^{2} b^{4} c^{2} d^{4} - 256 \, a^{3} b^{2} c^{3} d^{4} + 256 \, a^{4} c^{4} d^{4}\right )} {\left (c x^{2} + b x + a\right )}^{2}} + \frac {64 \, {\left (18 \, c^{4} x^{2} + 18 \, b c^{3} x + 5 \, b^{2} c^{2} - 2 \, a c^{3}\right )}}{3 \, {\left (b^{8} d^{4} - 16 \, a b^{6} c d^{4} + 96 \, a^{2} b^{4} c^{2} d^{4} - 256 \, a^{3} b^{2} c^{3} d^{4} + 256 \, a^{4} c^{4} d^{4}\right )} {\left (2 \, c x + b\right )}^{3}} \] Input:

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^3,x, algorithm="giac")
 

Output:

140*c^2*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^8*d^4 - 16*a*b^6*c*d^4 
+ 96*a^2*b^4*c^2*d^4 - 256*a^3*b^2*c^3*d^4 + 256*a^4*c^4*d^4)*sqrt(-b^2 + 
4*a*c)) + 1/2*(44*c^3*x^3 + 66*b*c^2*x^2 + 20*b^2*c*x + 52*a*c^2*x - b^3 + 
 26*a*b*c)/((b^8*d^4 - 16*a*b^6*c*d^4 + 96*a^2*b^4*c^2*d^4 - 256*a^3*b^2*c 
^3*d^4 + 256*a^4*c^4*d^4)*(c*x^2 + b*x + a)^2) + 64/3*(18*c^4*x^2 + 18*b*c 
^3*x + 5*b^2*c^2 - 2*a*c^3)/((b^8*d^4 - 16*a*b^6*c*d^4 + 96*a^2*b^4*c^2*d^ 
4 - 256*a^3*b^2*c^3*d^4 + 256*a^4*c^4*d^4)*(2*c*x + b)^3)
 

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 827, normalized size of antiderivative = 4.92 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx=\frac {140\,c^2\,\mathrm {atan}\left (\frac {\frac {70\,c^2\,\left (256\,a^4\,b\,c^4\,d^4-256\,a^3\,b^3\,c^3\,d^4+96\,a^2\,b^5\,c^2\,d^4-16\,a\,b^7\,c\,d^4+b^9\,d^4\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{9/2}}+\frac {140\,c^3\,x\,\left (256\,a^4\,c^4\,d^4-256\,a^3\,b^2\,c^3\,d^4+96\,a^2\,b^4\,c^2\,d^4-16\,a\,b^6\,c\,d^4+b^8\,d^4\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{9/2}}}{70\,c^2}\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{9/2}}-\frac {\frac {2800\,x^3\,\left (b^3\,c^3+2\,a\,b\,c^4\right )}{3\,\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}-\frac {256\,a^3\,c^3-640\,a^2\,b^2\,c^2-78\,a\,b^4\,c+3\,b^6}{6\,\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {7\,x^2\,\left (128\,a^2\,c^4+536\,a\,b^2\,c^3+83\,b^4\,c^2\right )}{3\,\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {560\,c^6\,x^6}{\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {7\,b\,x\,\left (128\,a^2\,c^3+136\,a\,b^2\,c^2+3\,b^4\,c\right )}{3\,\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {1680\,b\,c^5\,x^5}{\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {2800\,c\,x^4\,\left (2\,b^2\,c^3+a\,c^4\right )}{3\,\left (4\,a\,c-b^2\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}}{x^2\,\left (12\,a^2\,b\,c^2\,d^4+14\,a\,b^3\,c\,d^4+b^5\,d^4\right )+x^5\,\left (38\,b^2\,c^3\,d^4+16\,a\,c^4\,d^4\right )+x\,\left (6\,c\,a^2\,b^2\,d^4+2\,a\,b^4\,d^4\right )+x^3\,\left (8\,a^2\,c^3\,d^4+36\,a\,b^2\,c^2\,d^4+8\,b^4\,c\,d^4\right )+x^4\,\left (25\,b^3\,c^2\,d^4+40\,a\,b\,c^3\,d^4\right )+a^2\,b^3\,d^4+8\,c^5\,d^4\,x^7+28\,b\,c^4\,d^4\,x^6} \] Input:

int(1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^3),x)
 

