Integrand size = 32, antiderivative size = 395 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\frac {3 (b C d-b c D-2 a d D) \sqrt [3]{c+d x}}{b^3 d^2}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt [3]{c+d x}}{b^3 (b c-a d) (a+b x)}+\frac {3 D (c+d x)^{4/3}}{4 b^2 d^2}-\frac {\left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} b^{10/3} (b c-a d)^{5/3}}-\frac {\left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \log (a+b x)}{6 b^{10/3} (b c-a d)^{5/3}}+\frac {\left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 b^{10/3} (b c-a d)^{5/3}} \] Output:
3*(C*b*d-2*D*a*d-D*b*c)*(d*x+c)^(1/3)/b^3/d^2-(A*b^3-a*(B*b^2-C*a*b+D*a^2) )*(d*x+c)^(1/3)/b^3/(-a*d+b*c)/(b*x+a)+3/4*D*(d*x+c)^(4/3)/b^2/d^2-1/3*(b^ 3*(-2*A*d+3*B*c)-a*b^2*(B*d+6*C*c)-7*a^3*d*D+a^2*b*(4*C*d+9*D*c))*arctan(1 /3*(1+2*b^(1/3)*(d*x+c)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))*3^(1/2)/b^(10/3)/ (-a*d+b*c)^(5/3)-1/6*(b^3*(-2*A*d+3*B*c)-a*b^2*(B*d+6*C*c)-7*a^3*d*D+a^2*b *(4*C*d+9*D*c))*ln(b*x+a)/b^(10/3)/(-a*d+b*c)^(5/3)+1/2*(b^3*(-2*A*d+3*B*c )-a*b^2*(B*d+6*C*c)-7*a^3*d*D+a^2*b*(4*C*d+9*D*c))*ln((-a*d+b*c)^(1/3)-b^( 1/3)*(d*x+c)^(1/3))/b^(10/3)/(-a*d+b*c)^(5/3)
Time = 1.22 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\frac {\frac {3 \sqrt [3]{b} \sqrt [3]{c+d x} \left (-28 a^3 d^2 D+a^2 b d (16 C d+3 D (5 c-7 d x))+b^3 \left (4 A d^2-3 c x (4 C d-3 c D+d D x)\right )+a b^2 \left (9 c^2 D-12 c d (C-D x)+d^2 (-4 B+3 x (4 C+D x))\right )\right )}{d^2 (-b c+a d) (a+b x)}+\frac {4 \sqrt {3} \left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{-b c+a d}}}{\sqrt {3}}\right )}{(-b c+a d)^{5/3}}-\frac {4 \left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \log \left (\sqrt [3]{-b c+a d}+\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{(-b c+a d)^{5/3}}+\frac {2 \left (b^3 (3 B c-2 A d)-a b^2 (6 c C+B d)-7 a^3 d D+a^2 b (4 C d+9 c D)\right ) \log \left ((-b c+a d)^{2/3}-\sqrt [3]{b} \sqrt [3]{-b c+a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{(-b c+a d)^{5/3}}}{12 b^{10/3}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^2*(c + d*x)^(2/3)),x]
Output:
((3*b^(1/3)*(c + d*x)^(1/3)*(-28*a^3*d^2*D + a^2*b*d*(16*C*d + 3*D*(5*c - 7*d*x)) + b^3*(4*A*d^2 - 3*c*x*(4*C*d - 3*c*D + d*D*x)) + a*b^2*(9*c^2*D - 12*c*d*(C - D*x) + d^2*(-4*B + 3*x*(4*C + D*x)))))/(d^2*(-(b*c) + a*d)*(a + b*x)) + (4*Sqrt[3]*(b^3*(3*B*c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3*d *D + a^2*b*(4*C*d + 9*c*D))*ArcTan[(1 - (2*b^(1/3)*(c + d*x)^(1/3))/(-(b*c ) + a*d)^(1/3))/Sqrt[3]])/(-(b*c) + a*d)^(5/3) - (4*(b^3*(3*B*c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3*d*D + a^2*b*(4*C*d + 9*c*D))*Log[(-(b*c) + a* d)^(1/3) + b^(1/3)*(c + d*x)^(1/3)])/(-(b*c) + a*d)^(5/3) + (2*(b^3*(3*B*c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3*d*D + a^2*b*(4*C*d + 9*c*D))*Log[( -(b*c) + a*d)^(2/3) - b^(1/3)*(-(b*c) + a*d)^(1/3)*(c + d*x)^(1/3) + b^(2/ 3)*(c + d*x)^(2/3)])/(-(b*c) + a*d)^(5/3))/(12*b^(10/3))
Time = 0.90 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2124, 27, 1195, 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.
