3.212 \(\int \frac{(d+e x)^3}{\sqrt{a+c x^4}} \, dx\)

Optimal. Leaf size=295 \[ \frac{d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+\sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt{a+c x^4}}-\frac{3 \sqrt [4]{a} d e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{a+c x^4}}+\frac{3 d^2 e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{3 d e^2 x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{e^3 \sqrt{a+c x^4}}{2 c} \]

[Out]

(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d*e^2*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (3*d^2*e*ArcTanh
[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (3*a^(1/4)*d*e^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt
[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[a + c*x^4]) + (d*(Sqrt[c]*d
^2 + 3*Sqrt[a]*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^
(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(3/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.160057, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {1885, 1198, 220, 1196, 1248, 641, 217, 206} \[ \frac{d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+\sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt{a+c x^4}}-\frac{3 \sqrt [4]{a} d e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{a+c x^4}}+\frac{3 d^2 e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{3 d e^2 x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{e^3 \sqrt{a+c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/Sqrt[a + c*x^4],x]

[Out]

(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d*e^2*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (3*d^2*e*ArcTanh
[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (3*a^(1/4)*d*e^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt
[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[a + c*x^4]) + (d*(Sqrt[c]*d
^2 + 3*Sqrt[a]*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^
(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(3/4)*Sqrt[a + c*x^4])

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.132394, size = 157, normalized size = 0.53 \[ \frac{3 d^2 e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{d^3 x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right )}{\sqrt{a+c x^4}}+\frac{d e^2 x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^4}{a}\right )}{\sqrt{a+c x^4}}+\frac{e^3 \sqrt{a+c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) + (d^3*x*Sqrt[1 + (
c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)])/Sqrt[a + c*x^4] + (d*e^2*x^3*Sqrt[1 + (c*x^4)/a]*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)])/Sqrt[a + c*x^4]

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Maple [C]  time = 0.014, size = 218, normalized size = 0.7 \begin{align*}{\frac{{e}^{3}}{2\,c}\sqrt{c{x}^{4}+a}}+{3\,id{e}^{2}\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}}+{\frac{3\,e{d}^{2}}{2}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){\frac{1}{\sqrt{c}}}}+{{d}^{3}\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^3*(c*x^4+a)^(1/2)/c+3*I*d*e^2*a^(1/2)/(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)/c^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/
2)*c^(1/2))^(1/2),I))+3/2*e*d^2*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))/c^(1/2)+d^3/(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{{\left (e x + d\right )}^{3}}{\sqrt{c x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.19374, size = 141, normalized size = 0.48 \begin{align*} e^{3} \left (\begin{cases} \frac{x^{4}}{4 \sqrt{a}} & \text{for}\: c = 0 \\\frac{\sqrt{a + c x^{4}}}{2 c} & \text{otherwise} \end{cases}\right ) + \frac{3 d^{2} e \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{c}} + \frac{d^{3} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} + \frac{3 d e^{2} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**3*Piecewise((x**4/(4*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**4)/(2*c), True)) + 3*d**2*e*asinh(sqrt(c)*x**2/sqr
t(a))/(2*sqrt(c)) + d**3*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4
)) + 3*d*e**2*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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