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

Optimal. Leaf size=303 \[ -\frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{b} d-\sqrt{a} f\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{4 a^{3/4} b^{5/4} \sqrt{a+b x^4}}+\frac{d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt{a+b x^4}}-\frac{x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt{a+b x^4}}-\frac{d x \sqrt{a+b x^4}}{2 a \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{e \sqrt{a+b x^4}}{2 a b} \]

[Out]

-(x*(a*f - b*c*x - b*d*x^2 - b*e*x^3))/(2*a*b*Sqrt[a + b*x^4]) - (e*Sqrt[a + b*x^4])/(2*a*b) - (d*x*Sqrt[a + b
*x^4])/(2*a*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*
x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*b^(3/4)*Sqrt[a + b*x^4]) - ((Sqrt[b]*d - Sqr
t[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(
1/4)], 1/2])/(4*a^(3/4)*b^(5/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.192557, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1828, 1885, 261, 1198, 220, 1196} \[ -\frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{b} d-\sqrt{a} f\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{3/4} b^{5/4} \sqrt{a+b x^4}}+\frac{d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt{a+b x^4}}-\frac{x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt{a+b x^4}}-\frac{d x \sqrt{a+b x^4}}{2 a \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{e \sqrt{a+b x^4}}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(a*f - b*c*x - b*d*x^2 - b*e*x^3))/(2*a*b*Sqrt[a + b*x^4]) - (e*Sqrt[a + b*x^4])/(2*a*b) - (d*x*Sqrt[a + b
*x^4])/(2*a*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*
x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*b^(3/4)*Sqrt[a + b*x^4]) - ((Sqrt[b]*d - Sqr
t[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(
1/4)], 1/2])/(4*a^(3/4)*b^(5/4)*Sqrt[a + b*x^4])

Rule 1828

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = Pol
ynomialQuotient[b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] +
1)*x^m*Pq, a + b*x^n, x]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[
a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q
- 1)/n] + 1)), x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 0]

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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

Mathematica [C]  time = 0.0769455, size = 116, normalized size = 0.38 \[ \frac{2 b d x^3 \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{b x^4}{a}\right )+3 a f x \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )-3 a e-3 a f x+3 b c x^2}{6 a b \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.011, size = 250, normalized size = 0.8 \begin{align*} f \left ( -{\frac{x}{2\,b}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{1}{2\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \right ) -{\frac{e}{2\,b}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+d \left ({\frac{{x}^{3}}{2\,a}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{{\frac{i}{2}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}} \right ) +{\frac{c{x}^{2}}{2\,a}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*(-1/2/b*x/((x^4+1/b*a)*b)^(1/2)+1/2/b/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)
*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I))-1/2*e/b/(b*x^4+a)^(1/2)+d*(1/2*x
^3/a/((x^4+1/b*a)*b)^(1/2)-1/2*I/a^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2
)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)/b^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*
b^(1/2))^(1/2),I)))+1/2*c/(b*x^4+a)^(1/2)/a*x^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}{\left (f x^{4} + e x^{3} + d x^{2} + c x\right )}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^4 + a)*(f*x^4 + e*x^3 + d*x^2 + c*x)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

________________________________________________________________________________________

Sympy [A]  time = 14.6385, size = 133, normalized size = 0.44 \begin{align*} e \left (\begin{cases} - \frac{1}{2 b \sqrt{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{3}{2}}} & \text{otherwise} \end{cases}\right ) + \frac{c x^{2}}{2 a^{\frac{3}{2}} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{d x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{3}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{f x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), True)) + c*x**2/(2*a**(3/2)*sqrt(1 + b*
x**4/a)) + d*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7/4)) + f*x
**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(9/4))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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