3.9.51 \(\int \frac {(e x)^m}{(a+b x^4)^2 (c+d x^4)^{3/2}} \, dx\) [851]

Optimal. Leaf size=84 \[ \frac {(e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} F_1\left (\frac {1+m}{4};2,\frac {3}{2};\frac {5+m}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 c e (1+m) \sqrt {c+d x^4}} \]

[Out]

(e*x)^(1+m)*AppellF1(1/4+1/4*m,2,3/2,5/4+1/4*m,-b*x^4/a,-d*x^4/c)*(1+d*x^4/c)^(1/2)/a^2/c/e/(1+m)/(d*x^4+c)^(1
/2)

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Rubi [A]
time = 0.04, antiderivative size = 84, 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 = {525, 524} \begin {gather*} \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} F_1\left (\frac {m+1}{4};2,\frac {3}{2};\frac {m+5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 c e (m+1) \sqrt {c+d x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x]

[Out]

((e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*AppellF1[(1 + m)/4, 2, 3/2, (5 + m)/4, -((b*x^4)/a), -((d*x^4)/c)])/(a^2*c*
e*(1 + m)*Sqrt[c + d*x^4])

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])

Rubi steps

\begin {align*} \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx &=\frac {\sqrt {1+\frac {d x^4}{c}} \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (1+\frac {d x^4}{c}\right )^{3/2}} \, dx}{c \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 {3}{2};\frac {5+m}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 c e (1+m) \sqrt {c+d x^4}}\\ \end {align*}

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Mathematica [A]
time = 10.08, size = 77, normalized size = 0.92 \begin {gather*} \frac {x (e x)^m \left (1+\frac {d x^4}{c}\right )^{3/2} F_1\left (\frac {1+m}{4};2,\frac {3}{2};\frac {5+m}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 (1+m) \left (c+d x^4\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x]

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m}}{\left (b \,x^{4}+a \right )^{2} \left (d \,x^{4}+c \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x)

[Out]

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),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((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((x*e)^m/((b*x^4 + a)^2*(d*x^4 + c)^(3/2)), 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((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x^4 + c)*(x*e)^m/(b^2*d^2*x^16 + 2*(b^2*c*d + a*b*d^2)*x^12 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*
x^8 + 2*(a*b*c^2 + a^2*c*d)*x^4 + a^2*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((e*x)**m/(b*x**4+a)**2/(d*x**4+c)**(3/2),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((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="giac")

[Out]

integrate((x*e)^m/((b*x^4 + a)^2*(d*x^4 + c)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^m}{{\left (b\,x^4+a\right )}^2\,{\left (d\,x^4+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x)

[Out]

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)), x)

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