Integrand size = 24, antiderivative size = 203 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {1}{a^2 d e^2 (c+d x)}-\frac {b (c+d x)^2}{3 a^2 d e^2 \left (a+b (c+d x)^3\right )}+\frac {4 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} d e^2}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}-\frac {2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2} \] Output:
-1/a^2/d/e^2/(d*x+c)-1/3*b*(d*x+c)^2/a^2/d/e^2/(a+b*(d*x+c)^3)+4/9*b^(1/3) *arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/a^(7/3)/d /e^2+4/9*b^(1/3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(7/3)/d/e^2-2/9*b^(1/3)*ln( a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(7/3)/d/e^2
Time = 0.03 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {-\frac {9 \sqrt [3]{a}}{c+d x}-\frac {3 \sqrt [3]{a} b (c+d x)^2}{a+b (c+d x)^3}-4 \sqrt {3} \sqrt [3]{b} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2} \] Input:
Integrate[1/((c*e + d*e*x)^2*(a + b*(c + d*x)^3)^2),x]
Output:
((-9*a^(1/3))/(c + d*x) - (3*a^(1/3)*b*(c + d*x)^2)/(a + b*(c + d*x)^3) - 4*Sqrt[3]*b^(1/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3) )] + 4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] - 2*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d*e^2)
Time = 0.57 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {895, 819, 847, 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.
| \(\displaystyle \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 895 |
| \(\displaystyle \frac {\int \frac {1}{(c+d x)^2 \left (b (c+d x)^3+a\right )^2}d(c+d x)}{d e^2}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {\frac {4 \int \frac {1}{(c+d x)^2 \left (b (c+d x)^3+a\right )}d(c+d x)}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \int \frac {c+d x}{b (c+d x)^3+a}d(c+d x)}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {4 \left (-\frac {b \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a (c+d x)}\right )}{3 a}+\frac {1}{3 a (c+d x) \left (a+b (c+d x)^3\right )}}{d e^2}\) |
Input:
Int[1/((c*e + d*e*x)^2*(a + b*(c + d*x)^3)^2),x]
Output:
(1/(3*a*(c + d*x)*(a + b*(c + d*x)^3)) + (4*(-(1/(a*(c + d*x))) - (b*(-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))))/a))/(3*a))/(d*e^2)
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[((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]
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]
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 = 0.82 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.78
| method | result | size |
| default | \(\frac {-\frac {1}{a^{2} d \left (x d +c \right )}-\frac {b \left (\frac {\frac {x^{2} d}{3}+\frac {2 c x}{3}+\frac {c^{2}}{3 d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{9 b d}\right )}{a^{2}}}{e^{2}}\) | \(158\) |
| risch | \(\frac {-\frac {4 b \,d^{2} x^{3}}{3 a^{2}}-\frac {4 b c d \,x^{2}}{a^{2}}-\frac {4 b x \,c^{2}}{a^{2}}-\frac {4 c^{3} b +3 a}{3 a^{2} d}}{e^{2} \left (x d +c \right ) \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} d^{3} e^{6} \textit {\_Z}^{3}-b \right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{7} d^{4} e^{6} \textit {\_R}^{3}+3 b d \right ) x -4 a^{7} c \,d^{3} e^{6} \textit {\_R}^{3}-a^{5} d^{2} e^{4} \textit {\_R}^{2}+3 b c \right )\right )}{9}\) | \(181\) |
Input:
int(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3)^2,x,method=_RETURNVERBOSE)
Output:
1/e^2*(-1/a^2/d/(d*x+c)-b/a^2*((1/3*x^2*d+2/3*c*x+1/3*c^2/d)/(b*d^3*x^3+3* b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)+4/9/b/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))))
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (164) = 328\).
