3.1039 \(\int \frac {x^{-1+2 n} (a+b x^n)^3}{c+d x^n} \, dx\)

Optimal. Leaf size=130 \[ \frac {b x^{2 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{2 d^3 n}-\frac {b^2 x^{3 n} (b c-3 a d)}{3 d^2 n}+\frac {c (b c-a d)^3 \log \left (c+d x^n\right )}{d^5 n}-\frac {x^n (b c-a d)^3}{d^4 n}+\frac {b^3 x^{4 n}}{4 d n} \]

[Out]

-(-a*d+b*c)^3*x^n/d^4/n+1/2*b*(3*a^2*d^2-3*a*b*c*d+b^2*c^2)*x^(2*n)/d^3/n-1/3*b^2*(-3*a*d+b*c)*x^(3*n)/d^2/n+1
/4*b^3*x^(4*n)/d/n+c*(-a*d+b*c)^3*ln(c+d*x^n)/d^5/n

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Rubi [A]  time = 0.14, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {446, 77} \[ \frac {b x^{2 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{2 d^3 n}-\frac {b^2 x^{3 n} (b c-3 a d)}{3 d^2 n}-\frac {x^n (b c-a d)^3}{d^4 n}+\frac {c (b c-a d)^3 \log \left (c+d x^n\right )}{d^5 n}+\frac {b^3 x^{4 n}}{4 d n} \]

Antiderivative was successfully verified.

[In]

Int[(x^(-1 + 2*n)*(a + b*x^n)^3)/(c + d*x^n),x]

[Out]

-(((b*c - a*d)^3*x^n)/(d^4*n)) + (b*(b^2*c^2 - 3*a*b*c*d + 3*a^2*d^2)*x^(2*n))/(2*d^3*n) - (b^2*(b*c - 3*a*d)*
x^(3*n))/(3*d^2*n) + (b^3*x^(4*n))/(4*d*n) + (c*(b*c - a*d)^3*Log[c + d*x^n])/(d^5*n)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[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]]

Rubi steps

\begin {align*} \int \frac {x^{-1+2 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x (a+b x)^3}{c+d x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {(-b c+a d)^3}{d^4}+\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x}{d^3}-\frac {b^2 (b c-3 a d) x^2}{d^2}+\frac {b^3 x^3}{d}+\frac {c (b c-a d)^3}{d^4 (c+d x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {(b c-a d)^3 x^n}{d^4 n}+\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x^{2 n}}{2 d^3 n}-\frac {b^2 (b c-3 a d) x^{3 n}}{3 d^2 n}+\frac {b^3 x^{4 n}}{4 d n}+\frac {c (b c-a d)^3 \log \left (c+d x^n\right )}{d^5 n}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 115, normalized size = 0.88 \[ \frac {6 b d^2 x^{2 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )-4 b^2 d^3 x^{3 n} (b c-3 a d)+12 d x^n (a d-b c)^3+12 c (b c-a d)^3 \log \left (c+d x^n\right )+3 b^3 d^4 x^{4 n}}{12 d^5 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(-1 + 2*n)*(a + b*x^n)^3)/(c + d*x^n),x]

[Out]

(12*d*(-(b*c) + a*d)^3*x^n + 6*b*d^2*(b^2*c^2 - 3*a*b*c*d + 3*a^2*d^2)*x^(2*n) - 4*b^2*d^3*(b*c - 3*a*d)*x^(3*
n) + 3*b^3*d^4*x^(4*n) + 12*c*(b*c - a*d)^3*Log[c + d*x^n])/(12*d^5*n)

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fricas [A]  time = 0.75, size = 177, normalized size = 1.36 \[ \frac {3 \, b^{3} d^{4} x^{4 \, n} - 4 \, {\left (b^{3} c d^{3} - 3 \, a b^{2} d^{4}\right )} x^{3 \, n} + 6 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x^{2 \, n} - 12 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x^{n} + 12 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \log \left (d x^{n} + c\right )}{12 \, d^{5} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="fricas")

