\(\int \frac {(c+d x)^4}{(a+b (c+d x)^3)^3} \, dx\) [76]

Optimal result

Integrand size = 21, antiderivative size = 205 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {(c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {(c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{4/3} b^{5/3} d}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{4/3} b^{5/3} d} \] Output:

-1/6*(d*x+c)^2/b/d/(a+b*(d*x+c)^3)^2+1/9*(d*x+c)^2/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^(4/3 
)/b^(5/3)/d-1/27*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(4/3)/b^(5/3)/d+1/54*ln(a^( 
2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(4/3)/b^(5/3)/d
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {-\frac {9 b^{2/3} (c+d x)^2}{\left (a+b (c+d x)^3\right )^2}+\frac {6 b^{2/3} (c+d x)^2}{a \left (a+b (c+d x)^3\right )}+\frac {2 \sqrt {3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{4/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{4/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{4/3}}}{54 b^{5/3} d} \] Input:

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

Output:

((-9*b^(2/3)*(c + d*x)^2)/(a + b*(c + d*x)^3)^2 + (6*b^(2/3)*(c + d*x)^2)/ 
(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^(4/3) - (2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/a^(4 
/3) + Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]/a^(4/ 
3))/(54*b^(5/3)*d)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {895, 817, 819, 821, 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.

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

\(\Big \downarrow \) 895

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(-1/6*(c + d*x)^2/(b*(a + b*(c + d*x)^3)^2) + ((c + d*x)^2/(3*a*(a + b*(c 
+ d*x)^3)) + (-1/3*Log[a^(1/3) + b^(1/3)*(c + d*x)]/(a^(1/3)*b^(2/3)) + (- 
((Sqrt[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^(1/3)*b^(1/3)))/(3*a))/(3*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 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 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 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(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 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 = 0.60 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {d^{4} x^{5}}{9 a}+\frac {5 c \,d^{3} x^{4}}{9 a}+\frac {10 c^{2} d^{2} x^{3}}{9 a}-\frac {d \left (-20 c^{3} b +a \right ) x^{2}}{18 b a}-\frac {c \left (-5 c^{3} b +a \right ) x}{9 b a}-\frac {c^{2} \left (-2 c^{3} b +a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 a \,b^{2} d}\)	\(214\)
risch	\(\frac {\frac {d^{4} x^{5}}{9 a}+\frac {5 c \,d^{3} x^{4}}{9 a}+\frac {10 c^{2} d^{2} x^{3}}{9 a}-\frac {d \left (-20 c^{3} b +a \right ) x^{2}}{18 b a}-\frac {c \left (-5 c^{3} b +a \right ) x}{9 b a}-\frac {c^{2} \left (-2 c^{3} b +a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 a \,b^{2} d}\)	\(214\)

Input:

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

Output:

(1/9*d^4/a*x^5+5/9*c*d^3/a*x^4+10/9*c^2*d^2/a*x^3-1/18/b*d*(-20*b*c^3+a)/a 
*x^2-1/9/b*c*(-5*b*c^3+a)/a*x-1/18/b*c^2/d*(-2*b*c^3+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((_R*d+c)/(_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. 792 vs. \(2 (164) = 328\).

Time = 0.12 (sec) , antiderivative size = 1706, normalized size of antiderivative = 8.32 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/54*(6*a*b^3*d^5*x^5 + 30*a*b^3*c*d^4*x^4 + 60*a*b^3*c^2*d^3*x^3 + 6*a*b 
^3*c^5 - 3*a^2*b^2*c^2 + 3*(20*a*b^3*c^3 - a^2*b^2)*d^2*x^2 + 6*(5*a*b^3*c 
^4 - a^2*b^2*c)*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*b^2)^(1/3)/a)*log((2*b^2*d^3*x^3 + 6*b^2*c*d^2* 
x^2 + 6*b^2*c^2*d*x + 2*b^2*c^3 - a*b + 3*sqrt(1/3)*(a*b*d*x + a*b*c + 2*( 
d^2*x^2 + 2*c*d*x + c^2)*(-a*b^2)^(2/3) + (-a*b^2)^(1/3)*a)*sqrt((-a*b^2)^ 
(1/3)/a) - 3*(-a*b^2)^(2/3)*(d*x + 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*b^2)^(2/3)*log(b^2*d^2 
*x^2 + 2*b^2*c*d*x + b^2*c^2 + (-a*b^2)^(1/3)*(b*d*x + b*c) + (-a*b^2)^(2/ 
3)) - 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*b^2)^(2/3)*log(b*d*x + b*c - (-a*b^2 
)^(1/3)))/(a^2*b^5*d^7*x^6 + 6*a^2*b^5*c*d^6*x^5 + 15*a^2*b^5*c^2*d^5*x^4 
+ 2*(10*a^2*b^5*c^3 + a^3*b^4)*d^4*x^3 + 3*(5*a^2*b^5*c^4 + 2*a^3*b^4*c)*d 
^3*x^2 + 6*(a^2*b^5*c^5 + a^3*b^4*c^2)*d^2*x + (a^2*b^5*c^6 + 2*a^3*b^4*c^ 
3 + a^4*b^3)*d), 1/54*(6*a*b^3*d^5*x^5 + 30*a*b^3*c*d^4*x^4 + 60*a*b^3*...
 

