\(\int \frac {1}{(c e+d e x)^3 (a+b (c+d x)^3)^2} \, dx\) [99]

Optimal result

Integrand size = 24, antiderivative size = 203 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {1}{2 a^2 d e^3 (c+d x)^2}-\frac {b (c+d x)}{3 a^2 d e^3 \left (a+b (c+d x)^3\right )}+\frac {5 b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} d e^3}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3} \] Output:

-1/2/a^2/d/e^3/(d*x+c)^2-1/3*b*(d*x+c)/a^2/d/e^3/(a+b*(d*x+c)^3)+5/9*b^(2/ 
3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/a^(8/3) 
/d/e^3-5/9*b^(2/3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(8/3)/d/e^3+5/18*b^(2/3)* 
ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(8/3)/d/e^3
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {-\frac {9 a^{2/3}}{(c+d x)^2}-\frac {6 a^{2/3} b (c+d x)}{a+b (c+d x)^3}-10 \sqrt {3} b^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )+5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3} \] Input:

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

Output:

((-9*a^(2/3))/(c + d*x)^2 - (6*a^(2/3)*b*(c + d*x))/(a + b*(c + d*x)^3) - 
10*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3 
))] - 10*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] + 5*b^(2/3)*Log[a^(2/3) 
- a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d*e^3)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {895, 819, 847, 750, 16, 1142, 25, 27, 1082, 217, 1103}

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.

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

\(\Big \downarrow \) 895

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

\(\Big \downarrow \) 819

\(\displaystyle \frac {\frac {5 \int \frac {1}{(c+d x)^3 \left (b (c+d x)^3+a\right )}d(c+d x)}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d e^3}\)

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 750

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(1/(3*a*(c + d*x)^2*(a + b*(c + d*x)^3)) + (5*(-1/2*1/(a*(c + d*x)^2) - (b 
*(Log[a^(1/3) + b^(1/3)*(c + d*x)]/(3*a^(2/3)*b^(1/3)) + (-((Sqrt[3]*ArcTa 
n[(1 - (2*b^(1/3)*(c + d*x))/a^(1/3))/Sqrt[3]])/b^(1/3)) - Log[a^(2/3) - a 
^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]/(2*b^(1/3)))/(3*a^(2/3)))) 
/a))/(3*a))/(d*e^3)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 895

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Simp[u^m/(Coeff 
icient[v, x, 1]*v^m)   Subst[Int[x^m*(a + b*x^n)^p, x], x, v], x] /; FreeQ[ 
{a, b, m, n, p}, x] && LinearPairQ[u, v, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.86 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {-\frac {1}{2 a^{2} d \left (x d +c \right )^{2}}-\frac {b \left (\frac {\frac {x}{3}+\frac {c}{3 d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{9 b d}\right )}{a^{2}}}{e^{3}}\)	\(144\)
risch	\(\frac {-\frac {5 b \,d^{2} x^{3}}{6 a^{2}}-\frac {5 b c d \,x^{2}}{2 a^{2}}-\frac {5 b x \,c^{2}}{2 a^{2}}-\frac {5 c^{3} b +3 a}{6 d \,a^{2}}}{e^{3} \left (x d +c \right )^{2} \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{8} d^{3} e^{9} \textit {\_Z}^{3}+b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{8} d^{4} e^{9} \textit {\_R}^{3}-3 d \,b^{2}\right ) x -4 a^{8} c \,d^{3} e^{9} \textit {\_R}^{3}-a^{3} b d \,e^{3} \textit {\_R} -3 b^{2} c \right )\right )}{9}\)	\(182\)

Input:

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

Output:

1/e^3*(-1/2/a^2/d/(d*x+c)^2-b/a^2*((1/3*x+1/3*c/d)/(b*d^3*x^3+3*b*c*d^2*x^ 
2+3*b*c^2*d*x+b*c^3+a)+5/9/b/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=R 
ootOf(_Z^3*b*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (162) = 324\).

