\(\int \frac {(c e+d e x)^4}{(a+b (c+d x)^3)^3} \, dx\) [2902]

Optimal result

Integrand size = 24, antiderivative size = 220 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{4/3} b^{5/3} d}-\frac {e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{4/3} b^{5/3} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {379, 294, 296, 298, 31, 648, 631, 210, 642} \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {e^4 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{4/3} b^{5/3} d}-\frac {e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{4/3} b^{5/3} d}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2} \]

[In]

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

[Out]

-1/6*(e^4*(c + d*x)^2)/(b*d*(a + b*(c + d*x)^3)^2) + (e^4*(c + d*x)^2)/(9*a*b*d*(a + b*(c + d*x)^3)) - (e^4*Ar
cTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(4/3)*b^(5/3)*d) - (e^4*Log[a^(1/3) + b^
(1/3)*(c + d*x)])/(27*a^(4/3)*b^(5/3)*d) + (e^4*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]
)/(54*a^(4/3)*b^(5/3)*d)

Rule 31

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

Rule 210

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 294

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] - Dist[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 296

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] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[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 379

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = \frac {e^4 \text {Subst}\left (\int \frac {x^4}{\left (a+b x^3\right )^3} \, dx,x,c+d x\right )}{d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 \text {Subst}\left (\int \frac {x}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{3 b d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}+\frac {e^4 \text {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{9 a b d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{27 a^{4/3} b^{4/3} d}+\frac {e^4 \text {Subst}\left (\int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{27 a^{4/3} b^{4/3} d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {e^4 \text {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{54 a^{4/3} b^{5/3} d}+\frac {e^4 \text {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{18 a b^{4/3} d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {e^4 \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}+\frac {e^4 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{9 a^{4/3} b^{5/3} d} \\ & = -\frac {e^4 (c+d x)^2}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^4 (c+d x)^2}{9 a b d \left (a+b (c+d x)^3\right )}-\frac {e^4 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{9 \sqrt {3} a^{4/3} b^{5/3} d}-\frac {e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{4/3} b^{5/3} d}+\frac {e^4 \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.84 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {e^4 \left (-\frac {9 b^{2/3} (c+d x)^2}{\left (a+b (c+d x)^3\right )^2}+\frac {6 b^{2/3} (c+d x)^2}{a \left (a+b (c+d x)^3\right )}+\frac {2 \sqrt {3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{4/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{4/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{4/3}}\right )}{54 b^{5/3} d} \]

[In]

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

[Out]

(e^4*((-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*Sq
rt[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)

Maple [C] (verified)

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

Time = 4.06 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99

method	result	size
default	\(e^{4} \left (\frac {\frac {d^{4} x^{5}}{9 a}+\frac {5 c \,d^{3} x^{4}}{9 a}+\frac {10 c^{2} d^{2} x^{3}}{9 a}-\frac {d \left (-20 c^{3} b +a \right ) x^{2}}{18 b a}-\frac {c \left (-5 c^{3} b +a \right ) x}{9 b a}-\frac {c^{2} \left (-2 c^{3} b +a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b \,d^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 b^{2} a d}\right )\)	\(218\)
risch	\(\frac {\frac {d^{4} e^{4} x^{5}}{9 a}+\frac {5 c \,d^{3} e^{4} x^{4}}{9 a}+\frac {10 c^{2} d^{2} e^{4} x^{3}}{9 a}-\frac {d \,e^{4} \left (-20 c^{3} b +a \right ) x^{2}}{18 b a}-\frac {c \,e^{4} \left (-5 c^{3} b +a \right ) x}{9 b a}-\frac {c^{2} e^{4} \left (-2 c^{3} b +a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a \right )^{2}}+\frac {e^{4} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b \,d^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{27 a \,b^{2} d}\)	\(235\)

[In]

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

[Out]

e^4*((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/b^2/a/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. 882 vs. \(2 (179) = 358\).