Sympy [A] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {- a c^{2} + 2 b c^{5} + 20 b c^{2} d^{3} x^{3} + 10 b c d^{4} x^{4} + 2 b d^{5} x^{5} + x^{2} \left (- a d^{2} + 20 b c^{3} d^{2}\right ) + x \left (- 2 a c d + 10 b c^{4} d\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \cdot \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \cdot \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac {\operatorname {RootSum} {\left (19683 t^{3} a^{4} b^{5} + 1, \left ( t \mapsto t \log {\left (x + \frac {729 t^{2} a^{3} b^{3} + c}{d} \right )} \right )\right )}}{d} \] Input:

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

Output:

(-a*c**2 + 2*b*c**5 + 20*b*c**2*d**3*x**3 + 10*b*c*d**4*x**4 + 2*b*d**5*x* 
*5 + x**2*(-a*d**2 + 20*b*c**3*d**2) + x*(-2*a*c*d + 10*b*c**4*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**4*b**5 + 1, Lambda(_t, _t*log(x + (729*_t**2*a**3*b**3 + c)/d)))/d
 

Maxima [F]

\[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\int { \frac {{\left (d x + c\right )}^{4}}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{3}} \,d x } \] Input:

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

Output:

1/18*(2*b*d^5*x^5 + 10*b*c*d^4*x^4 + 20*b*c^2*d^3*x^3 + 2*b*c^5 + (20*b*c^ 
3 - a)*d^2*x^2 - a*c^2 + 2*(5*b*c^4 - a*c)*d*x)/(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((d*x + c)/(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 = 274, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {2 \, \sqrt {3} \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{54 \, a b} + \frac {2 \, b d^{5} x^{5} + 10 \, b c d^{4} x^{4} + 20 \, b c^{2} d^{3} x^{3} + 20 \, b c^{3} d^{2} x^{2} + 10 \, b c^{4} d x + 2 \, b c^{5} - a d^{2} x^{2} - 2 \, a c d x - a c^{2}}{18 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \] Input:

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

Output:

-1/54*(2*sqrt(3)*(-1/(a*b^2*d^3))^(1/3)*arctan(1/3*sqrt(3)*(2*a*b*d*x + 2* 
a*b*c - (-a^2*b)^(2/3))/(-a^2*b)^(2/3)) + (-1/(a*b^2*d^3))^(1/3)*log((2*a* 
b*d*x + 2*a*b*c - (-a^2*b)^(2/3))^2 + 3*(-a^2*b)^(4/3)) - 2*(-1/(a*b^2*d^3 
))^(1/3)*log(abs(a*b*d*x + a*b*c + (-a^2*b)^(2/3))))/(a*b) + 1/18*(2*b*d^5 
*x^5 + 10*b*c*d^4*x^4 + 20*b*c^2*d^3*x^3 + 20*b*c^3*d^2*x^2 + 10*b*c^4*d*x 
 + 2*b*c^5 - a*d^2*x^2 - 2*a*c*d*x - a*c^2)/((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 = 1.12 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.95 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {\frac {d^4\,x^5}{9\,a}-\frac {a\,c^2-2\,b\,c^5}{18\,a\,b\,d}+\frac {5\,c\,d^3\,x^4}{9\,a}+\frac {10\,c^2\,d^2\,x^3}{9\,a}-\frac {c\,x\,\left (a-5\,b\,c^3\right )}{9\,a\,b}-\frac {d\,x^2\,\left (a-20\,b\,c^3\right )}{18\,a\,b}}{x^3\,\left (20\,b^2\,c^3\,d^3+2\,a\,b\,d^3\right )+x^2\,\left (15\,b^2\,c^4\,d^2+6\,a\,b\,c\,d^2\right )+a^2+x\,\left (6\,d\,b^2\,c^5+6\,a\,d\,b\,c^2\right )+b^2\,c^6+b^2\,d^6\,x^6+2\,a\,b\,c^3+6\,b^2\,c\,d^5\,x^5+15\,b^2\,c^2\,d^4\,x^4}+\frac {\ln \left (a^{1/3}\,b+{\left (-b\right )}^{4/3}\,c+{\left (-b\right )}^{4/3}\,d\,x\right )}{27\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d}+\frac {\ln \left (\frac {c\,d^4}{81\,a^2\,b}+\frac {d^5\,x}{81\,a^2\,b}+\frac {d^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{324\,a^{5/3}\,{\left (-b\right )}^{4/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{54\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d}-\frac {\ln \left (\frac {c\,d^4}{81\,a^2\,b}+\frac {d^5\,x}{81\,a^2\,b}+\frac {d^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{324\,a^{5/3}\,{\left (-b\right )}^{4/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{54\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d} \] Input:

