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

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

[Out]

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

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Rubi [A]  time = 0.06, 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 = {511, 510} \[ \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} F_1\left (\frac {m+1}{4};1,\frac {3}{2};\frac {m+5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a c e (m+1) \sqrt {c+d x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*AppellF1[(1 + m)/4, 1, 3/2, (5 + m)/4, -((b*x^4)/a), -((d*x^4)/c)])/(a*c*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 ) \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 ) \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};1,\frac {3}{2};\frac {5+m}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a c e (1+m) \sqrt {c+d x^4}}\\ \end {align*}

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Mathematica [B]  time = 0.21, size = 169, normalized size = 2.01 \[ \frac {x \sqrt {c+d x^4} (e x)^m \left (b^2 c^2 F_1\left (\frac {m+1}{4};-\frac {1}{2},1;\frac {m+5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+a d \left ((a d-b c) \, _2F_1\left (\frac {3}{2},\frac {m+1}{4};\frac {m+5}{4};-\frac {d x^4}{c}\right )-b c \, _2F_1\left (\frac {1}{2},\frac {m+1}{4};\frac {m+5}{4};-\frac {d x^4}{c}\right )\right )\right )}{a c^2 (m+1) \sqrt {\frac {d x^4}{c}+1} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[c + d*x^4]*(b^2*c^2*AppellF1[(1 + m)/4, -1/2, 1, (5 + m)/4, -((d*x^4)/c), -((b*x^4)/a)] + a*d*
(-(b*c*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)]) + (-(b*c) + a*d)*Hypergeometric2F1[3/2, (1
+ m)/4, (5 + m)/4, -((d*x^4)/c)])))/(a*c^2*(b*c - a*d)^2*(1 + m)*Sqrt[1 + (d*x^4)/c])

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x^{4} + c} \left (e x\right )^{m}}{b d^{2} x^{12} + {\left (2 \, b c d + a d^{2}\right )} x^{8} + {\left (b c^{2} + 2 \, a c d\right )} x^{4} + a c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.66, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{m}}{\left (b \,x^{4}+a \right ) \left (d \,x^{4}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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