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

Optimal. Leaf size=76 \[ \frac {x \left (c+d x^{2 n}\right )^p \left (\frac {d x^{2 n}}{c}+1\right )^{-p} F_1\left (\frac {1}{2 n};1,-p;\frac {1}{2} \left (2+\frac {1}{n}\right );\frac {b^2 x^{2 n}}{a^2},-\frac {d x^{2 n}}{c}\right )}{a^2} \]

[Out]

x*(c+d*x^(2*n))^p*AppellF1(1/2/n,1,-p,1+1/2/n,b^2*x^(2*n)/a^2,-d*x^(2*n)/c)/a^2/((1+d*x^(2*n)/c)^p)

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {517, 430, 429} \[ \frac {x \left (c+d x^{2 n}\right )^p \left (\frac {d x^{2 n}}{c}+1\right )^{-p} F_1\left (\frac {1}{2 n};1,-p;\frac {1}{2} \left (2+\frac {1}{n}\right );\frac {b^2 x^{2 n}}{a^2},-\frac {d x^{2 n}}{c}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c + d*x^(2*n))^p*AppellF1[1/(2*n), 1, -p, (2 + n^(-1))/2, (b^2*x^(2*n))/a^2, -((d*x^(2*n))/c)])/(a^2*(1 +
(d*x^(2*n))/c)^p)

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 517

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rubi steps

\begin {align*} \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx &=\int \frac {\left (c+d x^{2 n}\right )^p}{a^2-b^2 x^{2 n}} \, dx\\ &=\left (\left (c+d x^{2 n}\right )^p \left (1+\frac {d x^{2 n}}{c}\right )^{-p}\right ) \int \frac {\left (1+\frac {d x^{2 n}}{c}\right )^p}{a^2-b^2 x^{2 n}} \, dx\\ &=\frac {x \left (c+d x^{2 n}\right )^p \left (1+\frac {d x^{2 n}}{c}\right )^{-p} F_1\left (\frac {1}{2 n};1,-p;\frac {1}{2} \left (2+\frac {1}{n}\right );\frac {b^2 x^{2 n}}{a^2},-\frac {d x^{2 n}}{c}\right )}{a^2}\\ \end {align*}

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Mathematica [B]  time = 0.43, size = 258, normalized size = 3.39 \[ \frac {a^2 c (2 n+1) x \left (c+d x^{2 n}\right )^p F_1\left (\frac {1}{2 n};-p,1;1+\frac {1}{2 n};-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )}{\left (a^2-b^2 x^{2 n}\right ) \left (2 a^2 d n p x^{2 n} F_1\left (1+\frac {1}{2 n};1-p,1;2+\frac {1}{2 n};-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )+2 b^2 c n x^{2 n} F_1\left (1+\frac {1}{2 n};-p,2;2+\frac {1}{2 n};-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )+a^2 c (2 n+1) F_1\left (\frac {1}{2 n};-p,1;1+\frac {1}{2 n};-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*c*(1 + 2*n)*x*(c + d*x^(2*n))^p*AppellF1[1/(2*n), -p, 1, 1 + 1/(2*n), -((d*x^(2*n))/c), (b^2*x^(2*n))/a^2
])/((a^2 - b^2*x^(2*n))*(2*a^2*d*n*p*x^(2*n)*AppellF1[1 + 1/(2*n), 1 - p, 1, 2 + 1/(2*n), -((d*x^(2*n))/c), (b
^2*x^(2*n))/a^2] + 2*b^2*c*n*x^(2*n)*AppellF1[1 + 1/(2*n), -p, 2, 2 + 1/(2*n), -((d*x^(2*n))/c), (b^2*x^(2*n))
/a^2] + a^2*c*(1 + 2*n)*AppellF1[1/(2*n), -p, 1, 1 + 1/(2*n), -((d*x^(2*n))/c), (b^2*x^(2*n))/a^2]))

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (d x^{2 \, n} + c\right )}^{p}}{b^{2} x^{2 \, n} - a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{2 n}+c \right )^{p}}{\left (-b \,x^{n}+a \right ) \left (b \,x^{n}+a \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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