Integrand size = 26, antiderivative size = 118 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=-\frac {c (b c-a d)^2 x^n}{d^4 n}+\frac {(b c-a d)^2 x^{2 n}}{2 d^3 n}-\frac {b (b c-2 a d) x^{3 n}}{3 d^2 n}+\frac {b^2 x^{4 n}}{4 d n}+\frac {c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{d^5 n} \]
-c*(-a*d+b*c)^2*x^n/d^4/n+1/2*(-a*d+b*c)^2*x^(2*n)/d^3/n-1/3*b*(-2*a*d+b*c )*x^(3*n)/d^2/n+1/4*b^2*x^(4*n)/d/n+c^2*(-a*d+b*c)^2*ln(c+d*x^n)/d^5/n
Time = 0.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\frac {d x^n \left (6 a^2 d^2 \left (-2 c+d x^n\right )+4 a b d \left (6 c^2-3 c d x^n+2 d^2 x^{2 n}\right )+b^2 \left (-12 c^3+6 c^2 d x^n-4 c d^2 x^{2 n}+3 d^3 x^{3 n}\right )\right )+12 c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{12 d^5 n} \]
(d*x^n*(6*a^2*d^2*(-2*c + d*x^n) + 4*a*b*d*(6*c^2 - 3*c*d*x^n + 2*d^2*x^(2 *n)) + b^2*(-12*c^3 + 6*c^2*d*x^n - 4*c*d^2*x^(2*n) + 3*d^3*x^(3*n))) + 12 *c^2*(b*c - a*d)^2*Log[c + d*x^n])/(12*d^5*n)
Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {948, 99, 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.
\(\displaystyle \int \frac {x^{3 n-1} \left (a+b x^n\right )^2}{c+d x^n} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\int \frac {x^{2 n} \left (b x^n+a\right )^2}{d x^n+c}dx^n}{n}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {(a d-b c)^2 x^n}{d^3}-\frac {b (b c-2 a d) x^{2 n}}{d^2}+\frac {b^2 x^{3 n}}{d}-\frac {c (b c-a d)^2}{d^4}+\frac {c^2 (b c-a d)^2}{d^4 \left (d x^n+c\right )}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{d^5}-\frac {c x^n (b c-a d)^2}{d^4}+\frac {x^{2 n} (b c-a d)^2}{2 d^3}-\frac {b x^{3 n} (b c-2 a d)}{3 d^2}+\frac {b^2 x^{4 n}}{4 d}}{n}\) |
(-((c*(b*c - a*d)^2*x^n)/d^4) + ((b*c - a*d)^2*x^(2*n))/(2*d^3) - (b*(b*c - 2*a*d)*x^(3*n))/(3*d^2) + (b^2*x^(4*n))/(4*d) + (c^2*(b*c - a*d)^2*Log[c + d*x^n])/d^5)/n
3.11.46.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.90 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.33
method | result | size |
norman | \(\frac {b^{2} {\mathrm e}^{4 n \ln \left (x \right )}}{4 d n}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) {\mathrm e}^{2 n \ln \left (x \right )}}{2 d^{3} n}+\frac {b \left (2 a d -b c \right ) {\mathrm e}^{3 n \ln \left (x \right )}}{3 d^{2} n}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) {\mathrm e}^{n \ln \left (x \right )}}{d^{4} n}+\frac {c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (c +d \,{\mathrm e}^{n \ln \left (x \right )}\right )}{d^{5} n}\) | \(157\) |
risch | \(\frac {b^{2} x^{4 n}}{4 d n}+\frac {2 b \,x^{3 n} a}{3 d n}-\frac {b^{2} x^{3 n} c}{3 d^{2} n}+\frac {x^{2 n} a^{2}}{2 d n}-\frac {x^{2 n} a b c}{d^{2} n}+\frac {x^{2 n} b^{2} c^{2}}{2 d^{3} n}-\frac {c \,x^{n} a^{2}}{d^{2} n}+\frac {2 c^{2} x^{n} a b}{d^{3} n}-\frac {c^{3} x^{n} b^{2}}{d^{4} n}+\frac {c^{2} \ln \left (x^{n}+\frac {c}{d}\right ) a^{2}}{d^{3} n}-\frac {2 c^{3} \ln \left (x^{n}+\frac {c}{d}\right ) a b}{d^{4} n}+\frac {c^{4} \ln \left (x^{n}+\frac {c}{d}\right ) b^{2}}{d^{5} n}\) | \(218\) |
1/4*b^2/d/n*exp(n*ln(x))^4+1/2/d^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)/n*exp(n*ln( x))^2+1/3*b*(2*a*d-b*c)/d^2/n*exp(n*ln(x))^3-c*(a^2*d^2-2*a*b*c*d+b^2*c^2) /d^4/n*exp(n*ln(x))+c^2/d^5*(a^2*d^2-2*a*b*c*d+b^2*c^2)/n*ln(c+d*exp(n*ln( x)))
Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\frac {3 \, b^{2} d^{4} x^{4 \, n} - 4 \, {\left (b^{2} c d^{3} - 2 \, a b d^{4}\right )} x^{3 \, n} + 6 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2 \, n} - 12 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{n} + 12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \log \left (d x^{n} + c\right )}{12 \, d^{5} n} \]
1/12*(3*b^2*d^4*x^(4*n) - 4*(b^2*c*d^3 - 2*a*b*d^4)*x^(3*n) + 6*(b^2*c^2*d ^2 - 2*a*b*c*d^3 + a^2*d^4)*x^(2*n) - 12*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2* c*d^3)*x^n + 12*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*log(d*x^n + c))/(d^5 *n)
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (99) = 198\).
