Integrand size = 24, antiderivative size = 65 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=-\frac {1}{3 a d e^4 (c+d x)^3}-\frac {b \log (c+d x)}{a^2 d e^4}+\frac {b \log \left (a+b (c+d x)^3\right )}{3 a^2 d e^4} \] Output:
-1/3/a/d/e^4/(d*x+c)^3-b*ln(d*x+c)/a^2/d/e^4+1/3*b*ln(a+b*(d*x+c)^3)/a^2/d /e^4
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=\frac {-\frac {a}{(c+d x)^3}-3 b \log (c+d x)+b \log \left (a+b (c+d x)^3\right )}{3 a^2 d e^4} \] Input:
Integrate[1/((c*e + d*e*x)^4*(a + b*(c + d*x)^3)),x]
Output:
(-(a/(c + d*x)^3) - 3*b*Log[c + d*x] + b*Log[a + b*(c + d*x)^3])/(3*a^2*d* e^4)
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {895, 798, 54, 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.
\(\displaystyle \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx\) |
\(\Big \downarrow \) 895 |
\(\displaystyle \frac {\int \frac {1}{(c+d x)^4 \left (b (c+d x)^3+a\right )}d(c+d x)}{d e^4}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {1}{(c+d x)^6 \left (b (c+d x)^3+a\right )}d(c+d x)^3}{3 d e^4}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {b^2}{a^2 \left (b (c+d x)^3+a\right )}-\frac {b}{a^2 (c+d x)^3}+\frac {1}{a (c+d x)^6}\right )d(c+d x)^3}{3 d e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {b \log \left ((c+d x)^3\right )}{a^2}+\frac {b \log \left (a+b (c+d x)^3\right )}{a^2}-\frac {1}{a (c+d x)^3}}{3 d e^4}\) |
Input:
Int[1/((c*e + d*e*x)^4*(a + b*(c + d*x)^3)),x]
Output:
(-(1/(a*(c + d*x)^3)) - (b*Log[(c + d*x)^3])/a^2 + (b*Log[a + b*(c + d*x)^ 3])/a^2)/(3*d*e^4)
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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]
Time = 0.75 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {-\frac {1}{3 a d \left (x d +c \right )^{3}}-\frac {b \ln \left (x d +c \right )}{a^{2} d}+\frac {b \ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )}{3 a^{2} d}}{e^{4}}\) | \(79\) |
risch | \(-\frac {1}{3 a d \,e^{4} \left (x d +c \right )^{3}}-\frac {b \ln \left (x d +c \right )}{a^{2} d \,e^{4}}+\frac {b \ln \left (-b \,d^{3} x^{3}-3 b c \,d^{2} x^{2}-3 d b x \,c^{2}-c^{3} b -a \right )}{3 a^{2} e^{4} d}\) | \(88\) |
norman | \(\frac {\frac {x}{c e a}+\frac {d \,x^{2}}{c^{2} e a}+\frac {d^{2} x^{3}}{3 e \,c^{3} a}}{e^{3} \left (x d +c \right )^{3}}-\frac {b \ln \left (x d +c \right )}{a^{2} d \,e^{4}}+\frac {b \ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right )}{3 a^{2} e^{4} d}\) | \(120\) |
parallelrisch | \(\frac {-3 \ln \left (x d +c \right ) x^{3} b \,d^{5}+\ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right ) x^{3} b \,d^{5}-9 \ln \left (x d +c \right ) x^{2} b c \,d^{4}+3 \ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right ) x^{2} b c \,d^{4}-9 \ln \left (x d +c \right ) x b \,c^{2} d^{3}+3 \ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right ) x b \,c^{2} d^{3}-3 \ln \left (x d +c \right ) b \,c^{3} d^{2}+\ln \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 d b x \,c^{2}+c^{3} b +a \right ) b \,c^{3} d^{2}-a \,d^{2}}{3 a^{2} d^{3} e^{4} \left (x d +c \right )^{3}}\) | \(260\) |
Input:
int(1/(d*e*x+c*e)^4/(a+b*(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/e^4*(-1/3/a/d/(d*x+c)^3-b*ln(d*x+c)/a^2/d+1/3*b/a^2/d*ln(b*d^3*x^3+3*b*c *d^2*x^2+3*b*c^2*d*x+b*c^3+a))
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=\frac {{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right ) - 3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (d x + c\right ) - a}{3 \, {\left (a^{2} d^{4} e^{4} x^{3} + 3 \, a^{2} c d^{3} e^{4} x^{2} + 3 \, a^{2} c^{2} d^{2} e^{4} x + a^{2} c^{3} d e^{4}\right )}} \] Input:
integrate(1/(d*e*x+c*e)^4/(a+b*(d*x+c)^3),x, algorithm="fricas")
Output:
1/3*((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(b*d^3*x^3 + 3*b *c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a) - 3*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b *c^2*d*x + b*c^3)*log(d*x + c) - a)/(a^2*d^4*e^4*x^3 + 3*a^2*c*d^3*e^4*x^2 + 3*a^2*c^2*d^2*e^4*x + a^2*c^3*d*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).