[Out]

1/12*(3*b^3*d^4*x^(4*n) - 4*(b^3*c*d^3 - 3*a*b^2*d^4)*x^(3*n) + 6*(b^3*c^2*d^2 - 3*a*b^2*c*d^3 + 3*a^2*b*d^4)*
x^(2*n) - 12*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*x^n + 12*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2
*b*c^2*d^2 - a^3*c*d^3)*log(d*x^n + c))/(d^5*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{n} + a\right )}^{3} x^{2 \, n - 1}}{d x^{n} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="giac")

[Out]

integrate((b*x^n + a)^3*x^(2*n - 1)/(d*x^n + c), x)

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maple [B]  time = 0.08, size = 284, normalized size = 2.18 \[ -\frac {a^{3} c \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{2} n}+\frac {a^{3} {\mathrm e}^{n \ln \relax (x )}}{d n}+\frac {3 a^{2} b \,c^{2} \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{3} n}-\frac {3 a^{2} b c \,{\mathrm e}^{n \ln \relax (x )}}{d^{2} n}+\frac {3 a^{2} b \,{\mathrm e}^{2 n \ln \relax (x )}}{2 d n}-\frac {3 a \,b^{2} c^{3} \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{4} n}+\frac {3 a \,b^{2} c^{2} {\mathrm e}^{n \ln \relax (x )}}{d^{3} n}-\frac {3 a \,b^{2} c \,{\mathrm e}^{2 n \ln \relax (x )}}{2 d^{2} n}+\frac {a \,b^{2} {\mathrm e}^{3 n \ln \relax (x )}}{d n}+\frac {b^{3} c^{4} \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{5} n}-\frac {b^{3} c^{3} {\mathrm e}^{n \ln \relax (x )}}{d^{4} n}+\frac {b^{3} c^{2} {\mathrm e}^{2 n \ln \relax (x )}}{2 d^{3} n}-\frac {b^{3} c \,{\mathrm e}^{3 n \ln \relax (x )}}{3 d^{2} n}+\frac {b^{3} {\mathrm e}^{4 n \ln \relax (x )}}{4 d n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n-1)*(b*x^n+a)^3/(d*x^n+c),x)

[Out]

1/d/n*exp(n*ln(x))*a^3-3/d^2/n*exp(n*ln(x))*a^2*b*c+3/d^3/n*exp(n*ln(x))*a*b^2*c^2-1/d^4/n*exp(n*ln(x))*b^3*c^
3+1/4*b^3/d/n*exp(n*ln(x))^4+3/2*b/d/n*exp(n*ln(x))^2*a^2-3/2*b^2/d^2/n*exp(n*ln(x))^2*a*c+1/2*b^3/d^3/n*exp(n
*ln(x))^2*c^2+b^2/d/n*exp(n*ln(x))^3*a-1/3*b^3/d^2/n*exp(n*ln(x))^3*c-c/d^2/n*ln(d*exp(n*ln(x))+c)*a^3+3*c^2/d
^3/n*ln(d*exp(n*ln(x))+c)*a^2*b-3*c^3/d^4/n*ln(d*exp(n*ln(x))+c)*a*b^2+c^4/d^5/n*ln(d*exp(n*ln(x))+c)*b^3

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maxima [A]  time = 0.53, size = 231, normalized size = 1.78 \[ a^{3} {\left (\frac {x^{n}}{d n} - \frac {c \log \left (\frac {d x^{n} + c}{d}\right )}{d^{2} n}\right )} + \frac {1}{12} \, b^{3} {\left (\frac {12 \, c^{4} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{5} n} + \frac {3 \, d^{3} x^{4 \, n} - 4 \, c d^{2} x^{3 \, n} + 6 \, c^{2} d x^{2 \, n} - 12 \, c^{3} x^{n}}{d^{4} n}\right )} - \frac {1}{2} \, a b^{2} {\left (\frac {6 \, c^{3} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{4} n} - \frac {2 \, d^{2} x^{3 \, n} - 3 \, c d x^{2 \, n} + 6 \, c^{2} x^{n}}{d^{3} n}\right )} + \frac {3}{2} \, a^{2} b {\left (\frac {2 \, c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{3} n} + \frac {d x^{2 \, n} - 2 \, c x^{n}}{d^{2} n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="maxima")

