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

Optimal. Leaf size=158 \[ \frac {b x^{3 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{3 d^3 n}-\frac {b^2 x^{4 n} (b c-3 a d)}{4 d^2 n}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6 n}+\frac {c x^n (b c-a d)^3}{d^5 n}-\frac {x^{2 n} (b c-a d)^3}{2 d^4 n}+\frac {b^3 x^{5 n}}{5 d n} \]

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Rubi [A]  time = 0.15, antiderivative size = 158, 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, 88} \begin {gather*} \frac {b x^{3 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{3 d^3 n}-\frac {b^2 x^{4 n} (b c-3 a d)}{4 d^2 n}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6 n}+\frac {c x^n (b c-a d)^3}{d^5 n}-\frac {x^{2 n} (b c-a d)^3}{2 d^4 n}+\frac {b^3 x^{5 n}}{5 d n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

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+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)^3}{c+d x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {c (b c-a d)^3}{d^5}+\frac {(-b c+a d)^3 x}{d^4}+\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x^2}{d^3}-\frac {b^2 (b c-3 a d) x^3}{d^2}+\frac {b^3 x^4}{d}-\frac {c^2 (b c-a d)^3}{d^5 (c+d x)}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {c (b c-a d)^3 x^n}{d^5 n}-\frac {(b c-a d)^3 x^{2 n}}{2 d^4 n}+\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x^{3 n}}{3 d^3 n}-\frac {b^2 (b c-3 a d) x^{4 n}}{4 d^2 n}+\frac {b^3 x^{5 n}}{5 d n}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6 n}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 138, normalized size = 0.87 \begin {gather*} \frac {20 b d^3 x^{3 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )-15 b^2 d^4 x^{4 n} (b c-3 a d)-60 c^2 (b c-a d)^3 \log \left (c+d x^n\right )+30 d^2 x^{2 n} (a d-b c)^3+60 c d x^n (b c-a d)^3+12 b^3 d^5 x^{5 n}}{60 d^6 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(60*c*d*(b*c - a*d)^3*x^n + 30*d^2*(-(b*c) + a*d)^3*x^(2*n) + 20*b*d^3*(b^2*c^2 - 3*a*b*c*d + 3*a^2*d^2)*x^(3*
n) - 15*b^2*d^4*(b*c - 3*a*d)*x^(4*n) + 12*b^3*d^5*x^(5*n) - 60*c^2*(b*c - a*d)^3*Log[c + d*x^n])/(60*d^6*n)

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IntegrateAlgebraic [A]  time = 0.13, size = 220, normalized size = 1.39 \begin {gather*} \frac {x^n \left (-60 a^3 c d^3+30 a^3 d^4 x^n+180 a^2 b c^2 d^2-90 a^2 b c d^3 x^n+60 a^2 b d^4 x^{2 n}-180 a b^2 c^3 d+90 a b^2 c^2 d^2 x^n-60 a b^2 c d^3 x^{2 n}+45 a b^2 d^4 x^{3 n}+60 b^3 c^4-30 b^3 c^3 d x^n+20 b^3 c^2 d^2 x^{2 n}-15 b^3 c d^3 x^{3 n}+12 b^3 d^4 x^{4 n}\right )}{60 d^5 n}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^n*(60*b^3*c^4 - 180*a*b^2*c^3*d + 180*a^2*b*c^2*d^2 - 60*a^3*c*d^3 - 30*b^3*c^3*d*x^n + 90*a*b^2*c^2*d^2*x^
n - 90*a^2*b*c*d^3*x^n + 30*a^3*d^4*x^n + 20*b^3*c^2*d^2*x^(2*n) - 60*a*b^2*c*d^3*x^(2*n) + 60*a^2*b*d^4*x^(2*
n) - 15*b^3*c*d^3*x^(3*n) + 45*a*b^2*d^4*x^(3*n) + 12*b^3*d^4*x^(4*n)))/(60*d^5*n) - (c^2*(b*c - a*d)^3*Log[c
+ d*x^n])/(d^6*n)