Output:

(140*c^2*atan(((70*c^2*(b^9*d^4 + 256*a^4*b*c^4*d^4 + 96*a^2*b^5*c^2*d^4 - 
 256*a^3*b^3*c^3*d^4 - 16*a*b^7*c*d^4))/(d^4*(4*a*c - b^2)^(9/2)) + (140*c 
^3*x*(b^8*d^4 + 256*a^4*c^4*d^4 + 96*a^2*b^4*c^2*d^4 - 256*a^3*b^2*c^3*d^4 
 - 16*a*b^6*c*d^4))/(d^4*(4*a*c - b^2)^(9/2)))/(70*c^2)))/(d^4*(4*a*c - b^ 
2)^(9/2)) - ((2800*x^3*(b^3*c^3 + 2*a*b*c^4))/(3*(4*a*c - b^2)*(b^6 - 64*a 
^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) - (3*b^6 + 256*a^3*c^3 - 640*a^2*b^ 
2*c^2 - 78*a*b^4*c)/(6*(4*a*c - b^2)*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 
12*a*b^4*c)) + (7*x^2*(128*a^2*c^4 + 83*b^4*c^2 + 536*a*b^2*c^3))/(3*(4*a* 
c - b^2)*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) + (560*c^6*x^6) 
/((4*a*c - b^2)*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) + (7*b*x 
*(3*b^4*c + 128*a^2*c^3 + 136*a*b^2*c^2))/(3*(4*a*c - b^2)*(b^6 - 64*a^3*c 
^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) + (1680*b*c^5*x^5)/((4*a*c - b^2)*(b^6 
- 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) + (2800*c*x^4*(a*c^4 + 2*b^2* 
c^3))/(3*(4*a*c - b^2)*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)))/ 
(x^2*(b^5*d^4 + 12*a^2*b*c^2*d^4 + 14*a*b^3*c*d^4) + x^5*(16*a*c^4*d^4 + 3 
8*b^2*c^3*d^4) + x*(2*a*b^4*d^4 + 6*a^2*b^2*c*d^4) + x^3*(8*b^4*c*d^4 + 8* 
a^2*c^3*d^4 + 36*a*b^2*c^2*d^4) + x^4*(25*b^3*c^2*d^4 + 40*a*b*c^3*d^4) + 
a^2*b^3*d^4 + 8*c^5*d^4*x^7 + 28*b*c^4*d^4*x^6)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1605, normalized size of antiderivative = 9.55 \[ \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^3,x)
 

Output:

(840*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**4*c** 
2 + 5040*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3 
*c**3*x + 10080*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a* 
*2*b**2*c**4*x**2 + 6720*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - 
b**2))*a**2*b*c**5*x**3 + 1680*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4* 
a*c - b**2))*a*b**5*c**2*x + 11760*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqr 
t(4*a*c - b**2))*a*b**4*c**3*x**2 + 30240*sqrt(4*a*c - b**2)*atan((b + 2*c 
*x)/sqrt(4*a*c - b**2))*a*b**3*c**4*x**3 + 33600*sqrt(4*a*c - b**2)*atan(( 
b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c**5*x**4 + 13440*sqrt(4*a*c - b**2) 
*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c**6*x**5 + 840*sqrt(4*a*c - b** 
2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**6*c**2*x**2 + 6720*sqrt(4*a*c - 
 b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**5*c**3*x**3 + 21000*sqrt(4* 
a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**4*c**4*x**4 + 31920*sq 
rt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c**5*x**5 + 235 
20*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c**6*x**6 
+ 6720*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**7*x**7 
 - 1024*a**4*b*c**4 + 2336*a**3*b**3*c**3 + 4288*a**3*b**2*c**4*x + 1408*a 
**3*b*c**5*x**2 - 3840*a**3*c**6*x**3 - 208*a**2*b**5*c**2 + 5584*a**2*b** 
4*c**3*x + 22944*a**2*b**3*c**4*x**2 + 28480*a**2*b**2*c**5*x**3 + 3200*a* 
*2*b*c**6*x**4 - 7680*a**2*c**7*x**5 - 90*a*b**7*c - 1496*a*b**6*c**2*x...