3.145 \(\int \frac{x (c+d x^2+e x^4+f x^6)}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=121 \[ \frac{\sqrt{a+b x^2} \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{b^4}+\frac{\left (a+b x^2\right )^{3/2} \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}+\frac{\left (a+b x^2\right )^{5/2} (b e-3 a f)}{5 b^4}+\frac{f \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[Out]

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

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Rubi [A]  time = 0.133137, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {1799, 1850} \[ \frac{\sqrt{a+b x^2} \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{b^4}+\frac{\left (a+b x^2\right )^{3/2} \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}+\frac{\left (a+b x^2\right )^{5/2} (b e-3 a f)}{5 b^4}+\frac{f \left (a+b x^2\right )^{7/2}}{7 b^4} \]

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x^2 + e*x^4 + f*x^6))/Sqrt[a + b*x^2],x]

[Out]

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

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin{align*} \int \frac{x \left (c+d x^2+e x^4+f x^6\right )}{\sqrt{a+b x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{c+d x+e x^2+f x^3}{\sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{b^3 c-a b^2 d+a^2 b e-a^3 f}{b^3 \sqrt{a+b x}}+\frac{\left (b^2 d-2 a b e+3 a^2 f\right ) \sqrt{a+b x}}{b^3}+\frac{(b e-3 a f) (a+b x)^{3/2}}{b^3}+\frac{f (a+b x)^{5/2}}{b^3}\right ) \, dx,x,x^2\right )\\ &=\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \sqrt{a+b x^2}}{b^4}+\frac{\left (b^2 d-2 a b e+3 a^2 f\right ) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac{(b e-3 a f) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac{f \left (a+b x^2\right )^{7/2}}{7 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0823981, size = 89, normalized size = 0.74 \[ \frac{\sqrt{a+b x^2} \left (8 a^2 b \left (7 e+3 f x^2\right )-48 a^3 f-2 a b^2 \left (35 d+14 e x^2+9 f x^4\right )+b^3 \left (105 c+35 d x^2+21 e x^4+15 f x^6\right )\right )}{105 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x^2 + e*x^4 + f*x^6))/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[a + b*x^2]*(-48*a^3*f + 8*a^2*b*(7*e + 3*f*x^2) - 2*a*b^2*(35*d + 14*e*x^2 + 9*f*x^4) + b^3*(105*c + 35*
d*x^2 + 21*e*x^4 + 15*f*x^6)))/(105*b^4)

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Maple [A]  time = 0.005, size = 99, normalized size = 0.8 \begin{align*} -{\frac{-15\,f{x}^{6}{b}^{3}+18\,a{b}^{2}f{x}^{4}-21\,{b}^{3}e{x}^{4}-24\,{a}^{2}bf{x}^{2}+28\,a{b}^{2}e{x}^{2}-35\,{b}^{3}d{x}^{2}+48\,{a}^{3}f-56\,{a}^{2}be+70\,a{b}^{2}d-105\,{b}^{3}c}{105\,{b}^{4}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x)

[Out]

-1/105*(b*x^2+a)^(1/2)*(-15*b^3*f*x^6+18*a*b^2*f*x^4-21*b^3*e*x^4-24*a^2*b*f*x^2+28*a*b^2*e*x^2-35*b^3*d*x^2+4
8*a^3*f-56*a^2*b*e+70*a*b^2*d-105*b^3*c)/b^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37243, size = 221, normalized size = 1.83 \begin{align*} \frac{{\left (15 \, b^{3} f x^{6} + 3 \,{\left (7 \, b^{3} e - 6 \, a b^{2} f\right )} x^{4} + 105 \, b^{3} c - 70 \, a b^{2} d + 56 \, a^{2} b e - 48 \, a^{3} f +{\left (35 \, b^{3} d - 28 \, a b^{2} e + 24 \, a^{2} b f\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{105 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*b^3*f*x^6 + 3*(7*b^3*e - 6*a*b^2*f)*x^4 + 105*b^3*c - 70*a*b^2*d + 56*a^2*b*e - 48*a^3*f + (35*b^3*d
- 28*a*b^2*e + 24*a^2*b*f)*x^2)*sqrt(b*x^2 + a)/b^4

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Sympy [A]  time = 1.6298, size = 238, normalized size = 1.97 \begin{align*} \begin{cases} - \frac{16 a^{3} f \sqrt{a + b x^{2}}}{35 b^{4}} + \frac{8 a^{2} e \sqrt{a + b x^{2}}}{15 b^{3}} + \frac{8 a^{2} f x^{2} \sqrt{a + b x^{2}}}{35 b^{3}} - \frac{2 a d \sqrt{a + b x^{2}}}{3 b^{2}} - \frac{4 a e x^{2} \sqrt{a + b x^{2}}}{15 b^{2}} - \frac{6 a f x^{4} \sqrt{a + b x^{2}}}{35 b^{2}} + \frac{c \sqrt{a + b x^{2}}}{b} + \frac{d x^{2} \sqrt{a + b x^{2}}}{3 b} + \frac{e x^{4} \sqrt{a + b x^{2}}}{5 b} + \frac{f x^{6} \sqrt{a + b x^{2}}}{7 b} & \text{for}\: b \neq 0 \\\frac{\frac{c x^{2}}{2} + \frac{d x^{4}}{4} + \frac{e x^{6}}{6} + \frac{f x^{8}}{8}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-16*a**3*f*sqrt(a + b*x**2)/(35*b**4) + 8*a**2*e*sqrt(a + b*x**2)/(15*b**3) + 8*a**2*f*x**2*sqrt(a
+ b*x**2)/(35*b**3) - 2*a*d*sqrt(a + b*x**2)/(3*b**2) - 4*a*e*x**2*sqrt(a + b*x**2)/(15*b**2) - 6*a*f*x**4*sqr
t(a + b*x**2)/(35*b**2) + c*sqrt(a + b*x**2)/b + d*x**2*sqrt(a + b*x**2)/(3*b) + e*x**4*sqrt(a + b*x**2)/(5*b)
+ f*x**6*sqrt(a + b*x**2)/(7*b), Ne(b, 0)), ((c*x**2/2 + d*x**4/4 + e*x**6/6 + f*x**8/8)/sqrt(a), True))

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Giac [A]  time = 1.17078, size = 207, normalized size = 1.71 \begin{align*} \frac{105 \, \sqrt{b x^{2} + a} b^{3} c + 35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{2} d - 105 \, \sqrt{b x^{2} + a} a b^{2} d + 15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} f - 63 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a f + 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2} f - 105 \, \sqrt{b x^{2} + a} a^{3} f + 21 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} b e - 70 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a b e + 105 \, \sqrt{b x^{2} + a} a^{2} b e}{105 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/105*(105*sqrt(b*x^2 + a)*b^3*c + 35*(b*x^2 + a)^(3/2)*b^2*d - 105*sqrt(b*x^2 + a)*a*b^2*d + 15*(b*x^2 + a)^(
7/2)*f - 63*(b*x^2 + a)^(5/2)*a*f + 105*(b*x^2 + a)^(3/2)*a^2*f - 105*sqrt(b*x^2 + a)*a^3*f + 21*(b*x^2 + a)^(
5/2)*b*e - 70*(b*x^2 + a)^(3/2)*a*b*e + 105*sqrt(b*x^2 + a)*a^2*b*e)/b^4