Integrand size = 21, antiderivative size = 189 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac {5 b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} d}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d} \]
-5/6/a^2/d/(d*x+c)^2+1/3/a/d/(d*x+c)^2/(a+b*(d*x+c)^3)-5/9*b^(2/3)*ln(a^(1 /3)+b^(1/3)*(d*x+c))/a^(8/3)/d+5/18*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d* x+c)+b^(2/3)*(d*x+c)^2)/a^(8/3)/d+5/9*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3 )*(d*x+c))/a^(1/3)*3^(1/2))/a^(8/3)/d*3^(1/2)
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {-\frac {9 a^{2/3}}{(c+d x)^2}-\frac {6 a^{2/3} b (c+d x)}{a+b (c+d x)^3}-10 \sqrt {3} b^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )+5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d} \]
((-9*a^(2/3))/(c + d*x)^2 - (6*a^(2/3)*b*(c + d*x))/(a + b*(c + d*x)^3) - 10*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3 ))] - 10*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] + 5*b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d)
Time = 0.33 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {895, 819, 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.
| \(\displaystyle \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 895 |
| \(\displaystyle \frac {\int \frac {1}{(c+d x)^3 \left (b (c+d x)^3+a\right )^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {\frac {5 \int \frac {1}{(c+d x)^3 \left (b (c+d x)^3+a\right )}d(c+d x)}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \int \frac {1}{b (c+d x)^3+a}d(c+d x)}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 a^{2/3}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {b \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a (c+d x)^2}\right )}{3 a}+\frac {1}{3 a (c+d x)^2 \left (a+b (c+d x)^3\right )}}{d}\) |
(1/(3*a*(c + d*x)^2*(a + b*(c + d*x)^3)) + (5*(-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]*ArcTa n[(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))/(3*a))/d
3.29.75.3.1 Defintions of rubi rules used
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*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[((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 = 3.78 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(-\frac {1}{2 a^{2} d \left (d x +c \right )^{2}}-\frac {b \left (\frac {\frac {x}{3}+\frac {c}{3 d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{9 b d}\right )}{a^{2}}\) | \(140\) |
| risch | \(\frac {-\frac {5 b \,d^{2} x^{3}}{6 a^{2}}-\frac {5 b c d \,x^{2}}{2 a^{2}}-\frac {5 b x \,c^{2}}{2 a^{2}}-\frac {5 c^{3} b +3 a}{6 d \,a^{2}}}{\left (d x +c \right )^{2} \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{8} d^{3} \textit {\_Z}^{3}+b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{8} d^{4} \textit {\_R}^{3}-3 b^{2} d \right ) x -4 a^{8} c \,d^{3} \textit {\_R}^{3}-a^{3} b d \textit {\_R} -3 b^{2} c \right )\right )}{9}\) | \(167\) |
-1/2/a^2/d/(d*x+c)^2-b/a^2*((1/3*x+1/3*c/d)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c ^2*d*x+b*c^3+a)+5/9/b/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. 509 vs. \(2 (148) = 296\).
