\(\int \frac {(1+b x^4)^p}{1-x^2} \, dx\) [186]

Optimal result

Integrand size = 19, antiderivative size = 50 \[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=x \operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},x^4,-b x^4\right )+\frac {1}{3} x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},x^4,-b x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1254, 440, 524} \[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=x \operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},x^4,-b x^4\right )+\frac {1}{3} x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},x^4,-b x^4\right ) \]

[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 440

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 524

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

Rule 1254

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]

Rubi steps \begin{align*} \text {integral}& = \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]

\[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=\int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx \]

[In]

Integrate[(1 + b*x^4)^p/(1 - x^2),x]

[Out]

Integrate[(1 + b*x^4)^p/(1 - x^2), x]

Maple [F]

\[\int \frac {\left (b \,x^{4}+1\right )^{p}}{-x^{2}+1}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1} \,d x } \]

[In]

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

[Out]

integral(-(b*x^4 + 1)^p/(x^2 - 1), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{x^{2} - 1} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{1-x^2} \, dx=-\int \frac {{\left (b\,x^4+1\right )}^p}{x^2-1} \,d x \]

[In]

int(-(b*x^4 + 1)^p/(x^2 - 1),x)

[Out]

-int((b*x^4 + 1)^p/(x^2 - 1), x)