Time = 0.09 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.97 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {12 \, b d^{3} x^{3} + 36 \, b c d^{2} x^{2} + 36 \, b c^{2} d x + 12 \, b c^{3} + 4 \, \sqrt {3} {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 2 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} - {\left (a d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 4 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d x + b c + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 9 \, a}{9 \, {\left (a^{2} b d^{5} e^{2} x^{4} + 4 \, a^{2} b c d^{4} e^{2} x^{3} + 6 \, a^{2} b c^{2} d^{3} e^{2} x^{2} + {\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} e^{2} x + {\left (a^{2} b c^{4} + a^{3} c\right )} d e^{2}\right )}} \] Input:
integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3)^2,x, algorithm="fricas")
Output:
-1/9*(12*b*d^3*x^3 + 36*b*c*d^2*x^2 + 36*b*c^2*d*x + 12*b*c^3 + 4*sqrt(3)* (b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*arctan(2/3*sqrt(3)*(d*x + c)*(b/a)^(1/3) - 1/3*sqrt(3)) + 2*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d *x + a*c)*(b/a)^(1/3)*log(b*d^2*x^2 + 2*b*c*d*x + b*c^2 - (a*d*x + a*c)*(b /a)^(2/3) + a*(b/a)^(1/3)) - 4*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^ 2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d*x + b*c + a*(b/a) ^(2/3)) + 9*a)/(a^2*b*d^5*e^2*x^4 + 4*a^2*b*c*d^4*e^2*x^3 + 6*a^2*b*c^2*d^ 3*e^2*x^2 + (4*a^2*b*c^3 + a^3)*d^2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^2)
Time = 1.31 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {- 3 a - 4 b c^{3} - 12 b c^{2} d x - 12 b c d^{2} x^{2} - 4 b d^{3} x^{3}}{3 a^{3} c d e^{2} + 3 a^{2} b c^{4} d e^{2} + 18 a^{2} b c^{2} d^{3} e^{2} x^{2} + 12 a^{2} b c d^{4} e^{2} x^{3} + 3 a^{2} b d^{5} e^{2} x^{4} + x \left (3 a^{3} d^{2} e^{2} + 12 a^{2} b c^{3} d^{2} e^{2}\right )} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{7} - 64 b, \left ( t \mapsto t \log {\left (x + \frac {81 t^{2} a^{5} + 16 b c}{16 b d} \right )} \right )\right )}}{d e^{2}} \] Input:
integrate(1/(d*e*x+c*e)**2/(a+b*(d*x+c)**3)**2,x)
Output:
(-3*a - 4*b*c**3 - 12*b*c**2*d*x - 12*b*c*d**2*x**2 - 4*b*d**3*x**3)/(3*a* *3*c*d*e**2 + 3*a**2*b*c**4*d*e**2 + 18*a**2*b*c**2*d**3*e**2*x**2 + 12*a* *2*b*c*d**4*e**2*x**3 + 3*a**2*b*d**5*e**2*x**4 + x*(3*a**3*d**2*e**2 + 12 *a**2*b*c**3*d**2*e**2)) + RootSum(729*_t**3*a**7 - 64*b, Lambda(_t, _t*lo g(x + (81*_t**2*a**5 + 16*b*c)/(16*b*d))))/(d*e**2)
\[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{2} {\left (d e x + c e\right )}^{2}} \,d x } \] Input:
integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3)^2,x, algorithm="maxima")
Output:
-1/3*(4*b*d^3*x^3 + 12*b*c*d^2*x^2 + 12*b*c^2*d*x + 4*b*c^3 + 3*a)/(a^2*b* d^5*e^2*x^4 + 4*a^2*b*c*d^4*e^2*x^3 + 6*a^2*b*c^2*d^3*e^2*x^2 + (4*a^2*b*c ^3 + a^3)*d^2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^2) - 4/3*b*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^2*e^2)