Time = 0.27 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.69 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {15 \, b d^{3} x^{3} + 45 \, b c d^{2} x^{2} + 45 \, b c^{2} d x + 15 \, b c^{3} - 10 \, \sqrt {3} {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (a d x + a c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) + 5 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + {\left (a b d x + a b c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b d x + b c - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 9 \, a}{18 \, {\left (a^{2} b d^{6} x^{5} + 5 \, a^{2} b c d^{5} x^{4} + 10 \, a^{2} b c^{2} d^{4} x^{3} + {\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} x^{2} + {\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} x + {\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d\right )}} \]
-1/18*(15*b*d^3*x^3 + 45*b*c*d^2*x^2 + 45*b*c^2*d*x + 15*b*c^3 - 10*sqrt(3 )*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d ^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqr t(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) + 5*(b*d^5*x^5 + 5*b*c *d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5* b*c^4 + 2*a*c)*d*x)*(-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)) - 10*(b*d^ 5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b*d*x + b*c - a*(-b^ 2/a^2)^(1/3)) + 9*a)/(a^2*b*d^6*x^5 + 5*a^2*b*c*d^5*x^4 + 10*a^2*b*c^2*d^4 *x^3 + (10*a^2*b*c^3 + a^3)*d^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*x + (a^2 *b*c^5 + a^3*c^2)*d)
Time = 1.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {- 3 a - 5 b c^{3} - 15 b c^{2} d x - 15 b c d^{2} x^{2} - 5 b d^{3} x^{3}}{6 a^{3} c^{2} d + 6 a^{2} b c^{5} d + 60 a^{2} b c^{2} d^{4} x^{3} + 30 a^{2} b c d^{5} x^{4} + 6 a^{2} b d^{6} x^{5} + x^{2} \cdot \left (6 a^{3} d^{3} + 60 a^{2} b c^{3} d^{3}\right ) + x \left (12 a^{3} c d^{2} + 30 a^{2} b c^{4} d^{2}\right )} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{8} + 125 b^{2}, \left ( t \mapsto t \log {\left (x + \frac {- 9 t a^{3} + 5 b c}{5 b d} \right )} \right )\right )}}{d} \]
(-3*a - 5*b*c**3 - 15*b*c**2*d*x - 15*b*c*d**2*x**2 - 5*b*d**3*x**3)/(6*a* *3*c**2*d + 6*a**2*b*c**5*d + 60*a**2*b*c**2*d**4*x**3 + 30*a**2*b*c*d**5* x**4 + 6*a**2*b*d**6*x**5 + x**2*(6*a**3*d**3 + 60*a**2*b*c**3*d**3) + x*( 12*a**3*c*d**2 + 30*a**2*b*c**4*d**2)) + RootSum(729*_t**3*a**8 + 125*b**2 , Lambda(_t, _t*log(x + (-9*_t*a**3 + 5*b*c)/(5*b*d))))/d
\[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{2} {\left (d x + c\right )}^{3}} \,d x } \]
-1/6*(5*b*d^3*x^3 + 15*b*c*d^2*x^2 + 15*b*c^2*d*x + 5*b*c^3 + 3*a)/(a^2*b* d^6*x^5 + 5*a^2*b*c*d^5*x^4 + 10*a^2*b*c^2*d^4*x^3 + (10*a^2*b*c^3 + a^3)* d^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*x + (a^2*b*c^5 + a^3*c^2)*d) - 5/3*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^2
Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {5 \, {\left (2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | -b d x - b c + \left (-a b^{2}\right )^{\frac {1}{3}} \right |}\right )\right )}}{18 \, a^{2}} - \frac {b d x + b c}{3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a^{2} d} - \frac {1}{2 \, {\left (d x + c\right )}^{2} a^{2} d} \]
5/18*(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^2 - 1/3*(b*d*x + b*c)/((b*d^3*x^3 + 3*b* c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*a^2*d) - 1/2/((d*x + c)^2*a^2*d)
Time = 6.24 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {\frac {5\,b\,c^3+3\,a}{6\,a^2\,d}+\frac {5\,b\,d^2\,x^3}{6\,a^2}+\frac {5\,b\,c^2\,x}{2\,a^2}+\frac {5\,b\,c\,d\,x^2}{2\,a^2}}{x^2\,\left (10\,b\,c^3\,d^2+a\,d^2\right )+a\,c^2+b\,c^5+x\,\left (5\,b\,d\,c^4+2\,a\,d\,c\right )+b\,d^5\,x^5+10\,b\,c^2\,d^3\,x^3+5\,b\,c\,d^4\,x^4}-\frac {5\,b^{2/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{8/3}\,d}+\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d}-\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d} \]
(5*b^(2/3)*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)*1i)/2 + 1/2))/(9*a^(8/3)*d) - (5*b^(2/3)*log(b^(1/3)*c + a^(1/3 ) + b^(1/3)*d*x))/(9*a^(8/3)*d) - ((3*a + 5*b*c^3)/(6*a^2*d) + (5*b*d^2*x^ 3)/(6*a^2) + (5*b*c^2*x)/(2*a^2) + (5*b*c*d*x^2)/(2*a^2))/(x^2*(a*d^2 + 10 *b*c^3*d^2) + a*c^2 + b*c^5 + x*(2*a*c*d + 5*b*c^4*d) + b*d^5*x^5 + 10*b*c ^2*d^3*x^3 + 5*b*c*d^4*x^4) - (5*b^(2/3)*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^(8/3)*d)