Integrand size = 24, antiderivative size = 216 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=-\frac {e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac {e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}-\frac {e^3 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{5/3} b^{4/3} d}+\frac {e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac {e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d} \] Output:
-1/6*e^3*(d*x+c)/b/d/(a+b*(d*x+c)^3)^2+1/18*e^3*(d*x+c)/a/b/d/(a+b*(d*x+c) ^3)-1/27*e^3*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*e^3*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(5/3)/b^(4/3)/ d-1/54*e^3*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
Time = 0.01 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.84 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {e^3 \left (-\frac {9 \sqrt [3]{b} (c+d x)}{\left (a+b (c+d x)^3\right )^2}+\frac {3 \sqrt [3]{b} (c+d x)}{a \left (a+b (c+d x)^3\right )}+\frac {2 \sqrt {3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{5/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{5/3}}\right )}{54 b^{4/3} d} \] Input:
Integrate[(c*e + d*e*x)^3/(a + b*(c + d*x)^3)^3,x]
Output:
(e^3*((-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)
Time = 0.54 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, 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.
| \(\displaystyle \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 895 |
| \(\displaystyle \frac {e^3 \int \frac {(c+d x)^3}{\left (b (c+d x)^3+a\right )^3}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {e^3 \left (\frac {\int \frac {1}{\left (b (c+d x)^3+a\right )^2}d(c+d x)}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \int \frac {1}{b (c+d x)^3+a}d(c+d x)}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 a^{2/3}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {e^3 \left (\frac {\frac {2 \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {c+d x}{3 a \left (a+b (c+d x)^3\right )}}{6 b}-\frac {c+d x}{6 b \left (a+b (c+d x)^3\right )^2}\right )}{d}\) |
Input:
Int[(c*e + d*e*x)^3/(a + b*(c + d*x)^3)^3,x]
Output:
(e^3*(-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)) + ( -((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^(2/3))))/(3*a))/(6*b)))/d
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.62 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.88
| method | result | size |
| default | \(e^{3} \left (\frac {\frac {d^{3} x^{4}}{18 a}+\frac {2 c \,d^{2} x^{3}}{9 a}+\frac {c^{2} d \,x^{2}}{3 a}-\frac {\left (-2 c^{3} b +a \right ) x}{9 b a}-\frac {c \left (-c^{3} b +2 a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{27 a \,b^{2} d}\right )\) | \(190\) |
| risch | \(\frac {\frac {d^{3} e^{3} x^{4}}{18 a}+\frac {2 c \,d^{2} e^{3} x^{3}}{9 a}+\frac {e^{3} c^{2} d \,x^{2}}{3 a}-\frac {e^{3} \left (-2 c^{3} b +a \right ) x}{9 b a}-\frac {c \,e^{3} \left (-c^{3} b +2 a \right )}{18 b d a}}{\left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )^{2}}+\frac {e^{3} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{27 a \,b^{2} d}\) | \(204\) |
Input:
int((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x,method=_RETURNVERBOSE)
Output:
e^3*((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-1/18/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)))
Leaf count of result is larger than twice the leaf count of optimal. 851 vs. \(2 (175) = 350\).
Time = 0.13 (sec) , antiderivative size = 1822, normalized size of antiderivative = 8.44 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x, algorithm="fricas")
Output:
[1/54*(3*a^2*b^2*d^4*e^3*x^4 + 12*a^2*b^2*c*d^3*e^3*x^3 + 18*a^2*b^2*c^2*d ^2*e^3*x^2 + 6*(2*a^2*b^2*c^3 - a^3*b)*d*e^3*x + 3*(a^2*b^2*c^4 - 2*a^3*b* c)*e^3 + 3*sqrt(1/3)*(a*b^3*d^6*e^3*x^6 + 6*a*b^3*c*d^5*e^3*x^5 + 15*a*b^3 *c^2*d^4*e^3*x^4 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*e^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^2*e^3*x^2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*e^3*x + (a*b^3 *c^6 + 2*a^2*b^2*c^3 + a^3*b)*e^3)*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*e^3*x^6 + 6*b^2*c*d ^5*e^3*x^5 + 15*b^2*c^2*d^4*e^3*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^3*x^3 + 3 *(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^3*x + (b^2* c^6 + 2*a*b*c^3 + a^2)*e^3)*(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*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + 15*b^2*c^2*d^4*e^3*x^4 + 2*(10*b^2*c^3 + a*b)*d^3* e^3*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^ 3*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^3)*(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*...