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

Output:

((d^4*x^5)/(9*a) - (a*c^2 - 2*b*c^5)/(18*a*b*d) + (5*c*d^3*x^4)/(9*a) + (1 
0*c^2*d^2*x^3)/(9*a) - (c*x*(a - 5*b*c^3))/(9*a*b) - (d*x^2*(a - 20*b*c^3) 
)/(18*a*b))/(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(a^(1/3)*b + (-b)^(4 
/3)*c + (-b)^(4/3)*d*x)/(27*a^(4/3)*(-b)^(5/3)*d) + (log((c*d^4)/(81*a^2*b 
) + (d^5*x)/(81*a^2*b) + (d^4*(3^(1/2)*1i - 1)^2)/(324*a^(5/3)*(-b)^(4/3)) 
)*(3^(1/2)*1i - 1))/(54*a^(4/3)*(-b)^(5/3)*d) - (log((c*d^4)/(81*a^2*b) + 
(d^5*x)/(81*a^2*b) + (d^4*(3^(1/2)*1i + 1)^2)/(324*a^(5/3)*(-b)^(4/3)))*(3 
^(1/2)*1i + 1))/(54*a^(4/3)*(-b)^(5/3)*d)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1825, normalized size of antiderivative = 8.90 \[ \int \frac {(c+d x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sq 
rt(3)))*a**2*b*c - 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d* 
x)/(a**(1/3)*sqrt(3)))*a*b**2*c**4 - 12*sqrt(3)*atan((a**(1/3) - 2*b**(1/3 
)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*a*b**2*c**3*d*x - 12*sqrt(3)*ata 
n((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*a*b**2*c* 
*2*d**2*x**2 - 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/( 
a**(1/3)*sqrt(3)))*a*b**2*c*d**3*x**3 - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1 
/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**3*c**7 - 12*sqrt(3)*atan((a 
**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**3*c**6*d*x 
 - 30*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sq 
rt(3)))*b**3*c**5*d**2*x**2 - 40*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2 
*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**3*c**4*d**3*x**3 - 30*sqrt(3)*atan(( 
a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**3*c**3*d* 
*4*x**4 - 12*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d*x)/(a**( 
1/3)*sqrt(3)))*b**3*c**2*d**5*x**5 - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3) 
*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b**3*c*d**6*x**6 - b**(2/3)*a**(1 
/3)*a**2 - 5*b**(2/3)*a**(1/3)*a*b*c**3 - 12*b**(2/3)*a**(1/3)*a*b*c**2*d* 
x - 9*b**(2/3)*a**(1/3)*a*b*c*d**2*x**2 - 2*b**(2/3)*a**(1/3)*a*b*d**3*x** 
3 + 5*b**(2/3)*a**(1/3)*b**2*c**6 + 24*b**(2/3)*a**(1/3)*b**2*c**5*d*x + 4 
5*b**(2/3)*a**(1/3)*b**2*c**4*d**2*x**2 + 40*b**(2/3)*a**(1/3)*b**2*c**...