Time = 0.30 (sec) , antiderivative size = 1886, normalized size of antiderivative = 8.57 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/54*(6*a*b^3*d^5*e^4*x^5 + 30*a*b^3*c*d^4*e^4*x^4 + 60*a*b^3*c^2*d^3*e^4*x^3 + 3*(20*a*b^3*c^3 - a^2*b^2)*d^
2*e^4*x^2 + 6*(5*a*b^3*c^4 - a^2*b^2*c)*d*e^4*x + 3*(2*a*b^3*c^5 - a^2*b^2*c^2)*e^4 + 3*sqrt(1/3)*(a*b^3*d^6*e
^4*x^6 + 6*a*b^3*c*d^5*e^4*x^5 + 15*a*b^3*c^2*d^4*e^4*x^4 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*e^4*x^3 + 3*(5*a*b^
3*c^4 + 2*a^2*b^2*c)*d^2*e^4*x^2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*e^4*x + (a*b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b)*e
^4)*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*e^4*x^6 + 6*b^2*c
*d^5*e^4*x^5 + 15*b^2*c^2*d^4*e^4*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^4*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^4*x^2
+ 6*(b^2*c^5 + a*b*c^2)*d*e^4*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^4)*(-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*e^4*x^6 + 6*b^2*c*d^5*e^4*x^5 + 15*
b^2*c^2*d^4*e^4*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^4*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^4*x^2 + 6*(b^2*c^5 + a*
b*c^2)*d*e^4*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^4)*(-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*e^4*x^5 + 30*a*b^3*c*d^4*e^4*x^4 + 60*a*b^3*c^2*d^3*e^4*x^3 + 3*(20*a*b^3*c^3 - a^2*b^2)*d^2*
e^4*x^2 + 6*(5*a*b^3*c^4 - a^2*b^2*c)*d*e^4*x + 3*(2*a*b^3*c^5 - a^2*b^2*c^2)*e^4 + 6*sqrt(1/3)*(a*b^3*d^6*e^4
*x^6 + 6*a*b^3*c*d^5*e^4*x^5 + 15*a*b^3*c^2*d^4*e^4*x^4 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*e^4*x^3 + 3*(5*a*b^3*
c^4 + 2*a^2*b^2*c)*d^2*e^4*x^2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*e^4*x + (a*b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b)*e^4
)*sqrt(-(-a*b^2)^(1/3)/a)*arctan(sqrt(1/3)*(2*b*d*x + 2*b*c + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) + (b^
2*d^6*e^4*x^6 + 6*b^2*c*d^5*e^4*x^5 + 15*b^2*c^2*d^4*e^4*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^4*x^3 + 3*(5*b^2*c^4
+ 2*a*b*c)*d^2*e^4*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^4*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^4)*(-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*e^4*x^6 + 6*
b^2*c*d^5*e^4*x^5 + 15*b^2*c^2*d^4*e^4*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^4*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^
4*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^4*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^4)*(-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)]

Sympy [A] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.51 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {- a c^{2} e^{4} + 2 b c^{5} e^{4} + 20 b c^{2} d^{3} e^{4} x^{3} + 10 b c d^{4} e^{4} x^{4} + 2 b d^{5} e^{4} x^{5} + x^{2} \left (- a d^{2} e^{4} + 20 b c^{3} d^{2} e^{4}\right ) + x \left (- 2 a c d e^{4} + 10 b c^{4} d e^{4}\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \cdot \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \cdot \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac {e^{4} \operatorname {RootSum} {\left (19683 t^{3} a^{4} b^{5} + 1, \left ( t \mapsto t \log {\left (x + \frac {729 t^{2} a^{3} b^{3} e^{8} + c e^{8}}{d e^{8}} \right )} \right )\right )}}{d} \]

[In]

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

[Out]

(-a*c**2*e**4 + 2*b*c**5*e**4 + 20*b*c**2*d**3*e**4*x**3 + 10*b*c*d**4*e**4*x**4 + 2*b*d**5*e**4*x**5 + x**2*(
-a*d**2*e**4 + 20*b*c**3*d**2*e**4) + x*(-2*a*c*d*e**4 + 10*b*c**4*d*e**4))/(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)) + e**4*RootSum(19683*_t**3*a**4*b**5 + 1, Lambda(_t, _t*log(x + (729*_t**2*a**
3*b**3*e**8 + c*e**8)/(d*e**8))))/d

Maxima [F]

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

[In]

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

[Out]