| \(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle -\frac {\int -\frac {3 \left (c-\frac {a d}{b}\right ) D x^2+\frac {3 (b c-a d) (b C-a D) x}{b^2}+\frac {-d D a^3+b (C d+3 c D) a^2-b^2 (3 c C+B d) a+b^3 (3 B c-2 A d)}{b^3}}{3 (a+b x) (c+d x)^{2/3}}dx}{b c-a d}-\frac {\sqrt [3]{c+d x} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {-\frac {d D a^3}{b^3}+\frac {(C d+3 c D) a^2}{b^2}-\frac {(3 c C+B d) a}{b}+3 \left (c-\frac {a d}{b}\right ) D x^2+3 B c-2 A d+\frac {3 (b c-a d) (b C-a D) x}{b^2}}{(a+b x) (c+d x)^{2/3}}dx}{3 (b c-a d)}-\frac {\sqrt [3]{c+d x} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \frac {\int \left (\frac {3 (b c-a d) \sqrt [3]{c+d x} D}{b^2 d}+\frac {3 (b c-a d) (b C d-2 a D d-b c D)}{b^3 d (c+d x)^{2/3}}+\frac {-7 d D a^3+b (4 C d+9 c D) a^2-b^2 (6 c C+B d) a+b^3 (3 B c-2 A d)}{b^3 (a+b x) (c+d x)^{2/3}}\right )dx}{3 (b c-a d)}-\frac {\sqrt [3]{c+d x} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt {3}}\right ) \left (-7 a^3 d D+a^2 b (9 c D+4 C d)-a b^2 (B d+6 c C)+b^3 (3 B c-2 A d)\right )}{b^{10/3} (b c-a d)^{2/3}}-\frac {\log (a+b x) \left (-7 a^3 d D+a^2 b (9 c D+4 C d)-a b^2 (B d+6 c C)+b^3 (3 B c-2 A d)\right )}{2 b^{10/3} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \left (-7 a^3 d D+a^2 b (9 c D+4 C d)-a b^2 (B d+6 c C)+b^3 (3 B c-2 A d)\right )}{2 b^{10/3} (b c-a d)^{2/3}}+\frac {9 \sqrt [3]{c+d x} (b c-a d) (-2 a d D-b c D+b C d)}{b^3 d^2}+\frac {9 D (c+d x)^{4/3} (b c-a d)}{4 b^2 d^2}}{3 (b c-a d)}-\frac {\sqrt [3]{c+d x} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^2*(c + d*x)^(2/3)),x]
Output:
-(((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c + d*x)^(1/3))/((b*c - a*d)*(a + b*x))) + ((9*(b*c - a*d)*(b*C*d - b*c*D - 2*a*d*D)*(c + d*x)^(1/3))/(b^3 *d^2) + (9*(b*c - a*d)*D*(c + d*x)^(4/3))/(4*b^2*d^2) - (Sqrt[3]*(b^3*(3*B *c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3*d*D + a^2*b*(4*C*d + 9*c*D))*Arc Tan[(1 + (2*b^(1/3)*(c + d*x)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(b^(10/3 )*(b*c - a*d)^(2/3)) - ((b^3*(3*B*c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3 *d*D + a^2*b*(4*C*d + 9*c*D))*Log[a + b*x])/(2*b^(10/3)*(b*c - a*d)^(2/3)) + (3*(b^3*(3*B*c - 2*A*d) - a*b^2*(6*c*C + B*d) - 7*a^3*d*D + a^2*b*(4*C* d + 9*c*D))*Log[(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)])/(2*b^(10/3)* (b*c - a*d)^(2/3)))/(3*(b*c - a*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : > With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px , a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - a*d))), x] + Simp[1/((m + 1)*(b*c - a*d)) Int[(a + b*x)^(m + 1)*(c + d*x )^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] || ! ILtQ[n, -1])
Time = 1.00 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.80
| method | result | size |
