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

Optimal. Leaf size=158 \[ \frac{a^{3/4} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{c} \sqrt{a+c x^4}}+\frac{1}{3} d x \sqrt{a+c x^4}+\frac{1}{4} e x^2 \sqrt{a+c x^4}+\frac{a e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{4 \sqrt{c}} \]

[Out]

(d*x*Sqrt[a + c*x^4])/3 + (e*x^2*Sqrt[a + c*x^4])/4 + (a*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(4*Sqrt[c])
+ (a^(3/4)*d*(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])/(3*c^(1/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.0890388, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {1885, 195, 220, 275, 217, 206} \[ \frac{a^{3/4} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{c} \sqrt{a+c x^4}}+\frac{1}{3} d x \sqrt{a+c x^4}+\frac{1}{4} e x^2 \sqrt{a+c x^4}+\frac{a e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Sqrt[a + c*x^4],x]

[Out]

(d*x*Sqrt[a + c*x^4])/3 + (e*x^2*Sqrt[a + c*x^4])/4 + (a*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(4*Sqrt[c])
+ (a^(3/4)*d*(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])/(3*c^(1/4)*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [C]  time = 0.0628603, size = 109, normalized size = 0.69 \[ \frac{\sqrt{a+c x^4} \left (4 \sqrt{c} d x \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^4}{a}\right )+\sqrt{c} e x^2 \sqrt{\frac{c x^4}{a}+1}+\sqrt{a} e \sinh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )\right )}{4 \sqrt{c} \sqrt{\frac{c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.004, size = 127, normalized size = 0.8 \begin{align*}{\frac{e{x}^{2}}{4}\sqrt{c{x}^{4}+a}}+{\frac{ae}{4}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{dx}{3}\sqrt{c{x}^{4}+a}}+{\frac{2\,ad}{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)*(c*x^4+a)^(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 2.88789, size = 88, normalized size = 0.56 \begin{align*} \frac{\sqrt{a} d x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} e x^{2} \sqrt{1 + \frac{c x^{4}}{a}}}{4} + \frac{a e \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{4 \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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