Integrand size = 17, antiderivative size = 224 \[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\frac {\sqrt [3]{b} c \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} c+b^{2/3} c^2\right )}+\frac {\log (x)}{a+b c^3}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )}-\frac {\left (2 \sqrt [3]{a}-\sqrt [3]{b} c\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} c+b^{2/3} c^2\right )} \]
ln(x)/(b*c^3+a)-1/3*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(2/3)/(a^(1/3)+b^(1/3)*c )-1/6*(2*a^(1/3)-b^(1/3)*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d* x+c)^2)/a^(2/3)/(a^(2/3)-a^(1/3)*b^(1/3)*c+b^(2/3)*c^2)+1/3*b^(1/3)*c*arct an(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/a^(1/3)*3^(1/2))/a^(2/3)/(a^(2/3)-a^(1/ 3)*b^(1/3)*c+b^(2/3)*c^2)*3^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=-\frac {-3 \log (x)+\text {RootSum}\left [a+b c^3+3 b c^2 d \text {$\#$1}+3 b c d^2 \text {$\#$1}^2+b d^3 \text {$\#$1}^3\&,\frac {3 c^2 \log (x-\text {$\#$1})+3 c d \log (x-\text {$\#$1}) \text {$\#$1}+d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2}{c^2+2 c d \text {$\#$1}+d^2 \text {$\#$1}^2}\&\right ]}{3 \left (a+b c^3\right )} \]
-1/3*(-3*Log[x] + RootSum[a + b*c^3 + 3*b*c^2*d*#1 + 3*b*c*d^2*#1^2 + b*d^ 3*#1^3 & , (3*c^2*Log[x - #1] + 3*c*d*Log[x - #1]*#1 + d^2*Log[x - #1]*#1^ 2)/(c^2 + 2*c*d*#1 + d^2*#1^2) & ])/(a + b*c^3)
Time = 0.52 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {896, 25, 7276, 2009}
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.
\(\displaystyle \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \int \frac {1}{d x \left (a+b (c+d x)^3\right )}d(c+d x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {1}{d x \left (b (c+d x)^3+a\right )}d(c+d x)\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\int \left (\frac {b \left (c^2+(c+d x) c+(c+d x)^2\right )}{\left (b c^3+a\right ) \left (b (c+d x)^3+a\right )}-\frac {1}{\left (b c^3+a\right ) d x}\right )d(c+d x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [3]{b} c \left (\sqrt [3]{a}+\sqrt [3]{b} c\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a+b c^3\right )}-\frac {\sqrt [3]{b} c \left (\sqrt [3]{a}-\sqrt [3]{b} c\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )}+\frac {\sqrt [3]{b} c \left (\sqrt [3]{a}-\sqrt [3]{b} c\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )}+\frac {\log (-d x)}{a+b c^3}-\frac {\log \left (a+b (c+d x)^3\right )}{3 \left (a+b c^3\right )}\) |
(b^(1/3)*c*(a^(1/3) + b^(1/3)*c)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(S qrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*(a + b*c^3)) + Log[-(d*x)]/(a + b*c^3) + (b^(1/3)*c*(a^(1/3) - b^(1/3)*c)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*a^ (2/3)*(a + b*c^3)) - (b^(1/3)*c*(a^(1/3) - b^(1/3)*c)*Log[a^(2/3) - a^(1/3 )*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(6*a^(2/3)*(a + b*c^3)) - Log[ a + b*(c + d*x)^3]/(3*(a + b*c^3))
3.2.7.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {\ln \left (x \right )}{b \,c^{3}+a}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +b \,c^{3}+a \right )}{\sum }\frac {\left (d^{2} \textit {\_R}^{2}+3 c d \textit {\_R} +3 c^{2}\right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{3 \left (b \,c^{3}+a \right )}\) | \(105\) |
risch | \(\frac {\ln \left (-x \right )}{b \,c^{3}+a}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (a^{2} b \,c^{3}+a^{3}\right ) \textit {\_Z}^{3}+3 a^{2} \textit {\_Z}^{2}+3 a \textit {\_Z} \right )}{\sum }\textit {\_R} \ln \left (\left (\left (2 a b \,c^{3} d -4 d \,a^{2}\right ) \textit {\_R}^{2}+\left (b \,c^{3} d -8 d a \right ) \textit {\_R} -4 d \right ) x +\left (a b \,c^{4}+c \,a^{2}\right ) \textit {\_R}^{2}+\left (b \,c^{4}-2 a c \right ) \textit {\_R} -3 c \right )\right )}{3}\) | \(125\) |
ln(x)/(b*c^3+a)-1/3*sum((_R^2*d^2+3*_R*c*d+3*c^2)/(_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))/(b*c^3 +a)
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 4370, normalized size of antiderivative = 19.51 \[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\text {Too large to display} \]