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fricas [A]  time = 0.49, size = 230, normalized size = 1.46 \begin {gather*} \frac {12 \, b^{3} d^{5} x^{5 \, n} - 15 \, {\left (b^{3} c d^{4} - 3 \, a b^{2} d^{5}\right )} x^{4 \, n} + 20 \, {\left (b^{3} c^{2} d^{3} - 3 \, a b^{2} c d^{4} + 3 \, a^{2} b d^{5}\right )} x^{3 \, n} - 30 \, {\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{2 \, n} + 60 \, {\left (b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{n} - 60 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \log \left (d x^{n} + c\right )}{60 \, d^{6} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*b^3*d^5*x^(5*n) - 15*(b^3*c*d^4 - 3*a*b^2*d^5)*x^(4*n) + 20*(b^3*c^2*d^3 - 3*a*b^2*c*d^4 + 3*a^2*b*d^
5)*x^(3*n) - 30*(b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 - a^3*d^5)*x^(2*n) + 60*(b^3*c^4*d - 3*a*b^2*c^
3*d^2 + 3*a^2*b*c^2*d^3 - a^3*c*d^4)*x^n - 60*(b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*log(d*
x^n + c))/(d^6*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 342, normalized size = 2.16 \begin {gather*} \frac {a^{3} c^{2} \ln \left (x^{n}+\frac {c}{d}\right )}{d^{3} n}-\frac {a^{3} c \,x^{n}}{d^{2} n}+\frac {a^{3} x^{2 n}}{2 d n}-\frac {3 a^{2} b \,c^{3} \ln \left (x^{n}+\frac {c}{d}\right )}{d^{4} n}+\frac {3 a^{2} b \,c^{2} x^{n}}{d^{3} n}-\frac {3 a^{2} b c \,x^{2 n}}{2 d^{2} n}+\frac {a^{2} b \,x^{3 n}}{d n}+\frac {3 a \,b^{2} c^{4} \ln \left (x^{n}+\frac {c}{d}\right )}{d^{5} n}-\frac {3 a \,b^{2} c^{3} x^{n}}{d^{4} n}+\frac {3 a \,b^{2} c^{2} x^{2 n}}{2 d^{3} n}-\frac {a \,b^{2} c \,x^{3 n}}{d^{2} n}+\frac {3 a \,b^{2} x^{4 n}}{4 d n}-\frac {b^{3} c^{5} \ln \left (x^{n}+\frac {c}{d}\right )}{d^{6} n}+\frac {b^{3} c^{4} x^{n}}{d^{5} n}-\frac {b^{3} c^{3} x^{2 n}}{2 d^{4} n}+\frac {b^{3} c^{2} x^{3 n}}{3 d^{3} n}-\frac {b^{3} c \,x^{4 n}}{4 d^{2} n}+\frac {b^{3} x^{5 n}}{5 d n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 286, normalized size = 1.81 \begin {gather*} -\frac {1}{60} \, b^{3} {\left (\frac {60 \, c^{5} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{6} n} - \frac {12 \, d^{4} x^{5 \, n} - 15 \, c d^{3} x^{4 \, n} + 20 \, c^{2} d^{2} x^{3 \, n} - 30 \, c^{3} d x^{2 \, n} + 60 \, c^{4} x^{n}}{d^{5} n}\right )} + \frac {1}{4} \, a 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}{2} \, a^{2} 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^{3} {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*b^3*(60*c^5*log((d*x^n + c)/d)/(d^6*n) - (12*d^4*x^(5*n) - 15*c*d^3*x^(4*n) + 20*c^2*d^2*x^(3*n) - 30*c^
3*d*x^(2*n) + 60*c^4*x^n)/(d^5*n)) + 1/4*a*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)) - 1/2*a^2*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^3*(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 \begin {gather*} \int \frac {x^{3\,n-1}\,{\left (a+b\,x^n\right )}^3}{c+d\,x^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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