Integrand size = 28, antiderivative size = 161 \[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=-\frac {x^{-3-n} (c x)^{-((1+n) p)} \left (a x^n+b x^{1+n}\right )^{1+p}}{a c^4 (3+p)}+\frac {2 b x^{-2-n} (c x)^{-((1+n) p)} \left (a x^n+b x^{1+n}\right )^{1+p}}{a^2 c^4 (2+p) (3+p)}-\frac {2 b^2 x^{-1-n} (c x)^{-((1+n) p)} \left (a x^n+b x^{1+n}\right )^{1+p}}{a^3 c^4 (1+p) (2+p) (3+p)} \] Output:
-x^(-3-n)*(a*x^n+b*x^(1+n))^(p+1)/a/c^4/(3+p)/((c*x)^((1+n)*p))+2*b*x^(-2- n)*(a*x^n+b*x^(1+n))^(p+1)/a^2/c^4/(2+p)/(3+p)/((c*x)^((1+n)*p))-2*b^2*x^( -1-n)*(a*x^n+b*x^(1+n))^(p+1)/a^3/c^4/(p+1)/(2+p)/(3+p)/((c*x)^((1+n)*p))
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.50 \[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=-\frac {(c x)^{-((1+n) p)} (a+b x) \left (x^n (a+b x)\right )^p \left (a^2 \left (2+3 p+p^2\right )-2 a b (1+p) x+2 b^2 x^2\right )}{a^3 c^4 (1+p) (2+p) (3+p) x^3} \] Input:
Integrate[(c*x)^(-4 - (1 + n)*p)*(a*x^n + b*x^(1 + n))^p,x]
Output:
-(((a + b*x)*(x^n*(a + b*x))^p*(a^2*(2 + 3*p + p^2) - 2*a*b*(1 + p)*x + 2* b^2*x^2))/(a^3*c^4*(1 + p)*(2 + p)*(3 + p)*x^3*(c*x)^((1 + n)*p)))
Time = 0.56 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1923, 1922, 1922, 1920}
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.
\(\displaystyle \int (c x)^{-((n+1) p)-4} \left (a x^n+b x^{n+1}\right )^p \, dx\) |
\(\Big \downarrow \) 1923 |
\(\displaystyle \frac {x^{(n+1) p} (c x)^{-((n+1) p)} \int x^{-((n+1) p)-4} \left (a x^n+b x^{n+1}\right )^pdx}{c^4}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {x^{(n+1) p} (c x)^{-((n+1) p)} \left (-\frac {2 b \int x^{-((n+1) p)-3} \left (a x^n+b x^{n+1}\right )^pdx}{a (p+3)}-\frac {x^{-n (p+1)-p-3} \left (a x^n+b x^{n+1}\right )^{p+1}}{a (p+3)}\right )}{c^4}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {x^{(n+1) p} (c x)^{-((n+1) p)} \left (-\frac {2 b \left (-\frac {b \int x^{-((n+1) p)-2} \left (a x^n+b x^{n+1}\right )^pdx}{a (p+2)}-\frac {x^{-n (p+1)-p-2} \left (a x^n+b x^{n+1}\right )^{p+1}}{a (p+2)}\right )}{a (p+3)}-\frac {x^{-n (p+1)-p-3} \left (a x^n+b x^{n+1}\right )^{p+1}}{a (p+3)}\right )}{c^4}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {x^{(n+1) p} (c x)^{-((n+1) p)} \left (-\frac {2 b \left (\frac {b x^{-((n+1) (p+1))} \left (a x^n+b x^{n+1}\right )^{p+1}}{a^2 (p+1) (p+2)}-\frac {x^{-n (p+1)-p-2} \left (a x^n+b x^{n+1}\right )^{p+1}}{a (p+2)}\right )}{a (p+3)}-\frac {x^{-n (p+1)-p-3} \left (a x^n+b x^{n+1}\right )^{p+1}}{a (p+3)}\right )}{c^4}\) |
Input:
Int[(c*x)^(-4 - (1 + n)*p)*(a*x^n + b*x^(1 + n))^p,x]
Output:
(x^((1 + n)*p)*(-((x^(-3 - p - n*(1 + p))*(a*x^n + b*x^(1 + n))^(1 + p))/( a*(3 + p))) - (2*b*((b*(a*x^n + b*x^(1 + n))^(1 + p))/(a^2*(1 + p)*(2 + p) *x^((1 + n)*(1 + p))) - (x^(-2 - p - n*(1 + p))*(a*x^n + b*x^(1 + n))^(1 + p))/(a*(2 + p))))/(a*(3 + p))))/(c^4*(c*x)^((1 + n)*p))
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ [n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0]
Time = 2.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.57
method | result | size |
