3.181 \(\int \frac {(a+b x^4)^p}{(c+e x^2)^2} \, dx\)

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

[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 = 0.19, 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 = {1240, 430, 429, 511, 510} \[ \frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {1}{4};-p,2;\frac {5}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{c^2}-\frac {2 e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {3}{4};-p,2;\frac {7}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{3 c^3}+\frac {e^2 x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} F_1\left (\frac {5}{4};-p,2;\frac {9}{4};-\frac {b x^4}{a},\frac {e^2 x^4}{c^2}\right )}{5 c^4} \]

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 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 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 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 (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 = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b x^4\right )^p}{\left (c+e x^2\right )^2} \, dx \]

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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{p}}{e^{2} x^{4} + 2 \, c e x^{2} + c^{2}}, x\right ) \]

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/(e^2*x^4 + 2*c*e*x^2 + c^2), x)

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

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/(e*x^2 + c)^2, x)

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maple [F]  time = 0.09, 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 \[ \int \frac {{\left (b x^{4} + a\right )}^{p}}{{\left (e x^{2} + c\right )}^{2}}\,{d x} \]

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/(e*x^2 + c)^2, x)

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

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

[Out]

Timed out

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