Integrand size = 17, antiderivative size = 280 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\frac {4 a b^3 x}{d}+\frac {b^4 x^2}{2 d}+\frac {\left (b^4 c^{4/3}+4 a b^3 c \sqrt [3]{d}-4 a^3 b \sqrt [3]{c} d-a^4 d^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{5/3}}+\frac {\left (b \sqrt [3]{c} \left (b^3 c-4 a^3 d\right )-a \sqrt [3]{d} \left (4 b^3 c-a^3 d\right )\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{5/3}}+\frac {\left (4 a b^3 c-a^4 d-\frac {b \sqrt [3]{c} \left (b^3 c-4 a^3 d\right )}{\sqrt [3]{d}}\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}+\frac {2 a^2 b^2 \log \left (c+d x^3\right )}{d} \] Output:
4*a*b^3*x/d+1/2*b^4*x^2/d+1/3*(b^4*c^(4/3)+4*a*b^3*c*d^(1/3)-4*a^3*b*c^(1/ 3)*d-a^4*d^(4/3))*arctan(1/3*(c^(1/3)-2*d^(1/3)*x)*3^(1/2)/c^(1/3))*3^(1/2 )/c^(2/3)/d^(5/3)+1/3*(b*c^(1/3)*(-4*a^3*d+b^3*c)-a*d^(1/3)*(-a^3*d+4*b^3* c))*ln(c^(1/3)+d^(1/3)*x)/c^(2/3)/d^(5/3)+1/6*(4*a*b^3*c-a^4*d-b*c^(1/3)*( -4*a^3*d+b^3*c)/d^(1/3))*ln(c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/c^(2/3) /d^(4/3)+2*a^2*b^2*ln(d*x^3+c)/d
Time = 0.17 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\frac {24 a b^3 d^{2/3} x+3 b^4 d^{2/3} x^2+\frac {2 \sqrt {3} \left (b^4 c^{4/3}+4 a b^3 c \sqrt [3]{d}-4 a^3 b \sqrt [3]{c} d-a^4 d^{4/3}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{c^{2/3}}+\frac {2 \left (b^4 c^{4/3}-4 a b^3 c \sqrt [3]{d}-4 a^3 b \sqrt [3]{c} d+a^4 d^{4/3}\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{2/3}}-\frac {\left (b^4 c^{4/3}-4 a b^3 c \sqrt [3]{d}-4 a^3 b \sqrt [3]{c} d+a^4 d^{4/3}\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{2/3}}+12 a^2 b^2 d^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}} \] Input:
Integrate[(a + b*x)^4/(c + d*x^3),x]
Output:
(24*a*b^3*d^(2/3)*x + 3*b^4*d^(2/3)*x^2 + (2*Sqrt[3]*(b^4*c^(4/3) + 4*a*b^ 3*c*d^(1/3) - 4*a^3*b*c^(1/3)*d - a^4*d^(4/3))*ArcTan[(1 - (2*d^(1/3)*x)/c ^(1/3))/Sqrt[3]])/c^(2/3) + (2*(b^4*c^(4/3) - 4*a*b^3*c*d^(1/3) - 4*a^3*b* c^(1/3)*d + a^4*d^(4/3))*Log[c^(1/3) + d^(1/3)*x])/c^(2/3) - ((b^4*c^(4/3) - 4*a*b^3*c*d^(1/3) - 4*a^3*b*c^(1/3)*d + a^4*d^(4/3))*Log[c^(2/3) - c^(1 /3)*d^(1/3)*x + d^(2/3)*x^2])/c^(2/3) + 12*a^2*b^2*d^(2/3)*Log[c + d*x^3]) /(6*d^(5/3))
Time = 1.00 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {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.