orering | \(-\frac {\left (a^{2} p^{2}-2 a b p x +2 b^{2} x^{2}+3 a^{2} p -2 a b x +2 a^{2}\right ) x \left (b x +a \right ) \left (c x \right )^{-4-\left (1+n \right ) p} \left (a \,x^{n}+b \,x^{1+n}\right )^{p}}{\left (3+p \right ) \left (2+p \right ) a^{3} \left (p +1\right )}\) | \(92\) |
Input:
int((c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p,x,method=_RETURNVERBOSE)
Output:
-(a^2*p^2-2*a*b*p*x+2*b^2*x^2+3*a^2*p-2*a*b*x+2*a^2)/(3+p)/(2+p)/a^3/(p+1) *x*(b*x+a)*(c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81 \[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\frac {{\left (2 \, a b^{2} p x^{3} - 2 \, b^{3} x^{4} - {\left (a^{2} b p^{2} + a^{2} b p\right )} x^{2} - {\left (a^{3} p^{2} + 3 \, a^{3} p + 2 \, a^{3}\right )} x\right )} \left (\frac {{\left (b x + a\right )} x^{n + 1}}{x}\right )^{p} e^{\left (-{\left ({\left (n + 1\right )} p + 4\right )} \log \left (c\right ) - {\left ({\left (n + 1\right )} p + 4\right )} \log \left (x\right )\right )}}{a^{3} p^{3} + 6 \, a^{3} p^{2} + 11 \, a^{3} p + 6 \, a^{3}} \] Input:
integrate((c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p,x, algorithm="fricas")
Output:
(2*a*b^2*p*x^3 - 2*b^3*x^4 - (a^2*b*p^2 + a^2*b*p)*x^2 - (a^3*p^2 + 3*a^3* p + 2*a^3)*x)*((b*x + a)*x^(n + 1)/x)^p*e^(-((n + 1)*p + 4)*log(c) - ((n + 1)*p + 4)*log(x))/(a^3*p^3 + 6*a^3*p^2 + 11*a^3*p + 6*a^3)
\[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\int \left (c x\right )^{- p \left (n + 1\right ) - 4} \left (a x^{n} + b x^{n + 1}\right )^{p}\, dx \] Input:
integrate((c*x)**(-4-(1+n)*p)*(a*x**n+b*x**(1+n))**p,x)
Output:
Integral((c*x)**(-p*(n + 1) - 4)*(a*x**n + b*x**(n + 1))**p, x)
\[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\int { {\left (b x^{n + 1} + a x^{n}\right )}^{p} \left (c x\right )^{-{\left (n + 1\right )} p - 4} \,d x } \] Input:
integrate((c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p,x, algorithm="maxima")
Output:
integrate((b*x^(n + 1) + a*x^n)^p*(c*x)^(-(n + 1)*p - 4), x)
\[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\int { {\left (b x^{n + 1} + a x^{n}\right )}^{p} \left (c x\right )^{-{\left (n + 1\right )} p - 4} \,d x } \] Input:
integrate((c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p,x, algorithm="giac")
Output:
integrate((b*x^(n + 1) + a*x^n)^p*(c*x)^(-(n + 1)*p - 4), x)
Timed out. \[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\int \frac {{\left (a\,x^n+b\,x^{n+1}\right )}^p}{{\left (c\,x\right )}^{p\,\left (n+1\right )+4}} \,d x \] Input:
int((a*x^n + b*x^(n + 1))^p/(c*x)^(p*(n + 1) + 4),x)
Output:
int((a*x^n + b*x^(n + 1))^p/(c*x)^(p*(n + 1) + 4), x)
Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.70 \[ \int (c x)^{-4-(1+n) p} \left (a x^n+b x^{1+n}\right )^p \, dx=\frac {\left (x^{n} a +x^{n} b x \right )^{p} \left (-a^{2} b \,p^{2} x +2 a \,b^{2} p \,x^{2}-2 b^{3} x^{3}-a^{3} p^{2}-a^{2} b p x -3 a^{3} p -2 a^{3}\right )}{x^{n p +p} c^{n p +p} a^{3} c^{4} x^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int((c*x)^(-4-(1+n)*p)*(a*x^n+b*x^(1+n))^p,x)
Output:
((x**n*a + x**n*b*x)**p*( - a**3*p**2 - 3*a**3*p - 2*a**3 - a**2*b*p**2*x - a**2*b*p*x + 2*a*b**2*p*x**2 - 2*b**3*x**3))/(x**(n*p + p)*c**(n*p + p)* a**3*c**4*x**3*(p**3 + 6*p**2 + 11*p + 6))