3.1040 \(\int \frac{x^{-1+2 n} (a+b x^n)^2}{c+d x^n} \, dx\)

Optimal. Leaf size=90 \[ \frac{x^n (b c-a d)^2}{d^3 n}-\frac{b x^{2 n} (b c-2 a d)}{2 d^2 n}-\frac{c (b c-a d)^2 \log \left (c+d x^n\right )}{d^4 n}+\frac{b^2 x^{3 n}}{3 d n} \]

[Out]

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

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Rubi [A]  time = 0.0893512, antiderivative size = 90, normalized size of antiderivative = 1., 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} \[ \frac{x^n (b c-a d)^2}{d^3 n}-\frac{b x^{2 n} (b c-2 a d)}{2 d^2 n}-\frac{c (b c-a d)^2 \log \left (c+d x^n\right )}{d^4 n}+\frac{b^2 x^{3 n}}{3 d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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]))))

Rubi steps

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

Mathematica [A]  time = 0.11322, size = 82, normalized size = 0.91 \[ \frac{\frac{x^n (b c-a d)^2}{d^3}-\frac{b x^{2 n} (b c-2 a d)}{2 d^2}-\frac{c (b c-a d)^2 \log \left (c+d x^n\right )}{d^4}+\frac{b^2 x^{3 n}}{3 d}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 173, normalized size = 1.9 \begin{align*}{\frac{{{\rm e}^{n\ln \left ( x \right ) }}{a}^{2}}{dn}}-2\,{\frac{{{\rm e}^{n\ln \left ( x \right ) }}abc}{{d}^{2}n}}+{\frac{{{\rm e}^{n\ln \left ( x \right ) }}{b}^{2}{c}^{2}}{{d}^{3}n}}+{\frac{{b}^{2} \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{3}}{3\,dn}}+{\frac{b \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}a}{dn}}-{\frac{{b}^{2} \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}c}{2\,{d}^{2}n}}-{\frac{c\ln \left ( c+d{{\rm e}^{n\ln \left ( x \right ) }} \right ){a}^{2}}{{d}^{2}n}}+2\,{\frac{{c}^{2}\ln \left ( c+d{{\rm e}^{n\ln \left ( x \right ) }} \right ) ab}{{d}^{3}n}}-{\frac{{c}^{3}\ln \left ( c+d{{\rm e}^{n\ln \left ( x \right ) }} \right ){b}^{2}}{{d}^{4}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.950214, size = 203, normalized size = 2.26 \begin{align*} a^{2}{\left (\frac{x^{n}}{d n} - \frac{c \log \left (\frac{d x^{n} + c}{d}\right )}{d^{2} n}\right )} - \frac{1}{6} \, 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 )} + a 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(x^n/(d*n) - c*log((d*x^n + c)/d)/(d^2*n)) - 1/6*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)) + a*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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Fricas [A]  time = 1.11146, size = 227, normalized size = 2.52 \begin{align*} \frac{2 \, b^{2} d^{3} x^{3 \, n} - 3 \,{\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{2 \, n} + 6 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{n} - 6 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \log \left (d x^{n} + c\right )}{6 \, d^{4} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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