3.845 \(\int \frac {(e x)^m}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=81 \[ \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} F_1\left (\frac {m+1}{4};2,\frac {1}{2};\frac {m+5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 e (m+1) \sqrt {c+d x^4}} \]

[Out]

(e*x)^(1+m)*AppellF1(1/4+1/4*m,2,1/2,5/4+1/4*m,-b*x^4/a,-d*x^4/c)*(1+d*x^4/c)^(1/2)/a^2/e/(1+m)/(d*x^4+c)^(1/2
)

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {511, 510} \[ \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} F_1\left (\frac {m+1}{4};2,\frac {1}{2};\frac {m+5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 e (m+1) \sqrt {c+d x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

((e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*AppellF1[(1 + m)/4, 2, 1/2, (5 + m)/4, -((b*x^4)/a), -((d*x^4)/c)])/(a^2*e*
(1 + m)*Sqrt[c + d*x^4])

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])

Rubi steps

\begin {align*} \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {\sqrt {1+\frac {d x^4}{c}} \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \sqrt {1+\frac {d x^4}{c}}} \, dx}{\sqrt {c+d x^4}}\\ &=\frac {(e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} F_1\left (\frac {1+m}{4};2,\frac {1}{2};\frac {5+m}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 e (1+m) \sqrt {c+d x^4}}\\ \end {align*}

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Mathematica [B]  time = 0.23, size = 179, normalized size = 2.21 \[ \frac {x \sqrt {c+d x^4} (e x)^m \left (a^2 d^2 \, _2F_1\left (\frac {1}{2},\frac {m+1}{4};\frac {m+5}{4};-\frac {d x^4}{c}\right )-a b c d F_1\left (\frac {m+1}{4};-\frac {1}{2},1;\frac {m+5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+b c (b c-a d) F_1\left (\frac {m+1}{4};2,-\frac {1}{2};\frac {m+5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}{a^2 c (m+1) \sqrt {\frac {d x^4}{c}+1} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[c + d*x^4]*(-(a*b*c*d*AppellF1[(1 + m)/4, -1/2, 1, (5 + m)/4, -((d*x^4)/c), -((b*x^4)/a)]) + b
*c*(b*c - a*d)*AppellF1[(1 + m)/4, 2, -1/2, (5 + m)/4, -((b*x^4)/a), -((d*x^4)/c)] + a^2*d^2*Hypergeometric2F1
[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)]))/(a^2*c*(b*c - a*d)^2*(1 + m)*Sqrt[1 + (d*x^4)/c])

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fricas [F]  time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x^{4} + c} \left (e x\right )^{m}}{b^{2} d x^{12} + {\left (b^{2} c + 2 \, a b d\right )} x^{8} + {\left (2 \, a b c + a^{2} d\right )} x^{4} + a^{2} c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x^4 + c)*(e*x)^m/(b^2*d*x^12 + (b^2*c + 2*a*b*d)*x^8 + (2*a*b*c + a^2*d)*x^4 + a^2*c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x)^m/((b*x^4 + a)^2*sqrt(d*x^4 + c)), x)

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maple [F]  time = 0.55, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{m}}{\left (b \,x^{4}+a \right )^{2} \sqrt {d \,x^{4}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x)^m/((b*x^4 + a)^2*sqrt(d*x^4 + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(1/2)), 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((e*x)**m/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Timed out

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