Time = 0.13 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {4 \, \left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {1}{3}} \log \left ({\left | -\left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {1}{3}} - \frac {1}{{\left (d e x + c e\right )} d e} \right |}\right )}{9 \, a^{2}} - \frac {4 \, \sqrt {3} \left (a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {1}{3}} - \frac {2}{{\left (d e x + c e\right )} d e}\right )}}{3 \, \left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} d e^{2}} - \frac {2 \, \left (a^{2} b\right )^{\frac {1}{3}} \log \left (\left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {2}{3}} - \frac {\left (\frac {b}{a d^{3} e^{6}}\right )^{\frac {1}{3}}}{{\left (d e x + c e\right )} d e} + \frac {1}{{\left (d e x + c e\right )}^{2} d^{2} e^{2}}\right )}{9 \, a^{3} d e^{2}} - \frac {1}{{\left (d e x + c e\right )} a^{2} d e} - \frac {b}{3 \, {\left (d e x + c e\right )} a^{2} {\left (b + \frac {a e^{3}}{{\left (d e x + c e\right )}^{3}}\right )} d e} \] Input:
integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3)^2,x, algorithm="giac")
Output:
4/9*(b/(a*d^3*e^6))^(1/3)*log(abs(-(b/(a*d^3*e^6))^(1/3) - 1/((d*e*x + c*e )*d*e)))/a^2 - 4/9*sqrt(3)*(a^2*b)^(1/3)*arctan(1/3*sqrt(3)*((b/(a*d^3*e^6 ))^(1/3) - 2/((d*e*x + c*e)*d*e))/(b/(a*d^3*e^6))^(1/3))/(a^3*d*e^2) - 2/9 *(a^2*b)^(1/3)*log((b/(a*d^3*e^6))^(2/3) - (b/(a*d^3*e^6))^(1/3)/((d*e*x + c*e)*d*e) + 1/((d*e*x + c*e)^2*d^2*e^2))/(a^3*d*e^2) - 1/((d*e*x + c*e)*a ^2*d*e) - 1/3*b/((d*e*x + c*e)*a^2*(b + a*e^3/(d*e*x + c*e)^3)*d*e)
Time = 1.15 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {4\,b^{1/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{7/3}\,d\,e^2}-\frac {\frac {4\,b\,c^3+3\,a}{3\,a^2\,d}+\frac {4\,b\,d^2\,x^3}{3\,a^2}+\frac {4\,b\,c^2\,x}{a^2}+\frac {4\,b\,c\,d\,x^2}{a^2}}{x\,\left (4\,b\,d\,c^3\,e^2+a\,d\,e^2\right )+b\,c^4\,e^2+a\,c\,e^2+b\,d^4\,e^2\,x^4+4\,b\,c\,d^3\,e^2\,x^3+6\,b\,c^2\,d^2\,e^2\,x^2}-\frac {4\,b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{7/3}\,d\,e^2}+\frac {b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {2}{9}+\frac {\sqrt {3}\,2{}\mathrm {i}}{9}\right )}{a^{7/3}\,d\,e^2} \] Input:
int(1/((c*e + d*e*x)^2*(a + b*(c + d*x)^3)^2),x)
Output:
(4*b^(1/3)*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(9*a^(7/3)*d*e^2) - ((3 *a + 4*b*c^3)/(3*a^2*d) + (4*b*d^2*x^3)/(3*a^2) + (4*b*c^2*x)/a^2 + (4*b*c *d*x^2)/a^2)/(x*(a*d*e^2 + 4*b*c^3*d*e^2) + b*c^4*e^2 + a*c*e^2 + b*d^4*e^ 2*x^4 + 4*b*c*d^3*e^2*x^3 + 6*b*c^2*d^2*e^2*x^2) - (4*b^(1/3)*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) )/(9*a^(7/3)*d*e^2) + (b^(1/3)*log(2*b^(1/3)*c - 3^(1/2)*a^(1/3)*1i - a^(1 /3) + 2*b^(1/3)*d*x)*((3^(1/2)*2i)/9 - 2/9))/(a^(7/3)*d*e^2)
Time = 0.25 (sec) , antiderivative size = 1012, normalized size of antiderivative = 4.99 \[ \int \frac {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3)^2,x)
Output:
(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*c**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*c*d*x + 4*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 + 16*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*x + 24*sq rt(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**2*x**2 + 16*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**3*x**3 + 4*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**4*x**4 - 6* b**(2/3)*a**(1/3)*a*c + 3*b**(2/3)*a**(1/3)*a*d*x - 9*b**(2/3)*a**(1/3)*b* c**4 - 24*b**(2/3)*a**(1/3)*b*c**3*d*x - 18*b**(2/3)*a**(1/3)*b*c**2*d**2* x**2 + 3*b**(2/3)*a**(1/3)*b*d**4*x**4 - 2*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*b*c**2 - 2*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*b*c*d* x - 2*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**2*c**5 - 8*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**2*c**4*d*x - 12*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 + ...