Time = 0.11 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.60 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {15 \, b d^{3} x^{3} + 45 \, b c d^{2} x^{2} + 45 \, b c^{2} d x + 15 \, b c^{3} - 10 \, \sqrt {3} {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (a d x + a c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) + 5 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + {\left (a b d x + a b c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b d x + b c - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 9 \, a}{18 \, {\left (a^{2} b d^{6} e^{3} x^{5} + 5 \, a^{2} b c d^{5} e^{3} x^{4} + 10 \, a^{2} b c^{2} d^{4} e^{3} x^{3} + {\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} e^{3} x^{2} + {\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} e^{3} x + {\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d e^{3}\right )}} \] Input:

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

Output:

-1/18*(15*b*d^3*x^3 + 45*b*c*d^2*x^2 + 45*b*c^2*d*x + 15*b*c^3 - 10*sqrt(3 
)*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d 
^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqr 
t(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) + 5*(b*d^5*x^5 + 5*b*c 
*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5* 
b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c 
^2 + a^2*(-b^2/a^2)^(2/3) + (a*b*d*x + a*b*c)*(-b^2/a^2)^(1/3)) - 10*(b*d^ 
5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 
+ a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b*d*x + b*c - a*(-b^ 
2/a^2)^(1/3)) + 9*a)/(a^2*b*d^6*e^3*x^5 + 5*a^2*b*c*d^5*e^3*x^4 + 10*a^2*b 
*c^2*d^4*e^3*x^3 + (10*a^2*b*c^3 + a^3)*d^3*e^3*x^2 + (5*a^2*b*c^4 + 2*a^3 
*c)*d^2*e^3*x + (a^2*b*c^5 + a^3*c^2)*d*e^3)
 

Sympy [A] (verification not implemented)

Time = 2.25 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.15 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {- 3 a - 5 b c^{3} - 15 b c^{2} d x - 15 b c d^{2} x^{2} - 5 b d^{3} x^{3}}{6 a^{3} c^{2} d e^{3} + 6 a^{2} b c^{5} d e^{3} + 60 a^{2} b c^{2} d^{4} e^{3} x^{3} + 30 a^{2} b c d^{5} e^{3} x^{4} + 6 a^{2} b d^{6} e^{3} x^{5} + x^{2} \cdot \left (6 a^{3} d^{3} e^{3} + 60 a^{2} b c^{3} d^{3} e^{3}\right ) + x \left (12 a^{3} c d^{2} e^{3} + 30 a^{2} b c^{4} d^{2} e^{3}\right )} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{8} + 125 b^{2}, \left ( t \mapsto t \log {\left (x + \frac {- 9 t a^{3} + 5 b c}{5 b d} \right )} \right )\right )}}{d e^{3}} \] Input:

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

Output:

(-3*a - 5*b*c**3 - 15*b*c**2*d*x - 15*b*c*d**2*x**2 - 5*b*d**3*x**3)/(6*a* 
*3*c**2*d*e**3 + 6*a**2*b*c**5*d*e**3 + 60*a**2*b*c**2*d**4*e**3*x**3 + 30 
*a**2*b*c*d**5*e**3*x**4 + 6*a**2*b*d**6*e**3*x**5 + x**2*(6*a**3*d**3*e** 
3 + 60*a**2*b*c**3*d**3*e**3) + x*(12*a**3*c*d**2*e**3 + 30*a**2*b*c**4*d* 
*2*e**3)) + RootSum(729*_t**3*a**8 + 125*b**2, Lambda(_t, _t*log(x + (-9*_ 
t*a**3 + 5*b*c)/(5*b*d))))/(d*e**3)
 

Maxima [F]

\[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{2} {\left (d e x + c e\right )}^{3}} \,d x } \] Input:

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

Output:

-1/6*(5*b*d^3*x^3 + 15*b*c*d^2*x^2 + 15*b*c^2*d*x + 5*b*c^3 + 3*a)/(a^2*b* 
d^6*e^3*x^5 + 5*a^2*b*c*d^5*e^3*x^4 + 10*a^2*b*c^2*d^4*e^3*x^3 + (10*a^2*b 
*c^3 + a^3)*d^3*e^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*e^3*x + (a^2*b*c^5 + 
 a^3*c^2)*d*e^3) - 5/3*b*integrate(1/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2* 
d*x + b*c^3 + a), x)/(a^2*e^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {5 \, {\left (2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}} \right |}\right )\right )}}{18 \, a^{2} e^{3}} - \frac {b d x + b c}{3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a^{2} d e^{3}} - \frac {1}{2 \, {\left (d x + c\right )}^{2} a^{2} d e^{3}} \] Input:

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

Output:

