3.207 \(\int (d+e x)^2 \sqrt{a+c x^4} \, dx\)

Optimal. Leaf size=326 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+5 \sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{a+c x^4}}-\frac{2 a^{5/4} e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{1}{15} x \sqrt{a+c x^4} \left (5 d^2+3 e^2 x^2\right )+\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{a d e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

[Out]

(d*e*x^2*Sqrt[a + c*x^4])/2 + (2*a*e^2*x*Sqrt[a + c*x^4])/(5*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (x*(5*d^2 + 3*
e^2*x^2)*Sqrt[a + c*x^4])/15 + (a*d*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (2*a^(5/4)*e^2*(Sq
rt[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]
)/(5*c^(3/4)*Sqrt[a + c*x^4]) + (a^(3/4)*(5*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])/(15*c^(3/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.189014, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {1885, 275, 195, 217, 206, 1177, 1198, 220, 1196} \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+5 \sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{a+c x^4}}-\frac{2 a^{5/4} e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{1}{15} x \sqrt{a+c x^4} \left (5 d^2+3 e^2 x^2\right )+\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{a d e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Sqrt[a + c*x^4],x]

[Out]

(d*e*x^2*Sqrt[a + c*x^4])/2 + (2*a*e^2*x*Sqrt[a + c*x^4])/(5*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (x*(5*d^2 + 3*
e^2*x^2)*Sqrt[a + c*x^4])/15 + (a*d*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (2*a^(5/4)*e^2*(Sq
rt[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]
)/(5*c^(3/4)*Sqrt[a + c*x^4]) + (a^(3/4)*(5*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])/(15*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 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 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 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])

Rule 1177

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*(a
+ c*x^4)^p)/((4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/((4*p + 1)*(4*p + 3)), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

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]

Rubi steps

\begin{align*} \int (d+e x)^2 \sqrt{a+c x^4} \, dx &=\int \left (2 d e x \sqrt{a+c x^4}+\left (d^2+e^2 x^2\right ) \sqrt{a+c x^4}\right ) \, dx\\ &=(2 d e) \int x \sqrt{a+c x^4} \, dx+\int \left (d^2+e^2 x^2\right ) \sqrt{a+c x^4} \, dx\\ &=\frac{1}{15} x \left (5 d^2+3 e^2 x^2\right ) \sqrt{a+c x^4}+\frac{1}{15} \int \frac{10 a d^2+6 a e^2 x^2}{\sqrt{a+c x^4}} \, dx+(d e) \operatorname{Subst}\left (\int \sqrt{a+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{1}{15} x \left (5 d^2+3 e^2 x^2\right ) \sqrt{a+c x^4}+\frac{1}{2} (a d e) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^2}} \, dx,x,x^2\right )-\frac{\left (2 a^{3/2} e^2\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{5 \sqrt{c}}+\frac{1}{15} \left (2 a \left (5 d^2+\frac{3 \sqrt{a} e^2}{\sqrt{c}}\right )\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx\\ &=\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{1}{15} x \left (5 d^2+3 e^2 x^2\right ) \sqrt{a+c x^4}-\frac{2 a^{5/4} 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 )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{a^{3/4} \left (5 \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 )}{15 c^{3/4} \sqrt{a+c x^4}}+\frac{1}{2} (a d e) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{a+c x^4}}\right )\\ &=\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{1}{15} x \left (5 d^2+3 e^2 x^2\right ) \sqrt{a+c x^4}+\frac{a d e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}-\frac{2 a^{5/4} 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 )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{a^{3/4} \left (5 \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 )}{15 c^{3/4} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.135089, size = 146, normalized size = 0.45 \[ \frac{\sqrt{a+c x^4} \left (6 \sqrt{c} d^2 x \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^4}{a}\right )+e \left (3 d \left (\sqrt{c} x^2 \sqrt{\frac{c x^4}{a}+1}+\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )\right )+2 \sqrt{c} e x^3 \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^4}{a}\right )\right )\right )}{6 \sqrt{c} \sqrt{\frac{c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^4]*(6*Sqrt[c]*d^2*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^4)/a)] + e*(3*d*(Sqrt[c]*x^2*Sqrt[1
+ (c*x^4)/a] + Sqrt[a]*ArcSinh[(Sqrt[c]*x^2)/Sqrt[a]]) + 2*Sqrt[c]*e*x^3*Hypergeometric2F1[-1/2, 3/4, 7/4, -(
(c*x^4)/a)])))/(6*Sqrt[c]*Sqrt[1 + (c*x^4)/a])

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Maple [C]  time = 0.006, size = 310, normalized size = 1. \begin{align*}{\frac{{e}^{2}{x}^{3}}{5}\sqrt{c{x}^{4}+a}}+{{\frac{2\,i}{5}}{e}^{2}{a}^{{\frac{3}{2}}}\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}}}{\frac{1}{\sqrt{c}}}}-{{\frac{2\,i}{5}}{e}^{2}{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \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}}}{\frac{1}{\sqrt{c}}}}+{\frac{de{x}^{2}}{2}\sqrt{c{x}^{4}+a}}+{\frac{ade}{2}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{{d}^{2}x}{3}\sqrt{c{x}^{4}+a}}+{\frac{2\,a{d}^{2}}{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)^2*(c*x^4+a)^(1/2),x)

[Out]

1/5*e^2*x^3*(c*x^4+a)^(1/2)+2/5*I*e^2*a^(3/2)/(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)-2/5*I*e^2*a^(3/2)/(
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)*EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+1/2*d*e*x^2*(c*x^4+a)^(1/2)+1/2*d*e*a/c^(1/2)*ln(x^2*c^(1/2)+(c*x^
4+a)^(1/2))+1/3*d^2*x*(c*x^4+a)^(1/2)+2/3*d^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 \sqrt{c x^{4} + a}{\left (e x + d\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + a)*(e*x + d)^2, 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^{2} x^{2} + 2 \, d e x + d^{2}\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 3.36225, size = 138, normalized size = 0.42 \begin{align*} \frac{\sqrt{a} d^{2} 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} d e x^{2} \sqrt{1 + \frac{c x^{4}}{a}}}{2} + \frac{\sqrt{a} 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 \Gamma \left (\frac{7}{4}\right )} + \frac{a d e \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*d**2*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)*d*e*x*
*2*sqrt(1 + c*x**4/a)/2 + sqrt(a)*e**2*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4
*gamma(7/4)) + a*d*e*asinh(sqrt(c)*x**2/sqrt(a))/(2*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 )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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