Integrand size = 24, antiderivative size = 168 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=-\frac {1}{2 a d e^3 (c+d x)^2}+\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} d e^3}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3} \] Output:
-1/2/a/d/e^3/(d*x+c)^2+1/3*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^(5/3)/d/e^3-1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*(d*x +c))/a^(5/3)/d/e^3+1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)* (d*x+c)^2)/a^(5/3)/d/e^3
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {-\frac {3 a^{2/3}}{(c+d x)^2}-2 \sqrt {3} b^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )-2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )+b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3} \] Input:
Integrate[1/((c*e + d*e*x)^3*(a + b*(c + d*x)^3)),x]
Output:
((-3*a^(2/3))/(c + d*x)^2 - 2*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3) *(c + d*x))/(Sqrt[3]*a^(1/3))] - 2*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x) ] + b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2] )/(6*a^(5/3)*d*e^3)
Time = 0.51 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {895, 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.
| \(\displaystyle \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx\) |
| \(\Big \downarrow \) 895 |
| \(\displaystyle \frac {\int \frac {1}{(c+d x)^3 \left (b (c+d x)^3+a\right )}d(c+d x)}{d e^3}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {-\frac {b \int \frac {1}{b (c+d x)^3+a}d(c+d x)}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 a^{2/3}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {-\frac {b \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}}{d e^3}\) |
Input:
Int[1/((c*e + d*e*x)^3*(a + b*(c + d*x)^3)),x]
Output:
(-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]*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))))/a)/(d*e^3)
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_)^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*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]
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 = 0.78 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.54
| method | result | size |
| default | \(\frac {-\frac {1}{2 a d \left (x d +c \right )^{2}}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{3 d a}}{e^{3}}\) | \(91\) |
| risch | \(-\frac {1}{2 a d \,e^{3} \left (x d +c \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} d^{3} e^{9} \textit {\_Z}^{3}+b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{5} d^{4} e^{9} \textit {\_R}^{3}-3 d \,b^{2}\right ) x -4 a^{5} c \,d^{3} e^{9} \textit {\_R}^{3}-a^{2} b d \,e^{3} \textit {\_R} -3 b^{2} c \right )\right )}{3}\) | \(102\) |
Input:
int(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/e^3*(-1/2/a/d/(d*x+c)^2-1/3/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))/a)
Time = 0.09 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {2 \, \sqrt {3} {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (a d x + a c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + {\left (a b d x + a b c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b d x + b c - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 3}{6 \, {\left (a d^{3} e^{3} x^{2} + 2 \, a c d^{2} e^{3} x + a c^{2} d e^{3}\right )}} \] Input:
integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3),x, algorithm="fricas")
Output:
1/6*(2*sqrt(3)*(d^2*x^2 + 2*c*d*x + c^2)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sq rt(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) - (d^2*x^2 + 2*c*d*x + c^2)*(-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)) + 2*(d^2*x^2 + 2*c*d*x + c^2)*(-b^2/a^2)^(1/3)*log(b*d*x + b*c - a*(-b^2/a^2)^(1/3)) - 3)/(a*d^3* e^3*x^2 + 2*a*c*d^2*e^3*x + a*c^2*d*e^3)
Time = 0.69 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.45 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=- \frac {1}{2 a c^{2} d e^{3} + 4 a c d^{2} e^{3} x + 2 a d^{3} e^{3} x^{2}} + \frac {\operatorname {RootSum} {\left (27 t^{3} a^{5} + b^{2}, \left ( t \mapsto t \log {\left (x + \frac {- 3 t a^{2} + b c}{b d} \right )} \right )\right )}}{d e^{3}} \] Input:
integrate(1/(d*e*x+c*e)**3/(a+b*(d*x+c)**3),x)
Output:
-1/(2*a*c**2*d*e**3 + 4*a*c*d**2*e**3*x + 2*a*d**3*e**3*x**2) + RootSum(27 *_t**3*a**5 + b**2, Lambda(_t, _t*log(x + (-3*_t*a**2 + b*c)/(b*d))))/(d*e **3)
\[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )} {\left (d e x + c e\right )}^{3}} \,d x } \] Input:
integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3),x, algorithm="maxima")
Output:
-1/2/(a*d^3*e^3*x^2 + 2*a*c*d^2*e^3*x + a*c^2*d*e^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*e^3)
Time = 0.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | -b d x - b c + \left (-a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{6 \, a e^{3}} - \frac {1}{2 \, {\left (d x + c\right )}^{2} a d e^{3}} \] Input:
integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3),x, algorithm="giac")
Output:
1/6*(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*e^3) - 1/2/((d*x + c)^2*a*d*e^3)
Time = 0.50 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {b^{2/3}\,\ln \left (a^2\,b^{1/3}\,c-{\left (-a\right )}^{7/3}+a^2\,b^{1/3}\,d\,x\right )}{3\,{\left (-a\right )}^{5/3}\,d\,e^3}-\frac {1}{2\,a\,d\,\left (c^2\,e^3+2\,c\,d\,e^3\,x+d^2\,e^3\,x^2\right )}-\frac {b^{2/3}\,\ln \left ({\left (-a\right )}^{7/3}+2\,a^2\,b^{1/3}\,c+2\,a^2\,b^{1/3}\,d\,x+\sqrt {3}\,{\left (-a\right )}^{7/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,{\left (-a\right )}^{5/3}\,d\,e^3}+\frac {b^{2/3}\,\ln \left ({\left (-a\right )}^{7/3}+2\,a^2\,b^{1/3}\,c+2\,a^2\,b^{1/3}\,d\,x-\sqrt {3}\,{\left (-a\right )}^{7/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{{\left (-a\right )}^{5/3}\,d\,e^3} \] Input:
int(1/((c*e + d*e*x)^3*(a + b*(c + d*x)^3)),x)
Output:
(b^(2/3)*log(a^2*b^(1/3)*c - (-a)^(7/3) + a^2*b^(1/3)*d*x))/(3*(-a)^(5/3)* d*e^3) - 1/(2*a*d*(c^2*e^3 + d^2*e^3*x^2 + 2*c*d*e^3*x)) - (b^(2/3)*log((- a)^(7/3) + 3^(1/2)*(-a)^(7/3)*1i + 2*a^2*b^(1/3)*c + 2*a^2*b^(1/3)*d*x)*(( 3^(1/2)*1i)/2 + 1/2))/(3*(-a)^(5/3)*d*e^3) + (b^(2/3)*log((-a)^(7/3) - 3^( 1/2)*(-a)^(7/3)*1i + 2*a^2*b^(1/3)*c + 2*a^2*b^(1/3)*d*x)*((3^(1/2)*1i)/6 - 1/6))/((-a)^(5/3)*d*e^3)
Time = 0.21 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.42 \[ \int \frac {1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {2 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} c -2 b^{\frac {1}{3}} d x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b \,c^{2}+4 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} c -2 b^{\frac {1}{3}} d x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b c d x +2 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} c -2 b^{\frac {1}{3}} d x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b \,d^{2} x^{2}+a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b \,c^{2}+2 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b c d x +a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b \,d^{2} x^{2}-2 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b \,c^{2}-4 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b c d x -2 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b \,d^{2} x^{2}-3 b^{\frac {1}{3}} a}{6 b^{\frac {1}{3}} a^{2} d \,e^{3} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^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)))*b*c**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)))*b*c*d*x + 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*d**2*x**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**(2/3)*c*d*x + b**(2/3)*d**2*x**2)*b*c**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**(2/3)*c*d*x + b**(2/3)*d**2*x**2)*b*c*d*x + 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)*b*d**2*x**2 - 2*a**(1/3)*log(a**(1/3) + b**(1/3)* c + b**(1/3)*d*x)*b*c**2 - 4*a**(1/3)*log(a**(1/3) + b**(1/3)*c + b**(1/3) *d*x)*b*c*d*x - 2*a**(1/3)*log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*b*d** 2*x**2 - 3*b**(1/3)*a)/(6*b**(1/3)*a**2*d*e**3*(c**2 + 2*c*d*x + d**2*x**2 ))