Time = 0.91 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=- \frac {1}{3 a c^{3} d e^{4} + 9 a c^{2} d^{2} e^{4} x + 9 a c d^{3} e^{4} x^{2} + 3 a d^{4} e^{4} x^{3}} - \frac {b \log {\left (\frac {c}{d} + x \right )}}{a^{2} d e^{4}} + \frac {b \log {\left (\frac {3 c^{2} x}{d^{2}} + \frac {3 c x^{2}}{d} + x^{3} + \frac {a + b c^{3}}{b d^{3}} \right )}}{3 a^{2} d e^{4}} \] Input:
integrate(1/(d*e*x+c*e)**4/(a+b*(d*x+c)**3),x)
Output:
-1/(3*a*c**3*d*e**4 + 9*a*c**2*d**2*e**4*x + 9*a*c*d**3*e**4*x**2 + 3*a*d* *4*e**4*x**3) - b*log(c/d + x)/(a**2*d*e**4) + b*log(3*c**2*x/d**2 + 3*c*x **2/d + x**3 + (a + b*c**3)/(b*d**3))/(3*a**2*d*e**4)
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=-\frac {1}{3 \, {\left (a d^{4} e^{4} x^{3} + 3 \, a c d^{3} e^{4} x^{2} + 3 \, a c^{2} d^{2} e^{4} x + a c^{3} d e^{4}\right )}} + \frac {b \log \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}{3 \, a^{2} d e^{4}} - \frac {b \log \left (d x + c\right )}{a^{2} d e^{4}} \] Input:
integrate(1/(d*e*x+c*e)^4/(a+b*(d*x+c)^3),x, algorithm="maxima")
Output:
-1/3/(a*d^4*e^4*x^3 + 3*a*c*d^3*e^4*x^2 + 3*a*c^2*d^2*e^4*x + a*c^3*d*e^4) + 1/3*b*log(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)/(a^2*d*e ^4) - b*log(d*x + c)/(a^2*d*e^4)
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=\frac {b \log \left ({\left | b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a \right |}\right )}{3 \, a^{2} d e^{4}} - \frac {b \log \left ({\left | d x + c \right |}\right )}{a^{2} d e^{4}} - \frac {1}{3 \, {\left (d x + c\right )}^{3} a d e^{4}} \] Input:
integrate(1/(d*e*x+c*e)^4/(a+b*(d*x+c)^3),x, algorithm="giac")
Output:
1/3*b*log(abs(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(a^2*d *e^4) - b*log(abs(d*x + c))/(a^2*d*e^4) - 1/3/((d*x + c)^3*a*d*e^4)
Time = 0.63 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.82 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=\frac {b\,\ln \left (b\,c^3+3\,b\,c^2\,d\,x+3\,b\,c\,d^2\,x^2+b\,d^3\,x^3+a\right )}{3\,a^2\,d\,e^4}-\frac {1}{3\,\left (a\,c^3\,d\,e^4+3\,a\,c^2\,d^2\,e^4\,x+3\,a\,c\,d^3\,e^4\,x^2+a\,d^4\,e^4\,x^3\right )}-\frac {b\,\ln \left (c+d\,x\right )}{a^2\,d\,e^4} \] Input:
int(1/((c*e + d*e*x)^4*(a + b*(c + d*x)^3)),x)
Output:
(b*log(a + b*c^3 + b*d^3*x^3 + 3*b*c^2*d*x + 3*b*c*d^2*x^2))/(3*a^2*d*e^4) - 1/(3*(a*d^4*e^4*x^3 + a*c^3*d*e^4 + 3*a*c^2*d^2*e^4*x + 3*a*c*d^3*e^4*x ^2)) - (b*log(c + d*x))/(a^2*d*e^4)
Time = 0.23 (sec) , antiderivative size = 423, normalized size of antiderivative = 6.51 \[ \int \frac {1}{(c e+d e x)^4 \left (a+b (c+d x)^3\right )} \, dx=\frac {\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b \,c^{3}+3 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b \,c^{2} d x +3 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b c \,d^{2} x^{2}+\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} c -b^{\frac {1}{3}} a^{\frac {1}{3}} d x +b^{\frac {2}{3}} c^{2}+2 b^{\frac {2}{3}} c d x +b^{\frac {2}{3}} d^{2} x^{2}\right ) b \,d^{3} x^{3}+\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b \,c^{3}+3 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b \,c^{2} d x +3 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b c \,d^{2} x^{2}+\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} c +b^{\frac {1}{3}} d x \right ) b \,d^{3} x^{3}-3 \,\mathrm {log}\left (d x +c \right ) b \,c^{3}-9 \,\mathrm {log}\left (d x +c \right ) b \,c^{2} d x -9 \,\mathrm {log}\left (d x +c \right ) b c \,d^{2} x^{2}-3 \,\mathrm {log}\left (d x +c \right ) b \,d^{3} x^{3}-a}{3 a^{2} d \,e^{4} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:
int(1/(d*e*x+c*e)^4/(a+b*(d*x+c)^3),x)
Output:
(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 + 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**2*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)*b*c*d**2*x**2 + 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*d**3*x**3 + log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*b*c**3 + 3*log(a**(1/3) + b** (1/3)*c + b**(1/3)*d*x)*b*c**2*d*x + 3*log(a**(1/3) + b**(1/3)*c + b**(1/3 )*d*x)*b*c*d**2*x**2 + log(a**(1/3) + b**(1/3)*c + b**(1/3)*d*x)*b*d**3*x* *3 - 3*log(c + d*x)*b*c**3 - 9*log(c + d*x)*b*c**2*d*x - 9*log(c + d*x)*b* c*d**2*x**2 - 3*log(c + d*x)*b*d**3*x**3 - a)/(3*a**2*d*e**4*(c**3 + 3*c** 2*d*x + 3*c*d**2*x**2 + d**3*x**3))