3.186 \(\int \frac{(1+b x^4)^p}{1-x^2} \, dx\)

Optimal. Leaf size=50 \[ \frac{1}{3} x^3 F_1\left (\frac{3}{4};1,-p;\frac{7}{4};x^4,-b x^4\right )+x F_1\left (\frac{1}{4};1,-p;\frac{5}{4};x^4,-b x^4\right ) \]

[Out]

x*AppellF1[1/4, 1, -p, 5/4, x^4, -(b*x^4)] + (x^3*AppellF1[3/4, 1, -p, 7/4, x^4, -(b*x^4)])/3

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Rubi [A]  time = 0.0539169, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1240, 429, 510} \[ \frac{1}{3} x^3 F_1\left (\frac{3}{4};1,-p;\frac{7}{4};x^4,-b x^4\right )+x F_1\left (\frac{1}{4};1,-p;\frac{5}{4};x^4,-b x^4\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + b*x^4)^p/(1 - x^2),x]

[Out]

x*AppellF1[1/4, 1, -p, 5/4, x^4, -(b*x^4)] + (x^3*AppellF1[3/4, 1, -p, 7/4, x^4, -(b*x^4)])/3

Rule 1240

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - (e*x^2)/(d^2 - e^2*x^4))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

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 510

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

Rubi steps

\begin{align*} \int \frac{\left (1+b x^4\right )^p}{1-x^2} \, dx &=\int \left (\frac{\left (1+b x^4\right )^p}{1-x^4}-\frac{x^2 \left (1+b x^4\right )^p}{-1+x^4}\right ) \, dx\\ &=\int \frac{\left (1+b x^4\right )^p}{1-x^4} \, dx-\int \frac{x^2 \left (1+b x^4\right )^p}{-1+x^4} \, dx\\ &=x F_1\left (\frac{1}{4};1,-p;\frac{5}{4};x^4,-b x^4\right )+\frac{1}{3} x^3 F_1\left (\frac{3}{4};1,-p;\frac{7}{4};x^4,-b x^4\right )\\ \end{align*}

Mathematica [F]  time = 0.0804455, size = 0, normalized size = 0. \[ \int \frac{\left (1+b x^4\right )^p}{1-x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(1 + b*x^4)^p/(1 - x^2),x]

[Out]

Integrate[(1 + b*x^4)^p/(1 - x^2), x]

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{4}+1 \right ) ^{p}}{-{x}^{2}+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+1)^p/(-x^2+1),x)

[Out]

int((b*x^4+1)^p/(-x^2+1),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1),x, algorithm="maxima")

[Out]

-integrate((b*x^4 + 1)^p/(x^2 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1),x, algorithm="fricas")

[Out]

integral(-(b*x^4 + 1)^p/(x^2 - 1), x)

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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((b*x**4+1)**p/(-x**2+1),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*x^4 + 1)^p/(x^2 - 1), x)