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

Optimal. Leaf size=114 \[ \frac{d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt{a+c x^4}}+\frac{x (d+e x)}{2 a \sqrt{a+c x^4}} \]

[Out]

(x*(d + e*x))/(2*a*Sqrt[a + c*x^4]) + (d*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E
llipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(1/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.0474485, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1855, 12, 220} \[ \frac{d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt{a+c x^4}}+\frac{x (d+e x)}{2 a \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a + c*x^4)^(3/2),x]

[Out]

(x*(d + e*x))/(2*a*Sqrt[a + c*x^4]) + (d*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E
llipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(1/4)*Sqrt[a + c*x^4])

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.0299536, size = 59, normalized size = 0.52 \[ \frac{x \left (d \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right )+d+e x\right )}{2 a \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.004, size = 115, normalized size = 1. \begin{align*}{\frac{e{x}^{2}}{2\,a}{\frac{1}{\sqrt{c{x}^{4}+a}}}}+d \left ({\frac{x}{2\,a}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{c}} \right ) c}}}}+{\frac{1}{2\,a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e/(c*x^4+a)^(1/2)/a*x^2+d*(1/2*x/a/((x^4+a/c)*c)^(1/2)+1/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2
)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x + d}{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + a}{\left (e x + d\right )}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 7.74928, size = 61, normalized size = 0.54 \begin{align*} \frac{d x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{e x^{2}}{2 a^{\frac{3}{2}} \sqrt{1 + \frac{c x^{4}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + e*x**2/(2*a**(3/2
)*sqrt(1 + c*x**4/a))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x + d}{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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