Time = 1.89 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.38 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {- 2 a c e^{3} + b c^{4} e^{3} + 6 b c^{2} d^{2} e^{3} x^{2} + 4 b c d^{3} e^{3} x^{3} + b d^{4} e^{3} x^{4} + x \left (- 2 a d e^{3} + 4 b c^{3} d e^{3}\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \cdot \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \cdot \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac {e^{3} \operatorname {RootSum} {\left (19683 t^{3} a^{5} b^{4} - 1, \left ( t \mapsto t \log {\left (x + \frac {27 t a^{2} b e^{3} + c e^{3}}{d e^{3}} \right )} \right )\right )}}{d} \] Input:
integrate((d*e*x+c*e)**3/(a+b*(d*x+c)**3)**3,x)
Output:
(-2*a*c*e**3 + b*c**4*e**3 + 6*b*c**2*d**2*e**3*x**2 + 4*b*c*d**3*e**3*x** 3 + b*d**4*e**3*x**4 + x*(-2*a*d*e**3 + 4*b*c**3*d*e**3))/(18*a**3*b*d + 3 6*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*(1 08*a**2*b**2*c**2*d**2 + 108*a*b**3*c**5*d**2)) + e**3*RootSum(19683*_t**3 *a**5*b**4 - 1, Lambda(_t, _t*log(x + (27*_t*a**2*b*e**3 + c*e**3)/(d*e**3 ))))/d
\[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\int { \frac {{\left (d e x + c e\right )}^{3}}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{3}} \,d x } \] Input:
integrate((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x, algorithm="maxima")
Output:
1/9*e^3*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) + 1/18*(b*d^4*e^3*x^4 + 4*b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 2*(2*b*c^3 - a)*d*e^3*x + (b*c^4 - 2*a*c)*e^3)/(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)
Time = 0.15 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.38 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {2 \, \sqrt {3} \left (\frac {e^{9}}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (\frac {e^{9}}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (\frac {e^{9}}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{54 \, a b} + \frac {b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 4 \, b c^{3} d e^{3} x + b c^{4} e^{3} - 2 \, a d e^{3} x - 2 \, a c e^{3}}{18 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \] Input:
integrate((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x, algorithm="giac")
Output:
1/54*(2*sqrt(3)*(e^9/(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))) - (e^9/(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*(e^9/(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*e^3*x^4 + 4*b*c*d^3*e^3*x ^3 + 6*b*c^2*d^2*e^3*x^2 + 4*b*c^3*d*e^3*x + b*c^4*e^3 - 2*a*d*e^3*x - 2*a *c*e^3)/((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)
Time = 1.00 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.62 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx=\frac {\frac {b\,c^4\,e^3-2\,a\,c\,e^3}{18\,a\,b\,d}+\frac {d^3\,e^3\,x^4}{18\,a}-\frac {e^3\,x\,\left (a-2\,b\,c^3\right )}{9\,a\,b}+\frac {c^2\,d\,e^3\,x^2}{3\,a}+\frac {2\,c\,d^2\,e^3\,x^3}{9\,a}}{x^3\,\left (20\,b^2\,c^3\,d^3+2\,a\,b\,d^3\right )+x^2\,\left (15\,b^2\,c^4\,d^2+6\,a\,b\,c\,d^2\right )+a^2+x\,\left (6\,d\,b^2\,c^5+6\,a\,d\,b\,c^2\right )+b^2\,c^6+b^2\,d^6\,x^6+2\,a\,b\,c^3+6\,b^2\,c\,d^5\,x^5+15\,b^2\,c^2\,d^4\,x^4}+\frac {e^3\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{27\,a^{5/3}\,b^{4/3}\,d}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,c-2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (e^3+\sqrt {3}\,e^3\,1{}\mathrm {i}\right )}{54\,a^{5/3}\,b^{4/3}\,d}+\frac {e^3\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,a^{5/3}\,b^{4/3}\,d} \] Input:
int((c*e + d*e*x)^3/(a + b*(c + d*x)^3)^3,x)
Output:
((b*c^4*e^3 - 2*a*c*e^3)/(18*a*b*d) + (d^3*e^3*x^4)/(18*a) - (e^3*x*(a - 2 *b*c^3))/(9*a*b) + (c^2*d*e^3*x^2)/(3*a) + (2*c*d^2*e^3*x^3)/(9*a))/(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) + (e^3*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d* x))/(27*a^(5/3)*b^(4/3)*d) - (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)*e^3*1i + e^3))/(54*a^(5/3)*b^(4/3)*d) + (e^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))/(27*a^(5/3)*b^(4/3)*d)
Time = 0.22 (sec) , antiderivative size = 1796, normalized size of antiderivative = 8.31 \[ \int \frac {(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx =\text {Too large to display} \] Input:
int((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x)
Output:
(e**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)))*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)*s qrt(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)*ata n((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 +...