| \(\displaystyle \int \frac {(a+b x)^4}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \int \left (-\frac {a^4 (-d)+b x \left (b^3 c-4 a^3 d\right )-6 a^2 b^2 d x^2+4 a b^3 c}{d \left (c+d x^3\right )}+\frac {4 a b^3}{d}+\frac {b^4 x}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 b^2 \log \left (c+d x^3\right )}{d}+\frac {\left (a^4 \left (-d^{4/3}\right )-4 a^3 b \sqrt [3]{c} d+4 a b^3 c \sqrt [3]{d}+b^4 c^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{5/3}}+\frac {\left (a^4 (-d)-\frac {b \sqrt [3]{c} \left (b^3 c-4 a^3 d\right )}{\sqrt [3]{d}}+4 a b^3 c\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}+\frac {\left (b \sqrt [3]{c} \left (b^3 c-4 a^3 d\right )-\sqrt [3]{d} \left (4 a b^3 c-a^4 d\right )\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{5/3}}+\frac {4 a b^3 x}{d}+\frac {b^4 x^2}{2 d}\) |
Input:
Int[(a + b*x)^4/(c + d*x^3),x]
Output:
(4*a*b^3*x)/d + (b^4*x^2)/(2*d) + ((b^4*c^(4/3) + 4*a*b^3*c*d^(1/3) - 4*a^ 3*b*c^(1/3)*d - a^4*d^(4/3))*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/ 3))])/(Sqrt[3]*c^(2/3)*d^(5/3)) + ((b*c^(1/3)*(b^3*c - 4*a^3*d) - d^(1/3)* (4*a*b^3*c - a^4*d))*Log[c^(1/3) + d^(1/3)*x])/(3*c^(2/3)*d^(5/3)) + ((4*a *b^3*c - a^4*d - (b*c^(1/3)*(b^3*c - 4*a^3*d))/d^(1/3))*Log[c^(2/3) - c^(1 /3)*d^(1/3)*x + d^(2/3)*x^2])/(6*c^(2/3)*d^(4/3)) + (2*a^2*b^2*Log[c + d*x ^3])/d
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 = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.32
| method | result | size |
| risch | \(\frac {b^{4} x^{2}}{2 d}+\frac {4 a \,b^{3} x}{d}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} d +c \right )}{\sum }\frac {\left (6 a^{2} b^{2} d \,\textit {\_R}^{2}+b \left (4 a^{3} d -b^{3} c \right ) \textit {\_R} +a^{4} d -4 a \,b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 d^{2}}\) | \(90\) |
| default | \(\frac {b^{3} \left (\frac {1}{2} b \,x^{2}+4 a x \right )}{d}+\frac {\left (a^{4} d -4 a \,b^{3} c \right ) \left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right )+\left (4 a^{3} b d -b^{4} c \right ) \left (-\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )+2 a^{2} b^{2} \ln \left (d \,x^{3}+c \right )}{d}\) | \(250\) |
Input:
int((b*x+a)^4/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/2*b^4*x^2/d+4*a*b^3*x/d+1/3/d^2*sum((6*a^2*b^2*d*_R^2+b*(4*a^3*d-b^3*c)* _R+a^4*d-4*a*b^3*c)/_R^2*ln(x-_R),_R=RootOf(_Z^3*d+c))
Result contains complex when optimal does not.
Time = 4.52 (sec) , antiderivative size = 8787, normalized size of antiderivative = 31.38 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\text {Too large to display} \] Input:
integrate((b*x+a)^4/(d*x^3+c),x, algorithm="fricas")
Output:
Too large to include
