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

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

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Rubi [A]  time = 0.09, antiderivative size = 90, 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 {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} \end {gather*}

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 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 )^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*}

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Mathematica [A]  time = 0.14, size = 82, normalized size = 0.91 \begin {gather*} \frac {-\frac {c (b c-a d)^2 \log \left (c+d x^n\right )}{d^4}+\frac {x^n (b c-a d)^2}{d^3}-\frac {b x^{2 n} (b c-2 a d)}{2 d^2}+\frac {b^2 x^{3 n}}{3 d}}{n} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.08, size = 97, normalized size = 1.08 \begin {gather*} \frac {x^n \left (6 a^2 d^2-12 a b c d+6 a b d^2 x^n+6 b^2 c^2-3 b^2 c d x^n+2 b^2 d^2 x^{2 n}\right )}{6 d^3 n}-\frac {c (b c-a d)^2 \log \left (c+d x^n\right )}{d^4 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 108, normalized size = 1.20 \begin {gather*} \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 {gather*}

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

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)

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maple [A]  time = 0.08, size = 173, normalized size = 1.92 \begin {gather*} -\frac {a^{2} c \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{2} n}+\frac {a^{2} {\mathrm e}^{n \ln \relax (x )}}{d n}+\frac {2 a b \,c^{2} \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{3} n}-\frac {2 a b c \,{\mathrm e}^{n \ln \relax (x )}}{d^{2} n}+\frac {a b \,{\mathrm e}^{2 n \ln \relax (x )}}{d n}-\frac {b^{2} c^{3} \ln \left (d \,{\mathrm e}^{n \ln \relax (x )}+c \right )}{d^{4} n}+\frac {b^{2} c^{2} {\mathrm e}^{n \ln \relax (x )}}{d^{3} n}-\frac {b^{2} c \,{\mathrm e}^{2 n \ln \relax (x )}}{2 d^{2} n}+\frac {b^{2} {\mathrm e}^{3 n \ln \relax (x )}}{3 d n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.70, size = 150, normalized size = 1.67 \begin {gather*} 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 {gather*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 63.51, size = 202, normalized size = 2.24 \begin {gather*} \begin {cases} \frac {\left (a + b\right )^{2} \log {\relax (x )}}{c} & \text {for}\: d = 0 \wedge n = 0 \\\frac {\frac {a^{2} x^{2 n}}{2 n} + \frac {2 a b x^{3 n}}{3 n} + \frac {b^{2} x^{4 n}}{4 n}}{c} & \text {for}\: d = 0 \\\frac {\left (a + b\right )^{2} \log {\relax (x )}}{c + d} & \text {for}\: n = 0 \\- \frac {a^{2} c \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{2} n} + \frac {a^{2} x^{n}}{d n} + \frac {2 a b c^{2} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{3} n} - \frac {2 a b c x^{n}}{d^{2} n} + \frac {a b x^{2 n}}{d n} - \frac {b^{2} c^{3} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{4} n} + \frac {b^{2} c^{2} x^{n}}{d^{3} n} - \frac {b^{2} c x^{2 n}}{2 d^{2} n} + \frac {b^{2} x^{3 n}}{3 d n} & \text {otherwise} \end {cases} \end {gather*}

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]

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

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