3.2.81 \(\int \frac {(a+b x^4)^p}{(c+e x^2)^2} \, dx\) [181]

Optimal. Leaf size=189 \[ \frac {x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} F_1\left (\frac {1}{4};-p,2;\frac {5}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{c^2}-\frac {2 e x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} F_1\left (\frac {3}{4};-p,2;\frac {7}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{3 c^3}+\frac {e^2 x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} F_1\left (\frac {5}{4};-p,2;\frac {9}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{5 c^4} \]

[Out]

x*(b*x^4+a)^p*AppellF1(1/4,2,-p,5/4,e^2*x^4/c^2,-b*x^4/a)/c^2/((1+b*x^4/a)^p)-2/3*e*x^3*(b*x^4+a)^p*AppellF1(3
/4,2,-p,7/4,e^2*x^4/c^2,-b*x^4/a)/c^3/((1+b*x^4/a)^p)+1/5*e^2*x^5*(b*x^4+a)^p*AppellF1(5/4,2,-p,9/4,e^2*x^4/c^
2,-b*x^4/a)/c^4/((1+b*x^4/a)^p)

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Rubi [A]
time = 0.12, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1254, 441, 440, 525, 524} \begin {gather*} \frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {1}{4};-p,2;\frac {5}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{c^2}+\frac {e^2 x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {5}{4};-p,2;\frac {9}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{5 c^4}-\frac {2 e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {3}{4};-p,2;\frac {7}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{3 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^p/(c + e*x^2)^2,x]

[Out]

(x*(a + b*x^4)^p*AppellF1[1/4, -p, 2, 5/4, -((b*x^4)/a), (e^2*x^4)/c^2])/(c^2*(1 + (b*x^4)/a)^p) - (2*e*x^3*(a
+ b*x^4)^p*AppellF1[3/4, -p, 2, 7/4, -((b*x^4)/a), (e^2*x^4)/c^2])/(3*c^3*(1 + (b*x^4)/a)^p) + (e^2*x^5*(a +
b*x^4)^p*AppellF1[5/4, -p, 2, 9/4, -((b*x^4)/a), (e^2*x^4)/c^2])/(5*c^4*(1 + (b*x^4)/a)^p)

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 441

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 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

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

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Mathematica [F]
time = 0.58, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^4\right )^p}{\left (c+e x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a + b*x^4)^p/(c + e*x^2)^2,x]

[Out]

Integrate[(a + b*x^4)^p/(c + e*x^2)^2, x]

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{p}}{\left (e \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^p/(e*x^2+c)^2,x)

[Out]

int((b*x^4+a)^p/(e*x^2+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^p/(e*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^p/(x^2*e + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^p/(e*x^2+c)^2,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^p/(x^4*e^2 + 2*c*x^2*e + c^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**p/(e*x**2+c)**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^p/(e*x^2+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^p/(x^2*e + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^4+a\right )}^p}{{\left (e\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^p/(c + e*x^2)^2,x)

[Out]

int((a + b*x^4)^p/(c + e*x^2)^2, x)

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