\(\int \frac {(c+d x)^3}{(a+b (c+d x)^3)^3} \, dx\) [77]

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

Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {c+d x}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {c+d x}{18 a b d \left (a+b (c+d x)^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{5/3} b^{4/3} d}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d} \] Output:

-1/6*(d*x+c)/b/d/(a+b*(d*x+c)^3)^2+1/18*(d*x+c)/a/b/d/(a+b*(d*x+c)^3)-1/27 
*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/a^(5/3)/b 
^(4/3)/d+1/27*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(5/3)/b^(4/3)/d-1/54*ln(a^(2/3 
)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(5/3)/b^(4/3)/d
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {-\frac {9 \sqrt [3]{b} (c+d x)}{\left (a+b (c+d x)^3\right )^2}+\frac {3 \sqrt [3]{b} (c+d x)}{a \left (a+b (c+d x)^3\right )}+\frac {2 \sqrt {3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{5/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{5/3}}}{54 b^{4/3} d} \] Input:

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

Output:

((-9*b^(1/3)*(c + d*x))/(a + b*(c + d*x)^3)^2 + (3*b^(1/3)*(c + d*x))/(a*( 
a + b*(c + d*x)^3)) + (2*Sqrt[3]*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/( 
Sqrt[3]*a^(1/3))])/a^(5/3) + (2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/a^(5/3) 
- Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]/a^(5/3))/ 
(54*b^(4/3)*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {895, 817, 749, 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.

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

\(\Big \downarrow \) 895

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

\(\Big \downarrow \) 817

\(\displaystyle \frac {\frac {\int \frac {1}{\left (b (c+d x)^3+a\right )^2}d(c+d x)}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {\frac {\frac {2 \int \frac {1}{b (c+d x)^3+a}d(c+d x)}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 a^{2/3}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {\frac {2 \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}}{d}\)

Input:

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

Output:

(-1/6*(c + d*x)/(b*(a + b*(c + d*x)^3)^2) + ((c + d*x)/(3*a*(a + b*(c + d* 
x)^3)) + (2*(Log[a^(1/3) + b^(1/3)*(c + d*x)]/(3*a^(2/3)*b^(1/3)) + (-((Sq 
rt[3]*ArcTan[(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))))/(3*a))/(6*b))/d
 

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 749
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

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 817
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && 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 = 0.59 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.93

method result size
default \(\frac {\frac {d^{3} x^{4}}{18 a}+\frac {2 c \,d^{2} x^{3}}{9 a}+\frac {c^{2} d \,x^{2}}{3 a}-\frac {\left (-2 c^{3} b +a \right ) x}{9 b a}-\frac {c \left (-c^{3} b +2 a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 a \,b^{2} d}\) \(186\)
risch \(\frac {\frac {d^{3} x^{4}}{18 a}+\frac {2 c \,d^{2} x^{3}}{9 a}+\frac {c^{2} d \,x^{2}}{3 a}-\frac {\left (-2 c^{3} b +a \right ) x}{9 b a}-\frac {c \left (-c^{3} b +2 a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 a \,b^{2} d}\) \(186\)

Input:

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

Output:

(1/18/a*d^3*x^4+2/9*c/a*d^2*x^3+1/3*c^2/a*d*x^2-1/9/b*(-2*b*c^3+a)/a*x-1/1 
8/b*c/d*(-b*c^3+2*a)/a)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2+1/ 
27/a/b^2/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=RootOf(_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. 765 vs. \(2 (160) = 320\).