Time = 26.15 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\frac {4 a b^{3} x}{d} + \frac {b^{4} x^{2}}{2 d} + \operatorname {RootSum} {\left (27 t^{3} c^{2} d^{5} - 162 t^{2} a^{2} b^{2} c^{2} d^{4} + t \left (36 a^{7} b c d^{4} + 171 a^{4} b^{4} c^{2} d^{3} + 36 a b^{7} c^{3} d^{2}\right ) - a^{12} d^{4} + 4 a^{9} b^{3} c d^{3} - 6 a^{6} b^{6} c^{2} d^{2} + 4 a^{3} b^{9} c^{3} d - b^{12} c^{4}, \left ( t \mapsto t \log {\left (x + \frac {36 t^{2} a^{3} b c^{2} d^{4} - 9 t^{2} b^{4} c^{3} d^{3} + 3 t a^{8} c d^{4} - 168 t a^{5} b^{3} c^{2} d^{3} + 84 t a^{2} b^{6} c^{3} d^{2} + 26 a^{10} b^{2} c d^{3} + 48 a^{7} b^{5} c^{2} d^{2} - 66 a^{4} b^{8} c^{3} d - 8 a b^{11} c^{4}}{a^{12} d^{4} + 52 a^{9} b^{3} c d^{3} - 52 a^{3} b^{9} c^{3} d - b^{12} c^{4}} \right )} \right )\right )} \] Input:
integrate((b*x+a)**4/(d*x**3+c),x)
Output:
4*a*b**3*x/d + b**4*x**2/(2*d) + RootSum(27*_t**3*c**2*d**5 - 162*_t**2*a* *2*b**2*c**2*d**4 + _t*(36*a**7*b*c*d**4 + 171*a**4*b**4*c**2*d**3 + 36*a* b**7*c**3*d**2) - a**12*d**4 + 4*a**9*b**3*c*d**3 - 6*a**6*b**6*c**2*d**2 + 4*a**3*b**9*c**3*d - b**12*c**4, Lambda(_t, _t*log(x + (36*_t**2*a**3*b* c**2*d**4 - 9*_t**2*b**4*c**3*d**3 + 3*_t*a**8*c*d**4 - 168*_t*a**5*b**3*c **2*d**3 + 84*_t*a**2*b**6*c**3*d**2 + 26*a**10*b**2*c*d**3 + 48*a**7*b**5 *c**2*d**2 - 66*a**4*b**8*c**3*d - 8*a*b**11*c**4)/(a**12*d**4 + 52*a**9*b **3*c*d**3 - 52*a**3*b**9*c**3*d - b**12*c**4))))
Time = 0.12 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=-\frac {\sqrt {3} {\left ({\left (b^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}} + 4 \, a b^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}} + 4 \, a^{2} b^{2}\right )} c - {\left (4 \, a^{3} b \left (\frac {c}{d}\right )^{\frac {2}{3}} + a^{4} \left (\frac {c}{d}\right )^{\frac {1}{3}} + \frac {4 \, a^{2} b^{2} c}{d}\right )} d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, c d} + \frac {b^{4} x^{2} + 8 \, a b^{3} x}{2 \, d} - \frac {{\left ({\left (b^{4} \left (\frac {c}{d}\right )^{\frac {1}{3}} - 4 \, a b^{3}\right )} c - {\left (12 \, a^{2} b^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} + 4 \, a^{3} b \left (\frac {c}{d}\right )^{\frac {1}{3}} - a^{4}\right )} d\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left ({\left (b^{4} \left (\frac {c}{d}\right )^{\frac {1}{3}} - 4 \, a b^{3}\right )} c + {\left (6 \, a^{2} b^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - 4 \, a^{3} b \left (\frac {c}{d}\right )^{\frac {1}{3}} + a^{4}\right )} d\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \] Input:
integrate((b*x+a)^4/(d*x^3+c),x, algorithm="maxima")
Output:
-1/3*sqrt(3)*((b^4*(c/d)^(2/3) + 4*a*b^3*(c/d)^(1/3) + 4*a^2*b^2)*c - (4*a ^3*b*(c/d)^(2/3) + a^4*(c/d)^(1/3) + 4*a^2*b^2*c/d)*d)*arctan(1/3*sqrt(3)* (2*x - (c/d)^(1/3))/(c/d)^(1/3))/(c*d) + 1/2*(b^4*x^2 + 8*a*b^3*x)/d - 1/6 *((b^4*(c/d)^(1/3) - 4*a*b^3)*c - (12*a^2*b^2*(c/d)^(2/3) + 4*a^3*b*(c/d)^ (1/3) - a^4)*d)*log(x^2 - x*(c/d)^(1/3) + (c/d)^(2/3))/(d^2*(c/d)^(2/3)) + 1/3*((b^4*(c/d)^(1/3) - 4*a*b^3)*c + (6*a^2*b^2*(c/d)^(2/3) - 4*a^3*b*(c/ d)^(1/3) + a^4)*d)*log(x + (c/d)^(1/3))/(d^2*(c/d)^(2/3))