| pseudoelliptic | \(\frac {\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}} \left (\left (b^{3} A -\left (-\frac {3}{4} D x^{2}-3 C x +B \right ) a \,b^{2}+4 \left (-\frac {21 D x}{16}+C \right ) a^{2} b -7 a^{3} D\right ) d^{2}-3 \left (\left (\frac {D x}{4}+C \right ) b -\frac {5 D a}{4}\right ) \left (b x +a \right ) c b d +\frac {9 D b^{2} c^{2} \left (b x +a \right )}{4}\right ) b \left (x d +c \right )^{\frac {1}{3}}-\frac {\left (\left (b^{3} A +\frac {1}{2} a \,b^{2} B -2 a^{2} b C +\frac {7}{2} a^{3} D\right ) d -\frac {3 b c \left (B \,b^{2}-2 C a b +3 D a^{2}\right )}{2}\right ) \left (-2 \arctan \left (\frac {2 \sqrt {3}\, \left (x d +c \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}}-\frac {\sqrt {3}}{3}\right ) \sqrt {3}+\ln \left (\left (x d +c \right )^{\frac {2}{3}}-\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} \left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}\right )\right ) d^{2} \left (b x +a \right )}{3}}{\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}} d^{2} b^{4} \left (a d -b c \right ) \left (b x +a \right )}\) | \(317\) |
| derivativedivides | \(\frac {\frac {3 \left (\frac {D \left (x d +c \right )^{\frac {4}{3}} b}{4}+C d b \left (x d +c \right )^{\frac {1}{3}}-2 D a d \left (x d +c \right )^{\frac {1}{3}}-D b c \left (x d +c \right )^{\frac {1}{3}}\right )}{b^{3}}+\frac {3 d^{2} \left (\frac {d \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \left (x d +c \right )^{\frac {1}{3}}}{3 \left (a d -b c \right ) \left (\left (x d +c \right ) b +a d -b c \right )}+\frac {\left (2 b^{3} d A +B a \,b^{2} d -3 B \,b^{3} c -4 C \,a^{2} b d +6 a \,b^{2} c C +7 a^{3} d D-9 a^{2} b c D\right ) \left (\frac {\ln \left (\left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\left (x d +c \right )^{\frac {2}{3}}-\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} \left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (x d +c \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}\right )}{3 a d -3 b c}\right )}{b^{3}}}{d^{2}}\) | \(354\) |
| default | \(\frac {\frac {3 \left (\frac {D \left (x d +c \right )^{\frac {4}{3}} b}{4}+C d b \left (x d +c \right )^{\frac {1}{3}}-2 D a d \left (x d +c \right )^{\frac {1}{3}}-D b c \left (x d +c \right )^{\frac {1}{3}}\right )}{b^{3}}+\frac {3 d^{2} \left (\frac {d \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \left (x d +c \right )^{\frac {1}{3}}}{3 \left (a d -b c \right ) \left (\left (x d +c \right ) b +a d -b c \right )}+\frac {\left (2 b^{3} d A +B a \,b^{2} d -3 B \,b^{3} c -4 C \,a^{2} b d +6 a \,b^{2} c C +7 a^{3} d D-9 a^{2} b c D\right ) \left (\frac {\ln \left (\left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\left (x d +c \right )^{\frac {2}{3}}-\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} \left (x d +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (x d +c \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}}\right )}{3 a d -3 b c}\right )}{b^{3}}}{d^{2}}\) | \(354\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(2/3),x,method=_RETURNVERBOSE)
Output:
(((a*d-b*c)/b)^(2/3)*((b^3*A-(-3/4*D*x^2-3*C*x+B)*a*b^2+4*(-21/16*D*x+C)*a ^2*b-7*a^3*D)*d^2-3*((1/4*D*x+C)*b-5/4*D*a)*(b*x+a)*c*b*d+9/4*D*b^2*c^2*(b *x+a))*b*(d*x+c)^(1/3)-1/3*((b^3*A+1/2*a*b^2*B-2*a^2*b*C+7/2*a^3*D)*d-3/2* b*c*(B*b^2-2*C*a*b+3*D*a^2))*(-2*arctan(2/3*3^(1/2)/((a*d-b*c)/b)^(1/3)*(d *x+c)^(1/3)-1/3*3^(1/2))*3^(1/2)+ln((d*x+c)^(2/3)-((a*d-b*c)/b)^(1/3)*(d*x +c)^(1/3)+((a*d-b*c)/b)^(2/3))-2*ln((d*x+c)^(1/3)+((a*d-b*c)/b)^(1/3)))*d^ 2*(b*x+a))/((a*d-b*c)/b)^(2/3)/d^2/b^4/(a*d-b*c)/(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 1264 vs. \(2 (359) = 718\).