5/18*(2*sqrt(3)*(-b^2/(a^2*d^3))^(1/3)*arctan(-(b*d*x + b*c - (-a*b^2)^(1/ 
3))/(sqrt(3)*b*d*x + sqrt(3)*b*c + sqrt(3)*(-a*b^2)^(1/3))) - (-b^2/(a^2*d 
^3))^(1/3)*log(4*(sqrt(3)*b*d*x + sqrt(3)*b*c + sqrt(3)*(-a*b^2)^(1/3))^2 
+ 4*(b*d*x + b*c - (-a*b^2)^(1/3))^2) + 2*(-b^2/(a^2*d^3))^(1/3)*log(abs(b 
*d*x + b*c - (-a*b^2)^(1/3))))/(a^2*e^3) - 1/3*(b*d*x + b*c)/((b*d^3*x^3 + 
 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*a^2*d*e^3) - 1/2/((d*x + c)^2*a^ 
2*d*e^3)
 

Mupad [B] (verification not implemented)

Time = 1.27 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {\frac {5\,b\,c^3+3\,a}{6\,a^2\,d}+\frac {5\,b\,d^2\,x^3}{6\,a^2}+\frac {5\,b\,c^2\,x}{2\,a^2}+\frac {5\,b\,c\,d\,x^2}{2\,a^2}}{x\,\left (5\,b\,d\,c^4\,e^3+2\,a\,d\,c\,e^3\right )+x^2\,\left (10\,b\,c^3\,d^2\,e^3+a\,d^2\,e^3\right )+a\,c^2\,e^3+b\,c^5\,e^3+b\,d^5\,e^3\,x^5+5\,b\,c\,d^4\,e^3\,x^4+10\,b\,c^2\,d^3\,e^3\,x^3}-\frac {5\,b^{2/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{8/3}\,d\,e^3}+\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d\,e^3}-\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d\,e^3} \] Input:

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

Output:

(5*b^(2/3)*log(2*b^(1/3)*c - 3^(1/2)*a^(1/3)*1i - a^(1/3) + 2*b^(1/3)*d*x) 
*((3^(1/2)*1i)/2 + 1/2))/(9*a^(8/3)*d*e^3) - (5*b^(2/3)*log(b^(1/3)*c + a^ 
(1/3) + b^(1/3)*d*x))/(9*a^(8/3)*d*e^3) - ((3*a + 5*b*c^3)/(6*a^2*d) + (5* 
b*d^2*x^3)/(6*a^2) + (5*b*c^2*x)/(2*a^2) + (5*b*c*d*x^2)/(2*a^2))/(x*(2*a* 
c*d*e^3 + 5*b*c^4*d*e^3) + x^2*(a*d^2*e^3 + 10*b*c^3*d^2*e^3) + a*c^2*e^3 
+ b*c^5*e^3 + b*d^5*e^3*x^5 + 5*b*c*d^4*e^3*x^4 + 10*b*c^2*d^3*e^3*x^3) - 
(5*b^(2/3)*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*c - a^(1/3) + 2*b^(1/3)*d*x) 
*((3^(1/2)*1i)/2 - 1/2))/(9*a^(8/3)*d*e^3)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1333, normalized size of antiderivative = 6.57 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(10*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**( 
1/3)*sqrt(3)))*a*b*c**2 + 20*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)* 
c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*a*b*c*d*x + 10*a**(1/3)*sqrt(3)*at 
an((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*a*b*d**2 
*x**2 + 10*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x 
)/(a**(1/3)*sqrt(3)))*b**2*c**5 + 50*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b 
**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**2*c**4*d*x + 100*a**(1/ 
3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt( 
3)))*b**2*c**3*d**2*x**2 + 100*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3 
)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**2*c**2*d**3*x**3 + 50*a**(1/3 
)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3 
)))*b**2*c*d**4*x**4 + 10*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 
 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**2*d**5*x**5 + 5*a**(1/3)*log(a**(2 
/3) - b**(1/3)*a**(1/3)*c - b**(1/3)*a**(1/3)*d*x + b**(2/3)*c**2 + 2*b**( 
2/3)*c*d*x + b**(2/3)*d**2*x**2)*a*b*c**2 + 10*a**(1/3)*log(a**(2/3) - b** 
(1/3)*a**(1/3)*c - b**(1/3)*a**(1/3)*d*x + b**(2/3)*c**2 + 2*b**(2/3)*c*d* 
x + b**(2/3)*d**2*x**2)*a*b*c*d*x + 5*a**(1/3)*log(a**(2/3) - b**(1/3)*a** 
(1/3)*c - b**(1/3)*a**(1/3)*d*x + b**(2/3)*c**2 + 2*b**(2/3)*c*d*x + b**(2 
/3)*d**2*x**2)*a*b*d**2*x**2 + 5*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3) 
*c - b**(1/3)*a**(1/3)*d*x + b**(2/3)*c**2 + 2*b**(2/3)*c*d*x + b**(2/3...