Integrand size = 17, antiderivative size = 234 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\frac {x}{b d^3}+\frac {\left (a-3 \sqrt [3]{a} b^{2/3} c^2+b c^3\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{4/3} d^4}-\frac {\left (a+3 \sqrt [3]{a} b^{2/3} c^2+b c^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} b^{4/3} d^4}+\frac {\left (a+3 \sqrt [3]{a} b^{2/3} c^2+b c^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} b^{4/3} d^4}-\frac {c \log \left (a+b (c+d x)^3\right )}{b d^4} \] Output:
x/b/d^3+1/3*(a-3*a^(1/3)*b^(2/3)*c^2+b*c^3)*arctan(1/3*(a^(1/3)-2*b^(1/3)* (d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/a^(2/3)/b^(4/3)/d^4-1/3*(a+3*a^(1/3)*b^( 2/3)*c^2+b*c^3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(2/3)/b^(4/3)/d^4+1/6*(a+3*a ^(1/3)*b^(2/3)*c^2+b*c^3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+ c)^2)/a^(2/3)/b^(4/3)/d^4-c*ln(a+b*(d*x+c)^3)/b/d^4
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.56 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=-\frac {-3 b d x+\text {RootSum}\left [a+b c^3+3 b c^2 d \text {$\#$1}+3 b c d^2 \text {$\#$1}^2+b d^3 \text {$\#$1}^3\&,\frac {a \log (x-\text {$\#$1})+b c^3 \log (x-\text {$\#$1})+3 b c^2 d \log (x-\text {$\#$1}) \text {$\#$1}+3 b c d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2}{c^2+2 c d \text {$\#$1}+d^2 \text {$\#$1}^2}\&\right ]}{3 b^2 d^4} \] Input:
Integrate[x^3/(a + b*(c + d*x)^3),x]
Output:
-1/3*(-3*b*d*x + RootSum[a + b*c^3 + 3*b*c^2*d*#1 + 3*b*c*d^2*#1^2 + b*d^3 *#1^3 & , (a*Log[x - #1] + b*c^3*Log[x - #1] + 3*b*c^2*d*Log[x - #1]*#1 + 3*b*c*d^2*Log[x - #1]*#1^2)/(c^2 + 2*c*d*#1 + d^2*#1^2) & ])/(b^2*d^4)
Time = 0.82 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {896, 25, 2426, 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.
| \(\displaystyle \int \frac {x^3}{a+b (c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 896 |
| \(\displaystyle \frac {\int \frac {d^3 x^3}{b (c+d x)^3+a}d(c+d x)}{d^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int -\frac {d^3 x^3}{b (c+d x)^3+a}d(c+d x)}{d^4}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle -\frac {\int \left (\frac {b c^3-3 b (c+d x) c^2+3 b (c+d x)^2 c+a}{b \left (b (c+d x)^3+a\right )}-\frac {1}{b}\right )d(c+d x)}{d^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (-3 \sqrt [3]{a} b^{2/3} c^2+a+b c^3\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{4/3}}-\frac {\left (3 \sqrt [3]{a} b^{2/3} c^2+a+b c^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} b^{4/3}}+\frac {\left (3 \sqrt [3]{a} b^{2/3} c^2+a+b c^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} b^{4/3}}-\frac {c \log \left (a+b (c+d x)^3\right )}{b}+\frac {c+d x}{b}}{d^4}\) |
Input:
Int[x^3/(a + b*(c + d*x)^3),x]
Output:
((c + d*x)/b + ((a - 3*a^(1/3)*b^(2/3)*c^2 + b*c^3)*ArcTan[(a^(1/3) - 2*b^ (1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*b^(4/3)) - ((a + 3*a ^(1/3)*b^(2/3)*c^2 + b*c^3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*a^(2/3)*b ^(4/3)) + ((a + 3*a^(1/3)*b^(2/3)*c^2 + b*c^3)*Log[a^(2/3) - a^(1/3)*b^(1/ 3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(6*a^(2/3)*b^(4/3)) - (c*Log[a + b*(c + d*x)^3])/b)/d^4
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[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.46
| method | result | size |
| default | \(\frac {x}{b \,d^{3}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 \textit {\_Z} b \,c^{2} d +c^{3} b +a \right )}{\sum }\frac {\left (-3 \textit {\_R}^{2} b c \,d^{2}-3 \textit {\_R} b \,c^{2} d -c^{3} b -a \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{3 b^{2} d^{4}}\) | \(108\) |
| risch | \(\frac {x}{b \,d^{3}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 \textit {\_Z} b \,c^{2} d +c^{3} b +a \right )}{\sum }\frac {\left (-3 \textit {\_R}^{2} b c \,d^{2}-3 \textit {\_R} b \,c^{2} d -c^{3} b -a \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{3 b^{2} d^{4}}\) | \(108\) |
Input:
int(x^3/(a+b*(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
x/b/d^3+1/3/b^2/d^4*sum((-3*_R^2*b*c*d^2-3*_R*b*c^2*d-b*c^3-a)/(_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))
Result contains complex when optimal does not.