1/12*(2*(b*c^3 + a)*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(1/(a*b*c^3 + a^2) - 1 /(b*c^3 + a)^2)/(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b *c^3 + a^2)*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - (1/2)^(1/3)*(I*sqrt(3) + 1)*(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c^3 + a^2 )*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - 2/(b*c^3 + a))*log(b*c^2*d*x + b *c^3 + 1/4*(a^2*b*c^3 + a^3)*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(1/(a*b*c^3 + a^2) - 1/(b*c^3 + a)^2)/(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c^3 + a^2)*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - (1/2)^(1/3)*( I*sqrt(3) + 1)*(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b* c^3 + a^2)*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - 2/(b*c^3 + a))^2 - 1/2* (a*b*c^3 - 2*a^2)*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(1/(a*b*c^3 + a^2) - 1/( b*c^3 + a)^2)/(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c ^3 + a^2)*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - (1/2)^(1/3)*(I*sqrt(3) + 1)*(b*c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c^3 + a^2)* (b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - 2/(b*c^3 + a)) + a) - ((b*c^3 + a) *(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(1/(a*b*c^3 + a^2) - 1/(b*c^3 + a)^2)/(b* c^3/((b*c^3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c^3 + a^2)*(b*c^3 + a)) - 2/(b*c^3 + a)^3)^(1/3) - (1/2)^(1/3)*(I*sqrt(3) + 1)*(b*c^3/((b*c^ 3 + a)^2*a^2) - 1/(a^2*b*c^3 + a^3) + 3/((a*b*c^3 + a^2)*(b*c^3 + a)) - 2/ (b*c^3 + a)^3)^(1/3) - 2/(b*c^3 + a)) + 3*sqrt(1/3)*(b*c^3 + a)*sqrt(-(...
Timed out. \[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )} x} \,d x } \]
-b*d*integrate((d^2*x^2 + 3*c*d*x + 3*c^2)/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3* b*c^2*d*x + b*c^3 + a), x)/(b*c^3 + a) + log(x)/(b*c^3 + a)
\[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )} x} \,d x } \]
Time = 0.07 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.47 \[ \int \frac {1}{x \left (a+b (c+d x)^3\right )} \, dx=\frac {\ln \left (x\right )}{b\,c^3+a}+\left (\sum _{k=1}^3\ln \left ({\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^2\,b^4\,c^4\,d^8\,3-\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )\,b^3\,c\,d^8\,3-\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )\,b^3\,d^9\,x\,4-{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^2\,a\,b^3\,c\,d^8\,6-{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^2\,a\,b^3\,d^9\,x\,24+{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^3\,a^2\,b^3\,c\,d^8\,9+{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^3\,a\,b^4\,c^4\,d^8\,9-{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^3\,a^2\,b^3\,d^9\,x\,36+{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^2\,b^4\,c^3\,d^9\,x\,3+{\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )}^3\,a\,b^4\,c^3\,d^9\,x\,18\right )\,\mathrm {root}\left (27\,a^2\,b\,c^3\,z^3+27\,a^3\,z^3+27\,a^2\,z^2+9\,a\,z+1,z,k\right )\right ) \]
log(x)/(a + b*c^3) + symsum(log(3*root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27* a^2*z^2 + 9*a*z + 1, z, k)^2*b^4*c^4*d^8 - 3*root(27*a^2*b*c^3*z^3 + 27*a^ 3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)*b^3*c*d^8 - 4*root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)*b^3*d^9*x - 6*root(27*a^2*b*c ^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)^2*a*b^3*c*d^8 - 24*roo t(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)^2*a*b^3*d^ 9*x + 9*root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k) ^3*a^2*b^3*c*d^8 + 9*root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a *z + 1, z, k)^3*a*b^4*c^4*d^8 - 36*root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27 *a^2*z^2 + 9*a*z + 1, z, k)^3*a^2*b^3*d^9*x + 3*root(27*a^2*b*c^3*z^3 + 27 *a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)^2*b^4*c^3*d^9*x + 18*root(27*a^2* b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k)^3*a*b^4*c^3*d^9*x)* root(27*a^2*b*c^3*z^3 + 27*a^3*z^3 + 27*a^2*z^2 + 9*a*z + 1, z, k), k, 1, 3)