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

Optimal. Leaf size=105 \[ \frac {a}{2 b n (b c-a d) \left (a+b x^n\right )^2}-\frac {c}{n (b c-a d)^2 \left (a+b x^n\right )}-\frac {c d \log \left (a+b x^n\right )}{n (b c-a d)^3}+\frac {c d \log \left (c+d x^n\right )}{n (b c-a d)^3} \]

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Rubi [A]  time = 0.09, antiderivative size = 105, 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} \begin {gather*} \frac {a}{2 b n (b c-a d) \left (a+b x^n\right )^2}-\frac {c}{n (b c-a d)^2 \left (a+b x^n\right )}-\frac {c d \log \left (a+b x^n\right )}{n (b c-a d)^3}+\frac {c d \log \left (c+d x^n\right )}{n (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a/(2*b*(b*c - a*d)*n*(a + b*x^n)^2) - c/((b*c - a*d)^2*n*(a + b*x^n)) - (c*d*Log[a + b*x^n])/((b*c - a*d)^3*n)
+ (c*d*Log[c + d*x^n])/((b*c - a*d)^3*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 \left (c+d x^n\right )} \, 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 {a}{(b c-a d) (a+b x)^3}+\frac {b c}{(b c-a d)^2 (a+b x)^2}-\frac {b c d}{(b c-a d)^3 (a+b x)}+\frac {c d^2}{(b c-a d)^3 (c+d x)}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {a}{2 b (b c-a d) n \left (a+b x^n\right )^2}-\frac {c}{(b c-a d)^2 n \left (a+b x^n\right )}-\frac {c d \log \left (a+b x^n\right )}{(b c-a d)^3 n}+\frac {c d \log \left (c+d x^n\right )}{(b c-a d)^3 n}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 97, normalized size = 0.92 \begin {gather*} \frac {\frac {a}{2 b (b c-a d) \left (a+b x^n\right )^2}-\frac {c}{(b c-a d)^2 \left (a+b x^n\right )}-\frac {c d \log \left (a+b x^n\right )}{(b c-a d)^3}+\frac {c d \log \left (c+d x^n\right )}{(b c-a d)^3}}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 100, normalized size = 0.95 \begin {gather*} \frac {-a^2 d-a b c-2 b^2 c x^n}{2 b n (b c-a d)^2 \left (a+b x^n\right )^2}-\frac {c d \log \left (a+b x^n\right )}{n (b c-a d)^3}+\frac {c d \log \left (c+d x^n\right )}{n (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 267, normalized size = 2.54 \begin {gather*} -\frac {a b^{2} c^{2} - a^{3} d^{2} + 2 \, {\left (b^{3} c^{2} - a b^{2} c d\right )} x^{n} + 2 \, {\left (b^{3} c d x^{2 \, n} + 2 \, a b^{2} c d x^{n} + a^{2} b c d\right )} \log \left (b x^{n} + a\right ) - 2 \, {\left (b^{3} c d x^{2 \, n} + 2 \, a b^{2} c d x^{n} + a^{2} b c d\right )} \log \left (d x^{n} + c\right )}{2 \, {\left ({\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} n x^{2 \, n} + 2 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} n x^{n} + {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} n\right )}} \end {gather*}

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]

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

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

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

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maple [A]  time = 0.11, size = 203, normalized size = 1.93 \begin {gather*} \frac {c d \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )}{\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) n}-\frac {c d \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) n}+\frac {-\frac {b c \,{\mathrm e}^{n \ln \relax (x )}}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) n}+\frac {\left (-a b d -b^{2} c \right ) a}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} n}}{\left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )^{2}} \end {gather*}

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]

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

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maxima [B]  time = 0.66, size = 243, normalized size = 2.31 \begin {gather*} -\frac {c d \log \left (\frac {b x^{n} + a}{b}\right )}{b^{3} c^{3} n - 3 \, a b^{2} c^{2} d n + 3 \, a^{2} b c d^{2} n - a^{3} d^{3} n} + \frac {c d \log \left (\frac {d x^{n} + c}{d}\right )}{b^{3} c^{3} n - 3 \, a b^{2} c^{2} d n + 3 \, a^{2} b c d^{2} n - a^{3} d^{3} n} - \frac {2 \, b^{2} c x^{n} + a b c + a^{2} d}{2 \, {\left (a^{2} b^{3} c^{2} n - 2 \, a^{3} b^{2} c d n + a^{4} b d^{2} n + {\left (b^{5} c^{2} n - 2 \, a b^{4} c d n + a^{2} b^{3} d^{2} n\right )} x^{2 \, n} + 2 \, {\left (a b^{4} c^{2} n - 2 \, a^{2} b^{3} c d n + a^{3} b^{2} d^{2} n\right )} x^{n}\right )}} \end {gather*}

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]

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

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

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 [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed

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