3.9.1 \(\int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx\) [801]

Optimal. Leaf size=71 \[ \frac {2 (e x)^{5/2} \sqrt {c+d x^4} F_1\left (\frac {5}{8};1,-\frac {1}{2};\frac {13}{8};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a e \sqrt {1+\frac {d x^4}{c}}} \]

[Out]

2/5*(e*x)^(5/2)*AppellF1(5/8,1,-1/2,13/8,-b*x^4/a,-d*x^4/c)*(d*x^4+c)^(1/2)/a/e/(1+d*x^4/c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {477, 525, 524} \begin {gather*} \frac {2 (e x)^{5/2} \sqrt {c+d x^4} F_1\left (\frac {5}{8};1,-\frac {1}{2};\frac {13}{8};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a e \sqrt {\frac {d x^4}{c}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*x)^(5/2)*Sqrt[c + d*x^4]*AppellF1[5/8, 1, -1/2, 13/8, -((b*x^4)/a), -((d*x^4)/c)])/(5*a*e*Sqrt[1 + (d*x^
4)/c])

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

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)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx &=\frac {2 \text {Subst}\left (\int \frac {x^4 \sqrt {c+\frac {d x^8}{e^4}}}{a+\frac {b x^8}{e^4}} \, dx,x,\sqrt {e x}\right )}{e}\\ &=\frac {\left (2 \sqrt {c+d x^4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt {1+\frac {d x^8}{c e^4}}}{a+\frac {b x^8}{e^4}} \, dx,x,\sqrt {e x}\right )}{e \sqrt {1+\frac {d x^4}{c}}}\\ &=\frac {2 (e x)^{5/2} \sqrt {c+d x^4} F_1\left (\frac {5}{8};1,-\frac {1}{2};\frac {13}{8};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a e \sqrt {1+\frac {d x^4}{c}}}\\ \end {align*}

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Mathematica [A]
time = 10.05, size = 70, normalized size = 0.99 \begin {gather*} \frac {2 x (e x)^{3/2} \sqrt {c+d x^4} F_1\left (\frac {5}{8};-\frac {1}{2},1;\frac {13}{8};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{5 a \sqrt {\frac {c+d x^4}{c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(e*x)^(3/2)*Sqrt[c + d*x^4]*AppellF1[5/8, -1/2, 1, 13/8, -((d*x^4)/c), -((b*x^4)/a)])/(5*a*Sqrt[(c + d*x^
4)/c])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x)

[Out]

int((e*x)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),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)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="maxima")

[Out]

e^(3/2)*integrate(sqrt(d*x^4 + c)*x^(3/2)/(b*x^4 + a), 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)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x\right )^{\frac {3}{2}} \sqrt {c + d x^{4}}}{a + b x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(d*x**4+c)**(1/2)/(b*x**4+a),x)

[Out]

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

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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)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(3/2)*(c + d*x^4)^(1/2))/(a + b*x^4),x)

[Out]

int(((e*x)^(3/2)*(c + d*x^4)^(1/2))/(a + b*x^4), x)

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