Time = 10.89 (sec) , antiderivative size = 6315, normalized size of antiderivative = 26.99 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\text {Too large to display} \] Input:
integrate(x^3/(a+b*(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
Time = 1.75 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\operatorname {RootSum} {\left (27 t^{3} a^{2} b^{4} d^{12} + 81 t^{2} a^{2} b^{3} c d^{8} + t \left (54 a^{2} b^{2} c^{2} d^{4} - 27 a b^{3} c^{5} d^{4}\right ) + a^{3} + 3 a^{2} b c^{3} + 3 a b^{2} c^{6} + b^{3} c^{9}, \left ( t \mapsto t \log {\left (x + \frac {- 27 t^{2} a^{2} b^{3} c^{2} d^{8} - 3 t a^{3} b d^{4} - 60 t a^{2} b^{2} c^{3} d^{4} - 3 t a b^{3} c^{6} d^{4} - 2 a^{3} c - 12 a^{2} b c^{4} - 9 a b^{2} c^{7} + b^{3} c^{10}}{a^{3} d + 3 a^{2} b c^{3} d - 24 a b^{2} c^{6} d + b^{3} c^{9} d} \right )} \right )\right )} + \frac {x}{b d^{3}} \] Input:
integrate(x**3/(a+b*(d*x+c)**3),x)
Output:
RootSum(27*_t**3*a**2*b**4*d**12 + 81*_t**2*a**2*b**3*c*d**8 + _t*(54*a**2 *b**2*c**2*d**4 - 27*a*b**3*c**5*d**4) + a**3 + 3*a**2*b*c**3 + 3*a*b**2*c **6 + b**3*c**9, Lambda(_t, _t*log(x + (-27*_t**2*a**2*b**3*c**2*d**8 - 3* _t*a**3*b*d**4 - 60*_t*a**2*b**2*c**3*d**4 - 3*_t*a*b**3*c**6*d**4 - 2*a** 3*c - 12*a**2*b*c**4 - 9*a*b**2*c**7 + b**3*c**10)/(a**3*d + 3*a**2*b*c**3 *d - 24*a*b**2*c**6*d + b**3*c**9*d)))) + x/(b*d**3)
\[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\int { \frac {x^{3}}{{\left (d x + c\right )}^{3} b + a} \,d x } \] Input:
integrate(x^3/(a+b*(d*x+c)^3),x, algorithm="maxima")
Output:
x/(b*d^3) - integrate((3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a), x)/(b*d^3)
\[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\int { \frac {x^{3}}{{\left (d x + c\right )}^{3} b + a} \,d x } \] Input:
integrate(x^3/(a+b*(d*x+c)^3),x, algorithm="giac")
Output:
integrate(x^3/((d*x + c)^3*b + a), x)
Time = 1.50 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.60 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\left (\sum _{k=1}^3\ln \left (\frac {3\,\left (b\,c^5+a\,c^2\right )}{d^2}-\mathrm {root}\left (27\,a^2\,b^4\,d^{12}\,z^3+81\,a^2\,b^3\,c\,d^8\,z^2+54\,a^2\,b^2\,c^2\,d^4\,z-27\,a\,b^3\,c^5\,d^4\,z+3\,a\,b^2\,c^6+3\,a^2\,b\,c^3+b^3\,c^9+a^3,z,k\right )\,\left (\frac {3\,\left (b^2\,c^4\,d^4-5\,a\,b\,c\,d^4\right )}{d^2}+\frac {3\,x\,\left (b^2\,c^3\,d^4+a\,b\,d^4\right )}{d}-\mathrm {root}\left (27\,a^2\,b^4\,d^{12}\,z^3+81\,a^2\,b^3\,c\,d^8\,z^2+54\,a^2\,b^2\,c^2\,d^4\,z-27\,a\,b^3\,c^5\,d^4\,z+3\,a\,b^2\,c^6+3\,a^2\,b\,c^3+b^3\,c^9+a^3,z,k\right )\,a\,b^2\,d^6\,9\right )-\frac {3\,x\,\left (a\,c-2\,b\,c^4\right )}{d}\right )\,\mathrm {root}\left (27\,a^2\,b^4\,d^{12}\,z^3+81\,a^2\,b^3\,c\,d^8\,z^2+54\,a^2\,b^2\,c^2\,d^4\,z-27\,a\,b^3\,c^5\,d^4\,z+3\,a\,b^2\,c^6+3\,a^2\,b\,c^3+b^3\,c^9+a^3,z,k\right )\right )+\frac {x}{b\,d^3} \] Input:
int(x^3/(a + b*(c + d*x)^3),x)
Output:
symsum(log((3*(a*c^2 + b*c^5))/d^2 - root(27*a^2*b^4*d^12*z^3 + 81*a^2*b^3 *c*d^8*z^2 + 54*a^2*b^2*c^2*d^4*z - 27*a*b^3*c^5*d^4*z + 3*a*b^2*c^6 + 3*a ^2*b*c^3 + b^3*c^9 + a^3, z, k)*((3*(b^2*c^4*d^4 - 5*a*b*c*d^4))/d^2 + (3* x*(b^2*c^3*d^4 + a*b*d^4))/d - 9*root(27*a^2*b^4*d^12*z^3 + 81*a^2*b^3*c*d ^8*z^2 + 54*a^2*b^2*c^2*d^4*z - 27*a*b^3*c^5*d^4*z + 3*a*b^2*c^6 + 3*a^2*b *c^3 + b^3*c^9 + a^3, z, k)*a*b^2*d^6) - (3*x*(a*c - 2*b*c^4))/d)*root(27* a^2*b^4*d^12*z^3 + 81*a^2*b^3*c*d^8*z^2 + 54*a^2*b^2*c^2*d^4*z - 27*a*b^3* c^5*d^4*z + 3*a*b^2*c^6 + 3*a^2*b*c^3 + b^3*c^9 + a^3, z, k), k, 1, 3) + x /(b*d^3)
Time = 0.24 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.97 \[ \int \frac {x^3}{a+b (c+d x)^3} \, dx=\frac {2 b^{\frac {1}{3}} a^{\frac {5}{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 )+2 b^{\frac {4}{3}} a^{\frac {2}{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 ) c^{3}-6 \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 ) a b \,c^{2}+b^{\frac {1}{3}} a^{\frac {5}{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^{\frac {4}{3}} a^{\frac {2}{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 ) c^{3}-2 b^{\frac {1}{3}} a^{\frac {5}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right )-2 b^{\frac {4}{3}} a^{\frac {2}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) c^{3}-6 b^{\frac {2}{3}} a^{\frac {4}{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 ) c -6 b^{\frac {2}{3}} a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) c +6 b^{\frac {2}{3}} a^{\frac {4}{3}} d x +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 ) a b \,c^{2}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) a b \,c^{2}}{6 b^{\frac {5}{3}} a^{\frac {4}{3}} d^{4}} \] Input:
int(x^3/(a+b*(d*x+c)^3),x)
Output:
(2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*c - 2*b**(1/3)*d* x)/(a**(1/3)*sqrt(3)))*a + 2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2* b**(1/3)*c - 2*b**(1/3)*d*x)/(a**(1/3)*sqrt(3)))*b*c**3 - 6*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 + b **(1/3)*a**(2/3)*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 + b**(1/3)*a* *(2/3)*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**3 - 2*b**(1/3)*a**(2 /3)*log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*a - 2*b**(1/3)*a**(2/3)*log( a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*b*c**3 - 6*b**(2/3)*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*c - 6*b**(2/3)*a**(1/3)*log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*a*c + 6*b**(2/3)*a**(1/3)*a*d*x + 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*b*c**2 - 6*log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*a*b*c**2)/(6*b**(2/3)*a**(1/3)*a*b*d**4)