Integrand size = 24, antiderivative size = 168 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\frac {e^4 (c+d x)^2}{2 b d}+\frac {a^{2/3} e^4 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d} \]
1/2*e^4*(d*x+c)^2/b/d+1/3*a^(2/3)*e^4*ln(a^(1/3)+b^(1/3)*(d*x+c))/b^(5/3)/ d-1/6*a^(2/3)*e^4*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/b^ (5/3)/d+1/3*a^(2/3)*e^4*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/a^(1/3)*3^( 1/2))/b^(5/3)/d*3^(1/2)
Time = 0.03 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.97 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=e^4 \left (\frac {(c+d x)^2}{2 b d}-\frac {a^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}\right ) \]
e^4*((c + d*x)^2/(2*b*d) - (a^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x) )/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(5/3)*d) + (a^(2/3)*Log[a^(1/3) + b^(1/3) *(c + d*x)])/(3*b^(5/3)*d) - (a^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d *x) + b^(2/3)*(c + d*x)^2])/(6*b^(5/3)*d))
Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {895, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 895 |
| \(\displaystyle \frac {e^4 \int \frac {(c+d x)^4}{b (c+d x)^3+a}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \int \frac {c+d x}{b (c+d x)^3+a}d(c+d x)}{b}\right )}{d}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} (c+d x)+\sqrt [3]{a}}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)\right )}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{b^{2/3} (c+d x)^2-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+a^{2/3}}d(c+d x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {e^4 \left (\frac {(c+d x)^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{d}\) |
(e^4*((c + d*x)^2/(2*b) - (a*(-1/3*Log[a^(1/3) + b^(1/3)*(c + d*x)]/(a^(1/ 3)*b^(2/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*(c + d*x))/a^(1/3))/Sqrt[ 3]])/b^(1/3)) + Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x )^2]/(2*b^(1/3)))/(3*a^(1/3)*b^(1/3))))/b))/d
3.29.86.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 *x^2), x], x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Simp[u^m/(Coeff icient[v, x, 1]*v^m) Subst[Int[x^m*(a + b*x^n)^p, x], x, v], x] /; FreeQ[ {a, b, m, n, p}, x] && LinearPairQ[u, v, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.87 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(e^{4} \left (\frac {c x +\frac {1}{2} d \,x^{2}}{b}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right ) a}{3 b^{2} d}\right )\) | \(96\) |
| risch | \(\frac {e^{4} c x}{b}+\frac {e^{4} d \,x^{2}}{2 b}+\frac {a \,e^{4} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (-\textit {\_R} d -c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{3 b^{2} d}\) | \(105\) |
e^4*(1/b*(c*x+1/2*d*x^2)-1/3/b^2/d*sum((_R*d+c)/(_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))*a)
Time = 0.26 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.07 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\frac {3 \, d^{2} e^{4} x^{2} + 6 \, c d e^{4} x - 2 \, \sqrt {3} e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d^{2} x^{2} + 2 \, a c d x + a c^{2} - {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d x + a c + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{6 \, b d} \]
1/6*(3*d^2*e^4*x^2 + 6*c*d*e^4*x - 2*sqrt(3)*e^4*(a^2/b^2)^(1/3)*arctan(1/ 3*(2*sqrt(3)*(b*d*x + b*c)*(a^2/b^2)^(1/3) - sqrt(3)*a)/a) - e^4*(a^2/b^2) ^(1/3)*log(a*d^2*x^2 + 2*a*c*d*x + a*c^2 - (b*d*x + b*c)*(a^2/b^2)^(2/3) + a*(a^2/b^2)^(1/3)) + 2*e^4*(a^2/b^2)^(1/3)*log(a*d*x + a*c + b*(a^2/b^2)^ (2/3)))/(b*d)
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.39 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\frac {e^{4} \operatorname {RootSum} {\left (27 t^{3} b^{5} - a^{2}, \left ( t \mapsto t \log {\left (x + \frac {9 t^{2} b^{3} e^{8} + a c e^{8}}{a d e^{8}} \right )} \right )\right )}}{d} + \frac {c e^{4} x}{b} + \frac {d e^{4} x^{2}}{2 b} \]
e**4*RootSum(27*_t**3*b**5 - a**2, Lambda(_t, _t*log(x + (9*_t**2*b**3*e** 8 + a*c*e**8)/(a*d*e**8))))/d + c*e**4*x/b + d*e**4*x**2/(2*b)
\[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\int { \frac {{\left (d e x + c e\right )}^{4}}{{\left (d x + c\right )}^{3} b + a} \,d x } \]
-a*e^4*integrate((d*x + c)/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^ 3 + a), x)/b + 1/2*(d*e^4*x^2 + 2*c*e^4*x)/b
Time = 0.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\frac {b d^{7} e^{4} x^{2} + 2 \, b c d^{6} e^{4} x}{2 \, b^{2} d^{6}} - \frac {2 \, \sqrt {3} \left (a^{2} b d^{15} e^{12}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (a^{2} b d^{15} e^{12}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (a^{2} b d^{15} e^{12}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{6 \, b^{2} d^{6}} \]
1/2*(b*d^7*e^4*x^2 + 2*b*c*d^6*e^4*x)/(b^2*d^6) - 1/6*(2*sqrt(3)*(a^2*b*d^ 15*e^12)^(1/3)*arctan(1/3*sqrt(3)*(2*a*b*d*x + 2*a*b*c - (a^2*b)^(2/3))/(a ^2*b)^(2/3)) + (a^2*b*d^15*e^12)^(1/3)*log((2*a*b*d*x + 2*a*b*c - (a^2*b)^ (2/3))^2 + 3*(a^2*b)^(4/3)) - 2*(a^2*b*d^15*e^12)^(1/3)*log(abs(a*b*d*x + a*b*c + (a^2*b)^(2/3))))/(b^2*d^6)
Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx=\frac {d\,e^4\,x^2}{2\,b}+\frac {c\,e^4\,x}{b}+\frac {a^{2/3}\,e^4\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{3\,b^{5/3}\,d}+\frac {a^{2/3}\,e^4\,\ln \left (a^{1/3}-2\,b^{1/3}\,c-2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^{5/3}\,d}-\frac {a^{2/3}\,e^4\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{5/3}\,d} \]
(d*e^4*x^2)/(2*b) + (c*e^4*x)/b + (a^(2/3)*e^4*log(b^(1/3)*c + a^(1/3) + b ^(1/3)*d*x))/(3*b^(5/3)*d) + (a^(2/3)*e^4*log(3^(1/2)*a^(1/3)*1i - 2*b^(1/ 3)*c + a^(1/3) - 2*b^(1/3)*d*x)*((3^(1/2)*1i)/6 - 1/6))/(b^(5/3)*d) - (a^( 2/3)*e^4*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*c - a^(1/3) + 2*b^(1/3)*d*x)*( (3^(1/2)*1i)/2 + 1/2))/(3*b^(5/3)*d)