Time = 0.12 (sec) , antiderivative size = 1650, normalized size of antiderivative = 8.21 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/54*(3*a^2*b^2*d^4*x^4 + 12*a^2*b^2*c*d^3*x^3 + 18*a^2*b^2*c^2*d^2*x^2 + 
 3*a^2*b^2*c^4 - 6*a^3*b*c + 6*(2*a^2*b^2*c^3 - a^3*b)*d*x + 3*sqrt(1/3)*( 
a*b^3*d^6*x^6 + 6*a*b^3*c*d^5*x^5 + 15*a*b^3*c^2*d^4*x^4 + a*b^3*c^6 + 2*a 
^2*b^2*c^3 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b 
^2*c)*d^2*x^2 + a^3*b + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*x)*sqrt(-(a^2*b)^(1/ 
3)/b)*log((2*a*b*d^3*x^3 + 6*a*b*c*d^2*x^2 + 6*a*b*c^2*d*x + 2*a*b*c^3 - a 
^2 + 3*sqrt(1/3)*(2*a*b*d^2*x^2 + 4*a*b*c*d*x + 2*a*b*c^2 + (a^2*b)^(2/3)* 
(d*x + c) - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b) - 3*(a^2*b)^(1/3)*(a*d 
*x + a*c))/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)) - (b^2*d 
^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + 
a*b)*d^3*x^3 + 2*a*b*c^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 + 
a*b*c^2)*d*x + a^2)*(a^2*b)^(2/3)*log(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 
- (a^2*b)^(2/3)*(d*x + c) + (a^2*b)^(1/3)*a) + 2*(b^2*d^6*x^6 + 6*b^2*c*d^ 
5*x^5 + 15*b^2*c^2*d^4*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + a*b)*d^3*x^3 + 2*a* 
b*c^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*x + a^2) 
*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^3*b^4*d^7*x^6 + 6* 
a^3*b^4*c*d^6*x^5 + 15*a^3*b^4*c^2*d^5*x^4 + 2*(10*a^3*b^4*c^3 + a^4*b^3)* 
d^4*x^3 + 3*(5*a^3*b^4*c^4 + 2*a^4*b^3*c)*d^3*x^2 + 6*(a^3*b^4*c^5 + a^4*b 
^3*c^2)*d^2*x + (a^3*b^4*c^6 + 2*a^4*b^3*c^3 + a^5*b^2)*d), 1/54*(3*a^2*b^ 
2*d^4*x^4 + 12*a^2*b^2*c*d^3*x^3 + 18*a^2*b^2*c^2*d^2*x^2 + 3*a^2*b^2*c...
 

Sympy [A] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.29 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {- 2 a c + b c^{4} + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4} + x \left (- 2 a d + 4 b c^{3} d\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \cdot \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \cdot \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac {\operatorname {RootSum} {\left (19683 t^{3} a^{5} b^{4} - 1, \left ( t \mapsto t \log {\left (x + \frac {27 t a^{2} b + c}{d} \right )} \right )\right )}}{d} \] Input:

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

Output:

(-2*a*c + b*c**4 + 6*b*c**2*d**2*x**2 + 4*b*c*d**3*x**3 + b*d**4*x**4 + x* 
(-2*a*d + 4*b*c**3*d))/(18*a**3*b*d + 36*a**2*b**2*c**3*d + 18*a*b**3*c**6 
*d + 270*a*b**3*c**2*d**5*x**4 + 108*a*b**3*c*d**6*x**5 + 18*a*b**3*d**7*x 
**6 + x**3*(36*a**2*b**2*d**4 + 360*a*b**3*c**3*d**4) + x**2*(108*a**2*b** 
2*c*d**3 + 270*a*b**3*c**4*d**3) + x*(108*a**2*b**2*c**2*d**2 + 108*a*b**3 
*c**5*d**2)) + RootSum(19683*_t**3*a**5*b**4 - 1, Lambda(_t, _t*log(x + (2 
7*_t*a**2*b + c)/d)))/d
 

Maxima [F]

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

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

Output:

1/18*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + 2*(2*b*c^3 - a 
)*d*x - 2*a*c)/(a*b^3*d^7*x^6 + 6*a*b^3*c*d^6*x^5 + 15*a*b^3*c^2*d^5*x^4 + 
 2*(10*a*b^3*c^3 + a^2*b^2)*d^4*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^3*x^ 
2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d^2*x + (a*b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b 
)*d) + 1/9*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*b)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.33 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {2 \, \sqrt {3} \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{54 \, a b} + \frac {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} - 2 \, a d x - 2 \, a c}{18 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \] Input:

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

Output:

1/54*(2*sqrt(3)*(1/(a^2*b*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))) - (1/(a^2*b*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*(1/(a^2*b*d^3))^(1/3)*log(abs(b*d*x + 
b*c + (a*b^2)^(1/3))))/(a*b) + 1/18*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d 
^2*x^2 + 4*b*c^3*d*x + b*c^4 - 2*a*d*x - 2*a*c)/((b*d^3*x^3 + 3*b*c*d^2*x^ 
2 + 3*b*c^2*d*x + b*c^3 + a)^2*a*b*d)
 

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.73 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {\frac {d^3\,x^4}{18\,a}+\frac {c^2\,d\,x^2}{3\,a}+\frac {2\,c\,d^2\,x^3}{9\,a}-\frac {x\,\left (a-2\,b\,c^3\right )}{9\,a\,b}-\frac {2\,a\,c-b\,c^4}{18\,a\,b\,d}}{x^3\,\left (20\,b^2\,c^3\,d^3+2\,a\,b\,d^3\right )+x^2\,\left (15\,b^2\,c^4\,d^2+6\,a\,b\,c\,d^2\right )+a^2+x\,\left (6\,d\,b^2\,c^5+6\,a\,d\,b\,c^2\right )+b^2\,c^6+b^2\,d^6\,x^6+2\,a\,b\,c^3+6\,b^2\,c\,d^5\,x^5+15\,b^2\,c^2\,d^4\,x^4}+\frac {\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{27\,a^{5/3}\,b^{4/3}\,d}+\frac {\ln \left (\frac {b\,c\,d^5}{3\,a}+\frac {b\,d^6\,x}{3\,a}+\frac {b^{2/3}\,d^5\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{54\,a^{5/3}\,b^{4/3}\,d}-\frac {\ln \left (\frac {b\,c\,d^5}{3\,a}+\frac {b\,d^6\,x}{3\,a}-\frac {b^{2/3}\,d^5\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{54\,a^{5/3}\,b^{4/3}\,d} \] Input:

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

Output:

((d^3*x^4)/(18*a) + (c^2*d*x^2)/(3*a) + (2*c*d^2*x^3)/(9*a) - (x*(a - 2*b* 
c^3))/(9*a*b) - (2*a*c - b*c^4)/(18*a*b*d))/(x^3*(20*b^2*c^3*d^3 + 2*a*b*d 
^3) + x^2*(15*b^2*c^4*d^2 + 6*a*b*c*d^2) + a^2 + x*(6*b^2*c^5*d + 6*a*b*c^ 
2*d) + b^2*c^6 + b^2*d^6*x^6 + 2*a*b*c^3 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^ 
4*x^4) + log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x)/(27*a^(5/3)*b^(4/3)*d) + ( 
log((b*c*d^5)/(3*a) + (b*d^6*x)/(3*a) + (b^(2/3)*d^5*(3^(1/2)*1i - 1))/(6* 
a^(2/3)))*(3^(1/2)*1i - 1))/(54*a^(5/3)*b^(4/3)*d) - (log((b*c*d^5)/(3*a) 
+ (b*d^6*x)/(3*a) - (b^(2/3)*d^5*(3^(1/2)*1i + 1))/(6*a^(2/3)))*(3^(1/2)*1 
i + 1))/(54*a^(5/3)*b^(4/3)*d)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1793, normalized size of antiderivative = 8.92 \[ \int \frac {(c+d x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*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**2 - 4*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**3 - 12*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*d*x 
 - 12*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**2*x**2 - 4*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*d**3*x**3 - 2*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**6 - 12*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*d*x - 30*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**2* 
x**2 - 40*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**3*x**3 - 30*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**4*x* 
*4 - 12*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**5*x**5 - 2*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**6*x**6 - 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**2 - 2*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**...