Integrand size = 38, antiderivative size = 357 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=-\frac {c}{2 a^3 x^2}-\frac {d}{a^3 x}-\frac {x \left (b c-a f+(b d-a g) x+(b e-a h) x^2\right )}{6 a^2 \left (a+b x^3\right )^2}-\frac {x \left (11 b c-5 a f+2 (5 b d-2 a g) x+3 (3 b e-a h) x^2\right )}{18 a^3 \left (a+b x^3\right )}+\frac {\left (20 b^{4/3} c+14 \sqrt [3]{a} b d-5 a \sqrt [3]{b} f-2 a^{4/3} g\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{11/3} b^{2/3}}+\frac {e \log (x)}{a^3}-\frac {\left (5 \sqrt [3]{b} (4 b c-a f)-2 \sqrt [3]{a} (7 b d-a g)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} b^{2/3}}+\frac {\left (20 b c-5 a f-\frac {2 \sqrt [3]{a} (7 b d-a g)}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{11/3} \sqrt [3]{b}}-\frac {e \log \left (a+b x^3\right )}{3 a^3} \] Output:
-1/2*c/a^3/x^2-d/a^3/x-1/6*x*(b*c-a*f+(-a*g+b*d)*x+(-a*h+b*e)*x^2)/a^2/(b* x^3+a)^2-1/18*x*(11*b*c-5*a*f+2*(-2*a*g+5*b*d)*x+3*(-a*h+3*b*e)*x^2)/a^3/( b*x^3+a)+1/27*(20*b^(4/3)*c+14*a^(1/3)*b*d-5*a*b^(1/3)*f-2*a^(4/3)*g)*arct an(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(11/3)/b^(2/3)+e*l n(x)/a^3-1/27*(5*b^(1/3)*(-a*f+4*b*c)-2*a^(1/3)*(-a*g+7*b*d))*ln(a^(1/3)+b ^(1/3)*x)/a^(11/3)/b^(2/3)+1/54*(20*b*c-5*a*f-2*a^(1/3)*(-a*g+7*b*d)/b^(1/ 3))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(11/3)/b^(1/3)-1/3*e*ln(b* x^3+a)/a^3
Time = 0.49 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=-\frac {\frac {27 a c}{x^2}+\frac {54 a d}{x}-\frac {3 a (6 a e-b x (11 c+10 d x)+a x (5 f+4 g x))}{a+b x^3}+\frac {9 a^2 \left (a^2 h+b^2 x (c+d x)-a b (e+x (f+g x))\right )}{b \left (a+b x^3\right )^2}+\frac {2 \sqrt {3} \sqrt [3]{a} \left (-20 b^{4/3} c-14 \sqrt [3]{a} b d+5 a \sqrt [3]{b} f+2 a^{4/3} g\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}-54 a e \log (x)+\frac {2 \sqrt [3]{a} \left (20 b^{4/3} c-14 \sqrt [3]{a} b d-5 a \sqrt [3]{b} f+2 a^{4/3} g\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}-\frac {\sqrt [3]{a} \left (20 b^{4/3} c-14 \sqrt [3]{a} b d-5 a \sqrt [3]{b} f+2 a^{4/3} g\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}+18 a e \log \left (a+b x^3\right )}{54 a^4} \] Input:
Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)^3),x]
Output:
-1/54*((27*a*c)/x^2 + (54*a*d)/x - (3*a*(6*a*e - b*x*(11*c + 10*d*x) + a*x *(5*f + 4*g*x)))/(a + b*x^3) + (9*a^2*(a^2*h + b^2*x*(c + d*x) - a*b*(e + x*(f + g*x))))/(b*(a + b*x^3)^2) + (2*Sqrt[3]*a^(1/3)*(-20*b^(4/3)*c - 14* a^(1/3)*b*d + 5*a*b^(1/3)*f + 2*a^(4/3)*g)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/ 3))/Sqrt[3]])/b^(2/3) - 54*a*e*Log[x] + (2*a^(1/3)*(20*b^(4/3)*c - 14*a^(1 /3)*b*d - 5*a*b^(1/3)*f + 2*a^(4/3)*g)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) - (a^(1/3)*(20*b^(4/3)*c - 14*a^(1/3)*b*d - 5*a*b^(1/3)*f + 2*a^(4/3)*g)*Lo g[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(2/3) + 18*a*e*Log[a + b*x ^3])/a^4