Time = 0.14 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\frac {2 \, a^{2} b^{2} \log \left ({\left | d x^{3} + c \right |}\right )}{d} + \frac {\sqrt {3} {\left (4 \, a b^{3} c d - a^{4} d^{2} - \left (-c d^{2}\right )^{\frac {1}{3}} b^{4} c + 4 \, \left (-c d^{2}\right )^{\frac {1}{3}} a^{3} b d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c d^{2}\right )^{\frac {2}{3}} d} + \frac {{\left (4 \, a b^{3} c d - a^{4} d^{2} + \left (-c d^{2}\right )^{\frac {1}{3}} b^{4} c - 4 \, \left (-c d^{2}\right )^{\frac {1}{3}} a^{3} b d\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c d^{2}\right )^{\frac {2}{3}} d} + \frac {b^{4} d x^{2} + 8 \, a b^{3} d x}{2 \, d^{2}} + \frac {{\left (b^{4} c d^{4} \left (-\frac {c}{d}\right )^{\frac {1}{3}} - 4 \, a^{3} b d^{5} \left (-\frac {c}{d}\right )^{\frac {1}{3}} + 4 \, a b^{3} c d^{4} - a^{4} d^{5}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d^{5}} \] Input:
integrate((b*x+a)^4/(d*x^3+c),x, algorithm="giac")
Output:
2*a^2*b^2*log(abs(d*x^3 + c))/d + 1/3*sqrt(3)*(4*a*b^3*c*d - a^4*d^2 - (-c *d^2)^(1/3)*b^4*c + 4*(-c*d^2)^(1/3)*a^3*b*d)*arctan(1/3*sqrt(3)*(2*x + (- c/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2/3)*d) + 1/6*(4*a*b^3*c*d - a^4*d^2 + (-c*d^2)^(1/3)*b^4*c - 4*(-c*d^2)^(1/3)*a^3*b*d)*log(x^2 + x*(-c/d)^(1/3 ) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*d) + 1/2*(b^4*d*x^2 + 8*a*b^3*d*x)/d^2 + 1/3*(b^4*c*d^4*(-c/d)^(1/3) - 4*a^3*b*d^5*(-c/d)^(1/3) + 4*a*b^3*c*d^4 - a^4*d^5)*(-c/d)^(1/3)*log(abs(x - (-c/d)^(1/3)))/(c*d^5)
Time = 6.07 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\left (\sum _{k=1}^3\ln \left (\mathrm {root}\left (27\,c^2\,d^5\,z^3-162\,a^2\,b^2\,c^2\,d^4\,z^2+171\,a^4\,b^4\,c^2\,d^3\,z+36\,a\,b^7\,c^3\,d^2\,z+36\,a^7\,b\,c\,d^4\,z-6\,a^6\,b^6\,c^2\,d^2+4\,a^9\,b^3\,c\,d^3+4\,a^3\,b^9\,c^3\,d-b^{12}\,c^4-a^{12}\,d^4,z,k\right )\,\left (\frac {x\,\left (3\,a^4\,d^3-12\,a\,b^3\,c\,d^2\right )}{d}+\mathrm {root}\left (27\,c^2\,d^5\,z^3-162\,a^2\,b^2\,c^2\,d^4\,z^2+171\,a^4\,b^4\,c^2\,d^3\,z+36\,a\,b^7\,c^3\,d^2\,z+36\,a^7\,b\,c\,d^4\,z-6\,a^6\,b^6\,c^2\,d^2+4\,a^9\,b^3\,c\,d^3+4\,a^3\,b^9\,c^3\,d-b^{12}\,c^4-a^{12}\,d^4,z,k\right )\,c\,d^2\,9-36\,a^2\,b^2\,c\,d\right )+\frac {4\,a^7\,b\,d^2+19\,a^4\,b^4\,c\,d+4\,a\,b^7\,c^2}{d}+\frac {x\,\left (10\,a^6\,b^2\,d^2+16\,a^3\,b^5\,c\,d+b^8\,c^2\right )}{d}\right )\,\mathrm {root}\left (27\,c^2\,d^5\,z^3-162\,a^2\,b^2\,c^2\,d^4\,z^2+171\,a^4\,b^4\,c^2\,d^3\,z+36\,a\,b^7\,c^3\,d^2\,z+36\,a^7\,b\,c\,d^4\,z-6\,a^6\,b^6\,c^2\,d^2+4\,a^9\,b^3\,c\,d^3+4\,a^3\,b^9\,c^3\,d-b^{12}\,c^4-a^{12}\,d^4,z,k\right )\right )+\frac {b^4\,x^2}{2\,d}+\frac {4\,a\,b^3\,x}{d} \] Input:
int((a + b*x)^4/(c + d*x^3),x)
Output:
symsum(log(root(27*c^2*d^5*z^3 - 162*a^2*b^2*c^2*d^4*z^2 + 171*a^4*b^4*c^2 *d^3*z + 36*a*b^7*c^3*d^2*z + 36*a^7*b*c*d^4*z - 6*a^6*b^6*c^2*d^2 + 4*a^9 *b^3*c*d^3 + 4*a^3*b^9*c^3*d - b^12*c^4 - a^12*d^4, z, k)*((x*(3*a^4*d^3 - 12*a*b^3*c*d^2))/d + 9*root(27*c^2*d^5*z^3 - 162*a^2*b^2*c^2*d^4*z^2 + 17 1*a^4*b^4*c^2*d^3*z + 36*a*b^7*c^3*d^2*z + 36*a^7*b*c*d^4*z - 6*a^6*b^6*c^ 2*d^2 + 4*a^9*b^3*c*d^3 + 4*a^3*b^9*c^3*d - b^12*c^4 - a^12*d^4, z, k)*c*d ^2 - 36*a^2*b^2*c*d) + (4*a*b^7*c^2 + 4*a^7*b*d^2 + 19*a^4*b^4*c*d)/d + (x *(b^8*c^2 + 10*a^6*b^2*d^2 + 16*a^3*b^5*c*d))/d)*root(27*c^2*d^5*z^3 - 162 *a^2*b^2*c^2*d^4*z^2 + 171*a^4*b^4*c^2*d^3*z + 36*a*b^7*c^3*d^2*z + 36*a^7 *b*c*d^4*z - 6*a^6*b^6*c^2*d^2 + 4*a^9*b^3*c*d^3 + 4*a^3*b^9*c^3*d - b^12* c^4 - a^12*d^4, z, k), k, 1, 3) + (b^4*x^2)/(2*d) + (4*a*b^3*x)/d
Time = 0.15 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^4}{c+d x^3} \, dx=\frac {-2 d^{\frac {4}{3}} c^{\frac {2}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a^{4}+8 d^{\frac {1}{3}} c^{\frac {5}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a \,b^{3}-8 \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a^{3} b c d +2 \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) b^{4} c^{2}-d^{\frac {4}{3}} c^{\frac {2}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{4}+4 d^{\frac {1}{3}} c^{\frac {5}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a \,b^{3}+2 d^{\frac {4}{3}} c^{\frac {2}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{4}-8 d^{\frac {1}{3}} c^{\frac {5}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a \,b^{3}+12 d^{\frac {2}{3}} c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{2} b^{2}+12 d^{\frac {2}{3}} c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{2} b^{2}+24 d^{\frac {2}{3}} c^{\frac {4}{3}} a \,b^{3} x +3 d^{\frac {2}{3}} c^{\frac {4}{3}} b^{4} x^{2}+4 \,\mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{3} b c d -\mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) b^{4} c^{2}-8 \,\mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{3} b c d +2 \,\mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) b^{4} c^{2}}{6 d^{\frac {5}{3}} c^{\frac {4}{3}}} \] Input:
int((b*x+a)^4/(d*x^3+c),x)
Output:
( - 2*d**(1/3)*c**(2/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*s qrt(3)))*a**4*d + 8*d**(1/3)*c**(2/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)* x)/(c**(1/3)*sqrt(3)))*a*b**3*c - 8*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x) /(c**(1/3)*sqrt(3)))*a**3*b*c*d + 2*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x) /(c**(1/3)*sqrt(3)))*b**4*c**2 - d**(1/3)*c**(2/3)*log(c**(2/3) - d**(1/3) *c**(1/3)*x + d**(2/3)*x**2)*a**4*d + 4*d**(1/3)*c**(2/3)*log(c**(2/3) - d **(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a*b**3*c + 2*d**(1/3)*c**(2/3)*log(c** (1/3) + d**(1/3)*x)*a**4*d - 8*d**(1/3)*c**(2/3)*log(c**(1/3) + d**(1/3)*x )*a*b**3*c + 12*d**(2/3)*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d** (2/3)*x**2)*a**2*b**2*c + 12*d**(2/3)*c**(1/3)*log(c**(1/3) + d**(1/3)*x)* a**2*b**2*c + 24*d**(2/3)*c**(1/3)*a*b**3*c*x + 3*d**(2/3)*c**(1/3)*b**4*c *x**2 + 4*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*b*c*d - log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*b**4*c**2 - 8*log(c** (1/3) + d**(1/3)*x)*a**3*b*c*d + 2*log(c**(1/3) + d**(1/3)*x)*b**4*c**2)/( 6*d**(2/3)*c**(1/3)*c*d)