3.11.49 \(\int \frac {x^{-1+3 n}}{(a+b x^n)^2 (c+d x^n)} \, dx\) [1049]

Optimal. Leaf size=95 \[ -\frac {a^2}{b^2 (b c-a d) n \left (a+b x^n\right )}-\frac {a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2 n}+\frac {c^2 \log \left (c+d x^n\right )}{d (b c-a d)^2 n} \]

[Out]

-a^2/b^2/(-a*d+b*c)/n/(a+b*x^n)-a*(-a*d+2*b*c)*ln(a+b*x^n)/b^2/(-a*d+b*c)^2/n+c^2*ln(c+d*x^n)/d/(-a*d+b*c)^2/n

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Rubi [A]
time = 0.06, antiderivative size = 95, 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 = {457, 90} \begin {gather*} -\frac {a^2}{b^2 n (b c-a d) \left (a+b x^n\right )}-\frac {a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 n (b c-a d)^2}+\frac {c^2 \log \left (c+d x^n\right )}{d n (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 90

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 457

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

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Mathematica [A]
time = 0.10, size = 93, normalized size = 0.98 \begin {gather*} -\frac {a^2}{b^2 (b c-a d) n \left (a+b x^n\right )}+\frac {a (-2 b c+a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2 n}+\frac {c^2 \log \left (c+d x^n\right )}{d (-b c+a d)^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.40, size = 125, normalized size = 1.32

method	result	size
norman	\(\frac {a^{2}}{\left (a d -b c \right ) b^{2} n \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}+\frac {c^{2} \ln \left (c +d \,{\mathrm e}^{n \ln \left (x \right )}\right )}{d n \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a \left (a d -2 b c \right ) \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} n}\)	\(125\)
risch	\(\frac {\ln \left (x \right )}{b^{2} d}-\frac {\ln \left (x \right ) c^{2}}{d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\ln \left (x \right ) a^{2} d}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2}}+\frac {2 \ln \left (x \right ) a c}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b}+\frac {a^{2}}{\left (a d -b c \right ) b^{2} n \left (a +b \,x^{n}\right )}+\frac {c^{2} \ln \left (x^{n}+\frac {c}{d}\right )}{d n \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a^{2} \ln \left (x^{n}+\frac {a}{b}\right ) d}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} n}-\frac {2 a \ln \left (x^{n}+\frac {a}{b}\right ) c}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b n}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 147, normalized size = 1.55 \begin {gather*} \frac {c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{b^{2} c^{2} d n - 2 \, a b c d^{2} n + a^{2} d^{3} n} - \frac {a^{2}}{a b^{3} c n - a^{2} b^{2} d n + {\left (b^{4} c n - a b^{3} d n\right )} x^{n}} - \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{4} c^{2} n - 2 \, a b^{3} c d n + a^{2} b^{2} d^{2} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.75, size = 166, normalized size = 1.75 \begin {gather*} -\frac {a^{2} b c d - a^{3} d^{2} + {\left (2 \, a^{2} b c d - a^{3} d^{2} + {\left (2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{n}\right )} \log \left (b x^{n} + a\right ) - {\left (b^{3} c^{2} x^{n} + a b^{2} c^{2}\right )} \log \left (d x^{n} + c\right )}{{\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} n x^{n} + {\left (a b^{4} c^{2} d - 2 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3}\right )} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^2\,\left (c+d\,x^n\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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