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

Optimal. Leaf size=115 \[ \frac {x (b c-a d) (a d (1-n)-b c (n+1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 d^2 n}-\frac {b x (a d-b c (n+1))}{c d^2 n}-\frac {x (b c-a d) \left (a+b x^n\right )}{c d n \left (c+d x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {413, 388, 245} \[ \frac {x (b c-a d) (a d (1-n)-b c (n+1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 d^2 n}-\frac {b x (a d-b c (n+1))}{c d^2 n}-\frac {x (b c-a d) \left (a+b x^n\right )}{c d n \left (c+d x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^2/(c + d*x^n)^2,x]

[Out]

-((b*(a*d - b*c*(1 + n))*x)/(c*d^2*n)) - ((b*c - a*d)*x*(a + b*x^n))/(c*d*n*(c + d*x^n)) + ((b*c - a*d)*(a*d*(
1 - n) - b*c*(1 + n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/(c^2*d^2*n)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^2} \, dx &=-\frac {(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac {\int \frac {a (b c-a d (1-n))-b (a d-b c (1+n)) x^n}{c+d x^n} \, dx}{c d n}\\ &=-\frac {b (a d-b c (1+n)) x}{c d^2 n}-\frac {(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac {((b c-a d) (a d (1-n)-b c (1+n))) \int \frac {1}{c+d x^n} \, dx}{c d^2 n}\\ &=-\frac {b (a d-b c (1+n)) x}{c d^2 n}-\frac {(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac {(b c-a d) (a d (1-n)-b c (1+n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 d^2 n}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 95, normalized size = 0.83 \[ \frac {x \left (\frac {c \left (a^2 d^2-2 a b c d+b^2 c \left (c n+c+d n x^n\right )\right )}{c+d x^n}-(b c-a d) (a d (n-1)+b c (n+1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )\right )}{c^2 d^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*((c*(-2*a*b*c*d + a^2*d^2 + b^2*c*(c + c*n + d*n*x^n)))/(c + d*x^n) - (b*c - a*d)*(a*d*(-1 + n) + b*c*(1 +
n))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)]))/(c^2*d^2*n)

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}{d^{2} x^{2 \, n} + 2 \, c d x^{n} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{n}+a \right )^{2}}{\left (d \,x^{n}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^2/(c + d*x^n)^2,x)

[Out]

int((a + b*x^n)^2/(c + d*x^n)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{n}\right )^{2}}{\left (c + d x^{n}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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