Time = 3.20 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.35 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\begin {cases} \frac {\left (a + b\right )^{2} \log {\left (x \right )}}{c} & \text {for}\: d = 0 \wedge n = 0 \\\frac {\frac {a^{2} x x^{3 n - 1}}{3 n} + \frac {a b x x^{n} x^{3 n - 1}}{2 n} + \frac {b^{2} x x^{2 n} x^{3 n - 1}}{5 n}}{c} & \text {for}\: d = 0 \\\frac {\left (a + b\right )^{2} \log {\left (x \right )}}{c + d} & \text {for}\: n = 0 \\\frac {a^{2} c^{2} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{3} n} - \frac {a^{2} c x^{n}}{d^{2} n} + \frac {a^{2} x^{2 n}}{2 d n} - \frac {2 a b c^{3} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{4} n} + \frac {2 a b c^{2} x^{n}}{d^{3} n} - \frac {a b c x^{2 n}}{d^{2} n} + \frac {2 a b x^{3 n}}{3 d n} + \frac {b^{2} c^{4} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{5} n} - \frac {b^{2} c^{3} x^{n}}{d^{4} n} + \frac {b^{2} c^{2} x^{2 n}}{2 d^{3} n} - \frac {b^{2} c x^{3 n}}{3 d^{2} n} + \frac {b^{2} x^{4 n}}{4 d n} & \text {otherwise} \end {cases} \]
Piecewise(((a + b)**2*log(x)/c, Eq(d, 0) & Eq(n, 0)), ((a**2*x*x**(3*n - 1 )/(3*n) + a*b*x*x**n*x**(3*n - 1)/(2*n) + b**2*x*x**(2*n)*x**(3*n - 1)/(5* n))/c, Eq(d, 0)), ((a + b)**2*log(x)/(c + d), Eq(n, 0)), (a**2*c**2*log(c/ d + x**n)/(d**3*n) - a**2*c*x**n/(d**2*n) + a**2*x**(2*n)/(2*d*n) - 2*a*b* c**3*log(c/d + x**n)/(d**4*n) + 2*a*b*c**2*x**n/(d**3*n) - a*b*c*x**(2*n)/ (d**2*n) + 2*a*b*x**(3*n)/(3*d*n) + b**2*c**4*log(c/d + x**n)/(d**5*n) - b **2*c**3*x**n/(d**4*n) + b**2*c**2*x**(2*n)/(2*d**3*n) - b**2*c*x**(3*n)/( 3*d**2*n) + b**2*x**(4*n)/(4*d*n), True))
Time = 0.22 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.63 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\frac {1}{12} \, b^{2} {\left (\frac {12 \, c^{4} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{5} n} + \frac {3 \, d^{3} x^{4 \, n} - 4 \, c d^{2} x^{3 \, n} + 6 \, c^{2} d x^{2 \, n} - 12 \, c^{3} x^{n}}{d^{4} n}\right )} - \frac {1}{3} \, a b {\left (\frac {6 \, c^{3} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{4} n} - \frac {2 \, d^{2} x^{3 \, n} - 3 \, c d x^{2 \, n} + 6 \, c^{2} x^{n}}{d^{3} n}\right )} + \frac {1}{2} \, a^{2} {\left (\frac {2 \, c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{3} n} + \frac {d x^{2 \, n} - 2 \, c x^{n}}{d^{2} n}\right )} \]
1/12*b^2*(12*c^4*log((d*x^n + c)/d)/(d^5*n) + (3*d^3*x^(4*n) - 4*c*d^2*x^( 3*n) + 6*c^2*d*x^(2*n) - 12*c^3*x^n)/(d^4*n)) - 1/3*a*b*(6*c^3*log((d*x^n + c)/d)/(d^4*n) - (2*d^2*x^(3*n) - 3*c*d*x^(2*n) + 6*c^2*x^n)/(d^3*n)) + 1 /2*a^2*(2*c^2*log((d*x^n + c)/d)/(d^3*n) + (d*x^(2*n) - 2*c*x^n)/(d^2*n))
\[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{2} x^{3 \, n - 1}}{d x^{n} + c} \,d x } \]
Timed out. \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^2}{c+d x^n} \, dx=\int \frac {x^{3\,n-1}\,{\left (a+b\,x^n\right )}^2}{c+d\,x^n} \,d x \]