Time = 1.90 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2368, 25, 2368, 27, 2373, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle -\frac {\int -\frac {-3 b^2 \left (\frac {b e}{a}-h\right ) x^5-4 b^2 \left (\frac {b d}{a}-g\right ) x^4-5 b^2 \left (\frac {b c}{a}-f\right ) x^3+6 b^2 e x^2+6 b^2 d x+6 b^2 c}{x^3 \left (b x^3+a\right )^2}dx}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-3 b^2 \left (\frac {b e}{a}-h\right ) x^5-4 b^2 \left (\frac {b d}{a}-g\right ) x^4-5 b^2 \left (\frac {b c}{a}-f\right ) x^3+6 b^2 e x^2+6 b^2 d x+6 b^2 c}{x^3 \left (b x^3+a\right )^2}dx}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (-\left (\left (\frac {5 b d}{a}-2 g\right ) x^4 b^4\right )-\left (\frac {11 b c}{a}-5 f\right ) x^3 b^4+9 e x^2 b^4+9 c b^4+9 d x b^4\right )}{x^3 \left (b x^3+a\right )}dx}{3 a b^2}-\frac {x \left (b^2 (11 b c-5 a f)+2 b^2 x (5 b d-2 a g)+3 b^2 x^2 (3 b e-a h)\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 \int \frac {-\left (\left (\frac {5 b d}{a}-2 g\right ) x^4 b^4\right )-\left (\frac {11 b c}{a}-5 f\right ) x^3 b^4+9 e x^2 b^4+9 c b^4+9 d x b^4}{x^3 \left (b x^3+a\right )}dx}{3 a b^2}-\frac {x \left (b^2 (11 b c-5 a f)+2 b^2 x (5 b d-2 a g)+3 b^2 x^2 (3 b e-a h)\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \frac {\frac {2 \int \left (\frac {9 e b^4}{a x}+\frac {\left (-9 b e x^2-2 (7 b d-a g) x-5 (4 b c-a f)\right ) b^4}{a \left (b x^3+a\right )}+\frac {9 d b^4}{a x^2}+\frac {9 c b^4}{a x^3}\right )dx}{3 a b^2}-\frac {x \left (b^2 (11 b c-5 a f)+2 b^2 x (5 b d-2 a g)+3 b^2 x^2 (3 b e-a h)\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 \left (\frac {b^{10/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-2 a^{4/3} g+14 \sqrt [3]{a} b d-5 a \sqrt [3]{b} f+20 b^{4/3} c\right )}{\sqrt {3} a^{5/3}}+\frac {b^{11/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {2 \sqrt [3]{a} (7 b d-a g)}{\sqrt [3]{b}}-5 a f+20 b c\right )}{6 a^{5/3}}-\frac {b^{10/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (5 \sqrt [3]{b} (4 b c-a f)-2 \sqrt [3]{a} (7 b d-a g)\right )}{3 a^{5/3}}-\frac {9 b^4 c}{2 a x^2}-\frac {9 b^4 d}{a x}-\frac {3 b^4 e \log \left (a+b x^3\right )}{a}+\frac {9 b^4 e \log (x)}{a}\right )}{3 a b^2}-\frac {x \left (b^2 (11 b c-5 a f)+2 b^2 x (5 b d-2 a g)+3 b^2 x^2 (3 b e-a h)\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
Input:
Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)^3),x]
Output:
-1/6*(x*(b*c - a*f + (b*d - a*g)*x + (b*e - a*h)*x^2))/(a^2*(a + b*x^3)^2) + (-1/3*(x*(b^2*(11*b*c - 5*a*f) + 2*b^2*(5*b*d - 2*a*g)*x + 3*b^2*(3*b*e - a*h)*x^2))/(a^2*(a + b*x^3)) + (2*((-9*b^4*c)/(2*a*x^2) - (9*b^4*d)/(a* x) + (b^(10/3)*(20*b^(4/3)*c + 14*a^(1/3)*b*d - 5*a*b^(1/3)*f - 2*a^(4/3)* g)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3)) + (9*b^4*e*Log[x])/a - (b^(10/3)*(5*b^(1/3)*(4*b*c - a*f) - 2*a^(1/3)*(7*b*d - a*g))*Log[a^(1/3) + b^(1/3)*x])/(3*a^(5/3)) + (b^(11/3)*(20*b*c - 5*a*f - (2*a^(1/3)*(7*b*d - a*g))/b^(1/3))*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^ (2/3)*x^2])/(6*a^(5/3)) - (3*b^4*e*Log[a + b*x^3])/a))/(3*a*b^2))/(6*a*b^2 )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1)*x^m *Pq, a + b*x^n, x], i}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^( Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[x^m*(a + b*x^n)^(p + 1)*ExpandToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x]]] /; F reeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[E xpandIntegrand[(c*x)^m*(Pq/(a + b*x^n)), x], x] /; FreeQ[{a, b, c, m}, x] & & PolyQ[Pq, x] && IntegerQ[n] && !IGtQ[m, 0]
Time = 0.15 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.95
| method | result | size |
| default | \(\frac {\frac {\left (\frac {2}{9} a b g -\frac {5}{9} b^{2} d \right ) x^{5}+\left (\frac {5}{18} a b f -\frac {11}{18} b^{2} c \right ) x^{4}+\frac {a b e \,x^{3}}{3}+\frac {a \left (7 a g -13 b d \right ) x^{2}}{18}+\frac {a \left (4 a f -7 c b \right ) x}{9}-\frac {a^{2} \left (a h -3 b e \right )}{6 b}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (5 a f -20 c b \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{9}+\frac {\left (2 a g -14 b d \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9}-\frac {e \ln \left (b \,x^{3}+a \right )}{3}}{a^{3}}-\frac {c}{2 a^{3} x^{2}}-\frac {d}{a^{3} x}+\frac {e \ln \left (x \right )}{a^{3}}\) | \(340\) |
| risch | \(\frac {\frac {2 b \left (a g -7 b d \right ) x^{7}}{9 a^{3}}+\frac {5 b \left (a f -4 c b \right ) x^{6}}{18 a^{3}}+\frac {e b \,x^{5}}{3 a^{2}}+\frac {7 \left (a g -7 b d \right ) x^{4}}{18 a^{2}}+\frac {4 \left (a f -4 c b \right ) x^{3}}{9 a^{2}}-\frac {\left (a h -3 b e \right ) x^{2}}{6 a b}-\frac {d x}{a}-\frac {c}{2 a}}{x^{2} \left (b \,x^{3}+a \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{11} b^{2} \textit {\_Z}^{3}+27 a^{8} b^{2} e \,\textit {\_Z}^{2}+\left (30 a^{6} b f g -120 a^{5} b^{2} c g -210 a^{5} b^{2} d f +243 a^{5} b^{2} e^{2}+840 a^{4} b^{3} c d \right ) \textit {\_Z} +8 a^{4} g^{3}-168 a^{3} b d \,g^{2}+270 a^{3} b e f g -125 a^{3} b \,f^{3}-1080 a^{2} b^{2} c e g +1500 a^{2} b^{2} c \,f^{2}+1176 a^{2} b^{2} d^{2} g -1890 a^{2} b^{2} d e f +729 a^{2} b^{2} e^{3}-6000 a \,b^{3} c^{2} f +7560 a \,b^{3} c d e -2744 a \,b^{3} d^{3}+8000 b^{4} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{11} b^{2}-72 \textit {\_R}^{2} a^{8} b^{2} e +\left (-100 a^{6} b f g +400 a^{5} b^{2} c g +700 a^{5} b^{2} d f -324 a^{5} b^{2} e^{2}-2800 a^{4} b^{3} c d \right ) \textit {\_R} -24 a^{4} g^{3}+504 a^{3} b d \,g^{2}-540 a^{3} b e f g +375 a^{3} b \,f^{3}+2160 a^{2} b^{2} c e g -4500 a^{2} b^{2} c \,f^{2}-3528 a^{2} b^{2} d^{2} g +3780 a^{2} b^{2} d e f +18000 a \,b^{3} c^{2} f -15120 a \,b^{3} c d e +8232 a \,b^{3} d^{3}-24000 b^{4} c^{3}\right ) x +\left (2 a^{9} b g -14 a^{8} b^{2} d \right ) \textit {\_R}^{2}+\left (-36 a^{6} b e g -25 a^{6} b \,f^{2}+200 a^{5} b^{2} c f +252 a^{5} b^{2} d e -400 a^{4} b^{3} c^{2}\right ) \textit {\_R} -486 a^{3} b \,e^{2} g +675 a^{3} b e \,f^{2}-5400 a^{2} b^{2} c e f +3402 a^{2} b^{2} d \,e^{2}+10800 a \,b^{3} c^{2} e \right )\right )}{27}+\frac {e \ln \left (-x \right )}{a^{3}}\) | \(668\) |
Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a)^3,x,method=_RETURNVERBOS E)
Output:
1/a^3*(((2/9*a*b*g-5/9*b^2*d)*x^5+(5/18*a*b*f-11/18*b^2*c)*x^4+1/3*a*b*e*x ^3+1/18*a*(7*a*g-13*b*d)*x^2+1/9*a*(4*a*f-7*b*c)*x-1/6*a^2*(a*h-3*b*e)/b)/ (b*x^3+a)^2+1/9*(5*a*f-20*b*c)*(1/3/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-1/6/b/ (a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/3/b/(a/b)^(2/3)*3^(1/2)*ar ctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))+1/9*(2*a*g-14*b*d)*(-1/3/b/(a/b)^(1 /3)*ln(x+(a/b)^(1/3))+1/6/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+ 1/3*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))-1/3*e*l n(b*x^3+a))-1/2*c/a^3/x^2-d/a^3/x+e*ln(x)/a^3
Result contains complex when optimal does not.
Time = 16.52 (sec) , antiderivative size = 12435, normalized size of antiderivative = 34.83 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a)^3,x, algorithm="fr icas")
Output:
Too large to include
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**3/(b*x**3+a)**3,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=\frac {6 \, a b^{2} e x^{5} - 4 \, {\left (7 \, b^{3} d - a b^{2} g\right )} x^{7} - 5 \, {\left (4 \, b^{3} c - a b^{2} f\right )} x^{6} - 18 \, a^{2} b d x - 7 \, {\left (7 \, a b^{2} d - a^{2} b g\right )} x^{4} - 9 \, a^{2} b c - 8 \, {\left (4 \, a b^{2} c - a^{2} b f\right )} x^{3} + 3 \, {\left (3 \, a^{2} b e - a^{3} h\right )} x^{2}}{18 \, {\left (a^{3} b^{3} x^{8} + 2 \, a^{4} b^{2} x^{5} + a^{5} b x^{2}\right )}} + \frac {e \log \left (x\right )}{a^{3}} - \frac {\sqrt {3} {\left (14 \, b d \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a g \left (\frac {a}{b}\right )^{\frac {2}{3}} + 20 \, b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a f \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{4}} - \frac {{\left (18 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} + 14 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} - 20 \, b c + 5 \, a f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, a^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (9 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - 14 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} + 20 \, b c - 5 \, a f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a)^3,x, algorithm="ma xima")
Output:
1/18*(6*a*b^2*e*x^5 - 4*(7*b^3*d - a*b^2*g)*x^7 - 5*(4*b^3*c - a*b^2*f)*x^ 6 - 18*a^2*b*d*x - 7*(7*a*b^2*d - a^2*b*g)*x^4 - 9*a^2*b*c - 8*(4*a*b^2*c - a^2*b*f)*x^3 + 3*(3*a^2*b*e - a^3*h)*x^2)/(a^3*b^3*x^8 + 2*a^4*b^2*x^5 + a^5*b*x^2) + e*log(x)/a^3 - 1/27*sqrt(3)*(14*b*d*(a/b)^(2/3) - 2*a*g*(a/b )^(2/3) + 20*b*c*(a/b)^(1/3) - 5*a*f*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/a^4 - 1/54*(18*b*e*(a/b)^(2/3) + 14*b*d*(a/b)^ (1/3) - 2*a*g*(a/b)^(1/3) - 20*b*c + 5*a*f)*log(x^2 - x*(a/b)^(1/3) + (a/b )^(2/3))/(a^3*b*(a/b)^(2/3)) - 1/27*(9*b*e*(a/b)^(2/3) - 14*b*d*(a/b)^(1/3 ) + 2*a*g*(a/b)^(1/3) + 20*b*c - 5*a*f)*log(x + (a/b)^(1/3))/(a^3*b*(a/b)^ (2/3))
Time = 0.14 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=-\frac {e \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3}} + \frac {e \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {\sqrt {3} {\left (20 \, b^{2} c - 5 \, a b f - 14 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{3}} + \frac {{\left (20 \, b^{2} c - 5 \, a b f + 14 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{3}} - \frac {28 \, b^{3} d x^{7} - 4 \, a b^{2} g x^{7} + 20 \, b^{3} c x^{6} - 5 \, a b^{2} f x^{6} - 6 \, a b^{2} e x^{5} + 49 \, a b^{2} d x^{4} - 7 \, a^{2} b g x^{4} + 32 \, a b^{2} c x^{3} - 8 \, a^{2} b f x^{3} - 9 \, a^{2} b e x^{2} + 3 \, a^{3} h x^{2} + 18 \, a^{2} b d x + 9 \, a^{2} b c}{18 \, {\left (b x^{4} + a x\right )}^{2} a^{3} b} + \frac {{\left (14 \, a^{3} b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a^{4} b g \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 20 \, a^{3} b^{2} c - 5 \, a^{4} b f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{7} b} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a)^3,x, algorithm="gi ac")
Output:
-1/3*e*log(abs(b*x^3 + a))/a^3 + e*log(abs(x))/a^3 + 1/27*sqrt(3)*(20*b^2* c - 5*a*b*f - 14*(-a*b^2)^(1/3)*b*d + 2*(-a*b^2)^(1/3)*a*g)*arctan(1/3*sqr t(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)*a^3) + 1/54*(20*b^ 2*c - 5*a*b*f + 14*(-a*b^2)^(1/3)*b*d - 2*(-a*b^2)^(1/3)*a*g)*log(x^2 + x* (-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a^3) - 1/18*(28*b^3*d*x^7 - 4 *a*b^2*g*x^7 + 20*b^3*c*x^6 - 5*a*b^2*f*x^6 - 6*a*b^2*e*x^5 + 49*a*b^2*d*x ^4 - 7*a^2*b*g*x^4 + 32*a*b^2*c*x^3 - 8*a^2*b*f*x^3 - 9*a^2*b*e*x^2 + 3*a^ 3*h*x^2 + 18*a^2*b*d*x + 9*a^2*b*c)/((b*x^4 + a*x)^2*a^3*b) + 1/27*(14*a^3 *b^2*d*(-a/b)^(1/3) - 2*a^4*b*g*(-a/b)^(1/3) + 20*a^3*b^2*c - 5*a^4*b*f)*( -a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^7*b)
Time = 6.84 (sec) , antiderivative size = 1697, normalized size of antiderivative = 4.75 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx=\text {Too large to display} \] Input:
int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)^3),x)
Output:
symsum(log((b^2*e*(400*b^2*c^2 + 25*a^2*f^2 - 18*a^2*e*g - 200*a*b*c*f + 1 26*a*b*d*e))/(81*a^8) - (root(19683*a^11*b^2*z^3 + 19683*a^8*b^2*e*z^2 + 8 10*a^6*b*f*g*z - 5670*a^5*b^2*d*f*z - 3240*a^5*b^2*c*g*z + 22680*a^4*b^3*c *d*z + 6561*a^5*b^2*e^2*z + 270*a^3*b*e*f*g + 7560*a*b^3*c*d*e - 1890*a^2* b^2*d*e*f - 1080*a^2*b^2*c*e*g - 168*a^3*b*d*g^2 - 6000*a*b^3*c^2*f + 1176 *a^2*b^2*d^2*g + 1500*a^2*b^2*c*f^2 + 729*a^2*b^2*e^3 - 125*a^3*b*f^3 - 27 44*a*b^3*d^3 + 8*a^4*g^3 + 8000*b^4*c^3, z, k)*b^2*(400*b^2*c^2 + 25*a^2*f ^2 - 54*root(19683*a^11*b^2*z^3 + 19683*a^8*b^2*e*z^2 + 810*a^6*b*f*g*z - 5670*a^5*b^2*d*f*z - 3240*a^5*b^2*c*g*z + 22680*a^4*b^3*c*d*z + 6561*a^5*b ^2*e^2*z + 270*a^3*b*e*f*g + 7560*a*b^3*c*d*e - 1890*a^2*b^2*d*e*f - 1080* a^2*b^2*c*e*g - 168*a^3*b*d*g^2 - 6000*a*b^3*c^2*f + 1176*a^2*b^2*d^2*g + 1500*a^2*b^2*c*f^2 + 729*a^2*b^2*e^3 - 125*a^3*b*f^3 - 2744*a*b^3*d^3 + 8* a^4*g^3 + 8000*b^4*c^3, z, k)*a^5*g + 36*a^2*e*g + 378*root(19683*a^11*b^2 *z^3 + 19683*a^8*b^2*e*z^2 + 810*a^6*b*f*g*z - 5670*a^5*b^2*d*f*z - 3240*a ^5*b^2*c*g*z + 22680*a^4*b^3*c*d*z + 6561*a^5*b^2*e^2*z + 270*a^3*b*e*f*g + 7560*a*b^3*c*d*e - 1890*a^2*b^2*d*e*f - 1080*a^2*b^2*c*e*g - 168*a^3*b*d *g^2 - 6000*a*b^3*c^2*f + 1176*a^2*b^2*d^2*g + 1500*a^2*b^2*c*f^2 + 729*a^ 2*b^2*e^3 - 125*a^3*b*f^3 - 2744*a*b^3*d^3 + 8*a^4*g^3 + 8000*b^4*c^3, z, k)*a^4*b*d + 324*a*b*e^2*x + 2800*b^2*c*d*x + 100*a^2*f*g*x + 2916*root(19 683*a^11*b^2*z^3 + 19683*a^8*b^2*e*z^2 + 810*a^6*b*f*g*z - 5670*a^5*b^2...
Time = 0.18 (sec) , antiderivative size = 1434, normalized size of antiderivative = 4.02 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^3} \, dx =\text {Too large to display} \] Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a)^3,x)
Output:
( - 10*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)* sqrt(3)))*a**3*b*f*x**2 + 40*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2* b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*c*x**2 - 20*b**(1/3)*a**(2/3)*sq rt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*f*x**5 + 80*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sq rt(3)))*a*b**3*c*x**5 - 10*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b* *(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**3*f*x**8 + 40*b**(1/3)*a**(2/3)*sqrt(3) *atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**4*c*x**8 - 4*sqrt(3 )*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**4*b*g*x**2 + 28*sq rt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b**2*d*x**2 - 8*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b**2*g *x**5 + 56*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2 *b**3*d*x**5 - 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)) )*a**2*b**3*g*x**8 + 28*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*s qrt(3)))*a*b**4*d*x**8 - 5*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1 /3)*x + b**(2/3)*x**2)*a**3*b*f*x**2 + 20*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*c*x**2 - 10*b**(1/3)*a**(2 /3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*f*x**5 + 40*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)* a*b**3*c*x**5 - 5*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x ...