[Out]

a^3*(x^n/(d*n) - c*log((d*x^n + c)/d)/(d^2*n)) + 1/12*b^3*(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)) - 1/2*a*b^2*(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)) + 3/2*a^2*b*(2*c^2*log((d*x^n + c)/d)/(d^3*n) + (d*x^(2*n) -
2*c*x^n)/(d^2*n))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{2\,n-1}\,{\left (a+b\,x^n\right )}^3}{c+d\,x^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(2*n - 1)*(a + b*x^n)^3)/(c + d*x^n),x)

[Out]

int((x^(2*n - 1)*(a + b*x^n)^3)/(c + d*x^n), x)

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sympy [A]  time = 135.87, size = 320, normalized size = 2.46 \[ \begin {cases} \frac {\left (a + b\right )^{3} \log {\relax (x )}}{c} & \text {for}\: d = 0 \wedge n = 0 \\\frac {\frac {a^{3} x^{2 n}}{2 n} + \frac {a^{2} b x^{3 n}}{n} + \frac {3 a b^{2} x^{4 n}}{4 n} + \frac {b^{3} x^{5 n}}{5 n}}{c} & \text {for}\: d = 0 \\\frac {\left (a + b\right )^{3} \log {\relax (x )}}{c + d} & \text {for}\: n = 0 \\- \frac {a^{3} c \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{2} n} + \frac {a^{3} x^{n}}{d n} + \frac {3 a^{2} b c^{2} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{3} n} - \frac {3 a^{2} b c x^{n}}{d^{2} n} + \frac {3 a^{2} b x^{2 n}}{2 d n} - \frac {3 a b^{2} c^{3} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{4} n} + \frac {3 a b^{2} c^{2} x^{n}}{d^{3} n} - \frac {3 a b^{2} c x^{2 n}}{2 d^{2} n} + \frac {a b^{2} x^{3 n}}{d n} + \frac {b^{3} c^{4} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{5} n} - \frac {b^{3} c^{3} x^{n}}{d^{4} n} + \frac {b^{3} c^{2} x^{2 n}}{2 d^{3} n} - \frac {b^{3} c x^{3 n}}{3 d^{2} n} + \frac {b^{3} x^{4 n}}{4 d n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)*(a+b*x**n)**3/(c+d*x**n),x)

[Out]

Piecewise(((a + b)**3*log(x)/c, Eq(d, 0) & Eq(n, 0)), ((a**3*x**(2*n)/(2*n) + a**2*b*x**(3*n)/n + 3*a*b**2*x**
(4*n)/(4*n) + b**3*x**(5*n)/(5*n))/c, Eq(d, 0)), ((a + b)**3*log(x)/(c + d), Eq(n, 0)), (-a**3*c*log(c/d + x**
n)/(d**2*n) + a**3*x**n/(d*n) + 3*a**2*b*c**2*log(c/d + x**n)/(d**3*n) - 3*a**2*b*c*x**n/(d**2*n) + 3*a**2*b*x
**(2*n)/(2*d*n) - 3*a*b**2*c**3*log(c/d + x**n)/(d**4*n) + 3*a*b**2*c**2*x**n/(d**3*n) - 3*a*b**2*c*x**(2*n)/(
2*d**2*n) + a*b**2*x**(3*n)/(d*n) + b**3*c**4*log(c/d + x**n)/(d**5*n) - b**3*c**3*x**n/(d**4*n) + b**3*c**2*x
**(2*n)/(2*d**3*n) - b**3*c*x**(3*n)/(3*d**2*n) + b**3*x**(4*n)/(4*d*n), True))

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