Time = 1.11 (sec) , antiderivative size = 2621, normalized size of antiderivative = 6.64 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\text {Too large to display} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(2/3),x, algorithm="fricas ")
Output:
[-1/12*(6*sqrt(1/3)*(3*(3*D*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*c^2*d^2 - 2*( 8*D*a^4*b^2 - 5*C*a^3*b^3 + 2*B*a^2*b^4 + A*a*b^5)*c*d^3 + (7*D*a^5*b - 4* C*a^4*b^2 + B*a^3*b^3 + 2*A*a^2*b^4)*d^4 + (3*(3*D*a^2*b^4 - 2*C*a*b^5 + B *b^6)*c^2*d^2 - 2*(8*D*a^3*b^3 - 5*C*a^2*b^4 + 2*B*a*b^5 + A*b^6)*c*d^3 + (7*D*a^4*b^2 - 4*C*a^3*b^3 + B*a^2*b^4 + 2*A*a*b^5)*d^4)*x)*sqrt(-(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)/b)*log(-(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2 + 2*(b^2*c*d - a*b*d^2)*x + 3*sqrt(1/3)*(2*(b^2*c - a*b*d)*(d*x + c)^(2/3 ) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d) - (b^3*c^2 - 2*a *b^2*c*d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3))*sqrt(-(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)/b) - 3*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d)*(d*x + c)^(1/3))/(b*x + a)) + 2*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^ (2/3)*(3*(3*D*a^3*b - 2*C*a^2*b^2 + B*a*b^3)*c*d^2 - (7*D*a^4 - 4*C*a^3*b + B*a^2*b^2 + 2*A*a*b^3)*d^3 + (3*(3*D*a^2*b^2 - 2*C*a*b^3 + B*b^4)*c*d^2 - (7*D*a^3*b - 4*C*a^2*b^2 + B*a*b^3 + 2*A*b^4)*d^3)*x)*log(-(b^2*c - a*b* d)*(d*x + c)^(2/3) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3)) - 4*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*(3*(3*D*a^3*b - 2*C*a^2*b^2 + B*a*b^3)*c *d^2 - (7*D*a^4 - 4*C*a^3*b + B*a^2*b^2 + 2*A*a*b^3)*d^3 + (3*(3*D*a^2*b^2 - 2*C*a*b^3 + B*b^4)*c*d^2 - (7*D*a^3*b - 4*C*a^2*b^2 + B*a*b^3 + 2*A*b^4 )*d^3)*x)*log(-(b^2*c - a*b*d)*(d*x + c)^(1/3) + (b^3*c^2 - 2*a*b^2*c*d...
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\text {Timed out} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**2/(d*x+c)**(2/3),x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(2/3),x, algorithm="maxima ")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.20 (sec) , antiderivative size = 552, normalized size of antiderivative = 1.40 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\frac {{\left (9 \, D a^{2} b c - 6 \, C a b^{2} c + 3 \, B b^{3} c - 7 \, D a^{3} d + 4 \, C a^{2} b d - B a b^{2} d - 2 \, A b^{3} d\right )} \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}} \log \left ({\left | {\left (d x + c\right )}^{\frac {1}{3}} - \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )}} + \frac {{\left (2 \, A b^{3} d - {\left (3 \, b^{3} c - a b^{2} d\right )} B + 2 \, {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} C - {\left (9 \, a^{2} b c - 7 \, a^{3} d\right )} D\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}}}\right )}{{\left (\sqrt {3} b^{3} c - \sqrt {3} a b^{2} d\right )} {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}}} + \frac {{\left (2 \, A b^{3} d - {\left (3 \, b^{3} c - a b^{2} d\right )} B + 2 \, {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} C - {\left (9 \, a^{2} b c - 7 \, a^{3} d\right )} D\right )} \log \left ({\left (d x + c\right )}^{\frac {2}{3}} + {\left (d x + c\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {5}{3}}} + \frac {{\left (d x + c\right )}^{\frac {1}{3}} D a^{3} d - {\left (d x + c\right )}^{\frac {1}{3}} C a^{2} b d + {\left (d x + c\right )}^{\frac {1}{3}} B a b^{2} d - {\left (d x + c\right )}^{\frac {1}{3}} A b^{3} d}{{\left (b^{4} c - a b^{3} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} + \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {4}{3}} D b^{6} d^{6} - 4 \, {\left (d x + c\right )}^{\frac {1}{3}} D b^{6} c d^{6} - 8 \, {\left (d x + c\right )}^{\frac {1}{3}} D a b^{5} d^{7} + 4 \, {\left (d x + c\right )}^{\frac {1}{3}} C b^{6} d^{7}\right )}}{4 \, b^{8} d^{8}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(2/3),x, algorithm="giac")
Output:
1/3*(9*D*a^2*b*c - 6*C*a*b^2*c + 3*B*b^3*c - 7*D*a^3*d + 4*C*a^2*b*d - B*a *b^2*d - 2*A*b^3*d)*((b*c - a*d)/b)^(1/3)*log(abs((d*x + c)^(1/3) - ((b*c - a*d)/b)^(1/3)))/(b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2) + (2*A*b^3*d - (3* b^3*c - a*b^2*d)*B + 2*(3*a*b^2*c - 2*a^2*b*d)*C - (9*a^2*b*c - 7*a^3*d)*D )*arctan(1/3*sqrt(3)*(2*(d*x + c)^(1/3) + ((b*c - a*d)/b)^(1/3))/((b*c - a *d)/b)^(1/3))/((sqrt(3)*b^3*c - sqrt(3)*a*b^2*d)*(b^3*c - a*b^2*d)^(2/3)) + 1/6*(2*A*b^3*d - (3*b^3*c - a*b^2*d)*B + 2*(3*a*b^2*c - 2*a^2*b*d)*C - ( 9*a^2*b*c - 7*a^3*d)*D)*log((d*x + c)^(2/3) + (d*x + c)^(1/3)*((b*c - a*d) /b)^(1/3) + ((b*c - a*d)/b)^(2/3))/(b^3*c - a*b^2*d)^(5/3) + ((d*x + c)^(1 /3)*D*a^3*d - (d*x + c)^(1/3)*C*a^2*b*d + (d*x + c)^(1/3)*B*a*b^2*d - (d*x + c)^(1/3)*A*b^3*d)/((b^4*c - a*b^3*d)*((d*x + c)*b - b*c + a*d)) + 3/4*( (d*x + c)^(4/3)*D*b^6*d^6 - 4*(d*x + c)^(1/3)*D*b^6*c*d^6 - 8*(d*x + c)^(1 /3)*D*a*b^5*d^7 + 4*(d*x + c)^(1/3)*C*b^6*d^7)/(b^8*d^8)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{2/3}} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^2*(c + d*x)^(2/3)),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^2*(c + d*x)^(2/3)), x)
Time = 0.21 (sec) , antiderivative size = 1644, normalized size of antiderivative = 4.16 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{2/3}} \, dx =\text {Too large to display} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(2/3),x)
Output:
( - 28*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a**3*d**2 + 24*sqrt(3)*atan(( b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/ 6)*(a*d - b*c)**(1/6)))*a**2*b*c*d - 28*sqrt(3)*atan((b**(1/6)*(a*d - b*c) **(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6 )))*a**2*b*d**2*x - 12*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a*b**3*d + 24 *sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)* *(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a*b**2*c*d*x - 12*sqrt(3)*atan((b** (1/6)*(a*d - b*c)**(1/6)*sqrt(3) - 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)* (a*d - b*c)**(1/6)))*b**4*d*x - 28*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/ 6)*sqrt(3) + 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a **3*d**2 + 24*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) + 2*b**(1/ 3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a**2*b*c*d - 28*sqrt(3 )*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) + 2*b**(1/3)*(c + d*x)**(1/6)) /(b**(1/6)*(a*d - b*c)**(1/6)))*a**2*b*d**2*x - 12*sqrt(3)*atan((b**(1/6)* (a*d - b*c)**(1/6)*sqrt(3) + 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a*b**3*d + 24*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqr t(3) + 2*b**(1/3)*(c + d*x)**(1/6))/(b**(1/6)*(a*d - b*c)**(1/6)))*a*b**2* c*d*x - 12*sqrt(3)*atan((b**(1/6)*(a*d - b*c)**(1/6)*sqrt(3) + 2*b**(1/...