1/9*e^4*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) + 1/18*(2*b*d^5*e^
4*x^5 + 10*b*c*d^4*e^4*x^4 + 20*b*c^2*d^3*e^4*x^3 + (20*b*c^3 - a)*d^2*e^4*x^2 + 2*(5*b*c^4 - a*c)*d*e^4*x + (
2*b*c^5 - a*c^2)*e^4)/(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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.41 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {2 \, \sqrt {3} \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{54 \, a b} + \frac {2 \, b d^{5} e^{4} x^{5} + 10 \, b c d^{4} e^{4} x^{4} + 20 \, b c^{2} d^{3} e^{4} x^{3} + 20 \, b c^{3} d^{2} e^{4} x^{2} + 10 \, b c^{4} d e^{4} x + 2 \, b c^{5} e^{4} - a d^{2} e^{4} x^{2} - 2 \, a c d e^{4} x - a c^{2} e^{4}}{18 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \]

[In]

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

[Out]

-1/54*(2*sqrt(3)*(-e^12/(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)) + (-e^12/(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*(-e^12
/(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*e^4*x^5 + 10*b*c*d^4*e^4
*x^4 + 20*b*c^2*d^3*e^4*x^3 + 20*b*c^3*d^2*e^4*x^2 + 10*b*c^4*d*e^4*x + 2*b*c^5*e^4 - a*d^2*e^4*x^2 - 2*a*c*d*
e^4*x - a*c^2*e^4)/((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 = 6.50 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.95 \[ \int \frac {(c e+d e x)^4}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {\frac {d^4\,e^4\,x^5}{9\,a}-\frac {a\,c^2\,e^4-2\,b\,c^5\,e^4}{18\,a\,b\,d}+\frac {10\,c^2\,d^2\,e^4\,x^3}{9\,a}+\frac {5\,c\,d^3\,e^4\,x^4}{9\,a}-\frac {c\,x\,\left (a\,e^4-5\,b\,c^3\,e^4\right )}{9\,a\,b}-\frac {d\,e^4\,x^2\,\left (a-20\,b\,c^3\right )}{18\,a\,b}}{x^3\,\left (20\,b^2\,c^3\,d^3+2\,a\,b\,d^3\right )+x^2\,\left (15\,b^2\,c^4\,d^2+6\,a\,b\,c\,d^2\right )+a^2+x\,\left (6\,d\,b^2\,c^5+6\,a\,d\,b\,c^2\right )+b^2\,c^6+b^2\,d^6\,x^6+2\,a\,b\,c^3+6\,b^2\,c\,d^5\,x^5+15\,b^2\,c^2\,d^4\,x^4}-\frac {\ln \left (2\,{\left (-b\right )}^{4/3}\,c-a^{1/3}\,b+2\,{\left (-b\right )}^{4/3}\,d\,x+\sqrt {3}\,a^{1/3}\,b\,1{}\mathrm {i}\right )\,\left (e^4+\sqrt {3}\,e^4\,1{}\mathrm {i}\right )}{54\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d}+\frac {e^4\,\ln \left (a^{1/3}\,b+{\left (-b\right )}^{4/3}\,c+{\left (-b\right )}^{4/3}\,d\,x\right )}{27\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d}+\frac {e^4\,\ln \left (\frac {d^4\,e^8\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{81\,a^{5/3}\,{\left (-b\right )}^{4/3}}+\frac {c\,d^4\,e^8}{81\,a^2\,b}+\frac {d^5\,e^8\,x}{81\,a^2\,b}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,a^{4/3}\,{\left (-b\right )}^{5/3}\,d} \]

[In]

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

[Out]

((d^4*e^4*x^5)/(9*a) - (a*c^2*e^4 - 2*b*c^5*e^4)/(18*a*b*d) + (10*c^2*d^2*e^4*x^3)/(9*a) + (5*c*d^3*e^4*x^4)/(
9*a) - (c*x*(a*e^4 - 5*b*c^3*e^4))/(9*a*b) - (d*e^4*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(2*(-b)^(4/3)*c - a^(1/3)*b + 2*(-b)^(4/3)*d*x + 3^(1/2)*
a^(1/3)*b*1i)*(3^(1/2)*e^4*1i + e^4))/(54*a^(4/3)*(-b)^(5/3)*d) + (e^4*log(a^(1/3)*b + (-b)^(4/3)*c + (-b)^(4/
3)*d*x))/(27*a^(4/3)*(-b)^(5/3)*d) + (e^4*log((d^4*e^8*((3^(1/2)*1i)/2 - 1/2)^2)/(81*a^(5/3)*(-b)^(4/3)) + (c*
d^4*e^8)/(81*a^2*b) + (d^5*e^8*x)/(81*a^2*b))*((3^(1/2)*1i)/2 - 1/2))/(27*a^(4/3)*(-b)^(5/3)*d)