3.188 \(\int \frac {(1+b x^4)^p}{(1-x^2)^3} \, dx\)

Optimal. Leaf size=101 \[ x F_1\left (\frac {1}{4};3,-p;\frac {5}{4};x^4,-b x^4\right )+\frac {1}{7} x^7 F_1\left (\frac {7}{4};3,-p;\frac {11}{4};x^4,-b x^4\right )+\frac {3}{5} x^5 F_1\left (\frac {5}{4};3,-p;\frac {9}{4};x^4,-b x^4\right )+x^3 F_1\left (\frac {3}{4};3,-p;\frac {7}{4};x^4,-b x^4\right ) \]

[Out]

x*AppellF1(1/4,3,-p,5/4,x^4,-b*x^4)+x^3*AppellF1(3/4,3,-p,7/4,x^4,-b*x^4)+3/5*x^5*AppellF1(5/4,3,-p,9/4,x^4,-b
*x^4)+1/7*x^7*AppellF1(7/4,3,-p,11/4,x^4,-b*x^4)

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Rubi [A]  time = 0.11, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1240, 429, 510} \[ \frac {1}{7} x^7 F_1\left (\frac {7}{4};3,-p;\frac {11}{4};x^4,-b x^4\right )+\frac {3}{5} x^5 F_1\left (\frac {5}{4};3,-p;\frac {9}{4};x^4,-b x^4\right )+x^3 F_1\left (\frac {3}{4};3,-p;\frac {7}{4};x^4,-b x^4\right )+x F_1\left (\frac {1}{4};3,-p;\frac {5}{4};x^4,-b x^4\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + b*x^4)^p/(1 - x^2)^3,x]

[Out]

x*AppellF1[1/4, 3, -p, 5/4, x^4, -(b*x^4)] + x^3*AppellF1[3/4, 3, -p, 7/4, x^4, -(b*x^4)] + (3*x^5*AppellF1[5/
4, 3, -p, 9/4, x^4, -(b*x^4)])/5 + (x^7*AppellF1[7/4, 3, -p, 11/4, x^4, -(b*x^4)])/7

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])

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]

Rubi steps

\begin {align*} \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx &=\int \left (-\frac {\left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {3 x^2 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {3 x^4 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {x^6 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}\right ) \, dx\\ &=-\left (3 \int \frac {x^2 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx\right )-3 \int \frac {x^4 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx-\int \frac {\left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx-\int \frac {x^6 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx\\ &=x F_1\left (\frac {1}{4};3,-p;\frac {5}{4};x^4,-b x^4\right )+x^3 F_1\left (\frac {3}{4};3,-p;\frac {7}{4};x^4,-b x^4\right )+\frac {3}{5} x^5 F_1\left (\frac {5}{4};3,-p;\frac {9}{4};x^4,-b x^4\right )+\frac {1}{7} x^7 F_1\left (\frac {7}{4};3,-p;\frac {11}{4};x^4,-b x^4\right )\\ \end {align*}

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Mathematica [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(1 + b*x^4)^p/(1 - x^2)^3,x]

[Out]

Integrate[(1 + b*x^4)^p/(1 - x^2)^3, x]

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b x^{4} + 1\right )}^{p}}{x^{6} - 3 \, x^{4} + 3 \, x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1)^3,x, algorithm="fricas")

[Out]

integral(-(b*x^4 + 1)^p/(x^6 - 3*x^4 + 3*x^2 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1)^3,x, algorithm="giac")

[Out]

integrate(-(b*x^4 + 1)^p/(x^2 - 1)^3, x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+1\right )^{p}}{\left (-x^{2}+1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+1)^p/(-x^2+1)^3,x)

[Out]

int((b*x^4+1)^p/(-x^2+1)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+1)^p/(-x^2+1)^3,x, algorithm="maxima")

[Out]

-integrate((b*x^4 + 1)^p/(x^2 - 1)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {{\left (b\,x^4+1\right )}^p}{{\left (x^2-1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b*x^4 + 1)^p/(x^2 - 1)^3,x)

[Out]

int(-(b*x^4 + 1)^p/(x^2 - 1)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+1)**p/(-x**2+1)**3,x)

[Out]

Timed out

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