3.144 \(\int \frac{x^3 (c+d x^2+e x^4+f x^6)}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{\left (a+b x^2\right )^{3/2} \left (3 a^2 b e-4 a^3 f-2 a b^2 d+b^3 c\right )}{3 b^5}-\frac{a \sqrt{a+b x^2} \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{b^5}+\frac{\left (a+b x^2\right )^{5/2} \left (6 a^2 f-3 a b e+b^2 d\right )}{5 b^5}+\frac{\left (a+b x^2\right )^{7/2} (b e-4 a f)}{7 b^5}+\frac{f \left (a+b x^2\right )^{9/2}}{9 b^5} \]

[Out]

-((a*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Sqrt[a + b*x^2])/b^5) + ((b^3*c - 2*a*b^2*d + 3*a^2*b*e - 4*a^3*f)*(a
+ b*x^2)^(3/2))/(3*b^5) + ((b^2*d - 3*a*b*e + 6*a^2*f)*(a + b*x^2)^(5/2))/(5*b^5) + ((b*e - 4*a*f)*(a + b*x^2
)^(7/2))/(7*b^5) + (f*(a + b*x^2)^(9/2))/(9*b^5)

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Rubi [A]  time = 0.194013, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {1799, 1620} \[ \frac{\left (a+b x^2\right )^{3/2} \left (3 a^2 b e-4 a^3 f-2 a b^2 d+b^3 c\right )}{3 b^5}-\frac{a \sqrt{a+b x^2} \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{b^5}+\frac{\left (a+b x^2\right )^{5/2} \left (6 a^2 f-3 a b e+b^2 d\right )}{5 b^5}+\frac{\left (a+b x^2\right )^{7/2} (b e-4 a f)}{7 b^5}+\frac{f \left (a+b x^2\right )^{9/2}}{9 b^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(c + d*x^2 + e*x^4 + f*x^6))/Sqrt[a + b*x^2],x]

[Out]

-((a*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Sqrt[a + b*x^2])/b^5) + ((b^3*c - 2*a*b^2*d + 3*a^2*b*e - 4*a^3*f)*(a
+ b*x^2)^(3/2))/(3*b^5) + ((b^2*d - 3*a*b*e + 6*a^2*f)*(a + b*x^2)^(5/2))/(5*b^5) + ((b*e - 4*a*f)*(a + b*x^2
)^(7/2))/(7*b^5) + (f*(a + b*x^2)^(9/2))/(9*b^5)

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{x^3 \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{x \left (c+d x+e x^2+f x^3\right )}{\sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{b^4 \sqrt{a+b x}}+\frac{\left (b^3 c-2 a b^2 d+3 a^2 b e-4 a^3 f\right ) \sqrt{a+b x}}{b^4}+\frac{\left (b^2 d-3 a b e+6 a^2 f\right ) (a+b x)^{3/2}}{b^4}+\frac{(b e-4 a f) (a+b x)^{5/2}}{b^4}+\frac{f (a+b x)^{7/2}}{b^4}\right ) \, dx,x,x^2\right )\\ &=-\frac{a \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \sqrt{a+b x^2}}{b^5}+\frac{\left (b^3 c-2 a b^2 d+3 a^2 b e-4 a^3 f\right ) \left (a+b x^2\right )^{3/2}}{3 b^5}+\frac{\left (b^2 d-3 a b e+6 a^2 f\right ) \left (a+b x^2\right )^{5/2}}{5 b^5}+\frac{(b e-4 a f) \left (a+b x^2\right )^{7/2}}{7 b^5}+\frac{f \left (a+b x^2\right )^{9/2}}{9 b^5}\\ \end{align*}

Mathematica [A]  time = 0.121023, size = 122, normalized size = 0.73 \[ \frac{\sqrt{a+b x^2} \left (24 a^2 b^2 \left (7 d+3 e x^2+2 f x^4\right )-16 a^3 b \left (9 e+4 f x^2\right )+128 a^4 f-2 a b^3 \left (105 c+42 d x^2+27 e x^4+20 f x^6\right )+b^4 x^2 \left (105 c+63 d x^2+45 e x^4+35 f x^6\right )\right )}{315 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(c + d*x^2 + e*x^4 + f*x^6))/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[a + b*x^2]*(128*a^4*f - 16*a^3*b*(9*e + 4*f*x^2) + 24*a^2*b^2*(7*d + 3*e*x^2 + 2*f*x^4) - 2*a*b^3*(105*c
+ 42*d*x^2 + 27*e*x^4 + 20*f*x^6) + b^4*x^2*(105*c + 63*d*x^2 + 45*e*x^4 + 35*f*x^6)))/(315*b^5)

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Maple [A]  time = 0.007, size = 145, normalized size = 0.9 \begin{align*}{\frac{35\,f{x}^{8}{b}^{4}-40\,a{b}^{3}f{x}^{6}+45\,{b}^{4}e{x}^{6}+48\,{a}^{2}{b}^{2}f{x}^{4}-54\,a{b}^{3}e{x}^{4}+63\,{b}^{4}d{x}^{4}-64\,{a}^{3}bf{x}^{2}+72\,{a}^{2}{b}^{2}e{x}^{2}-84\,a{b}^{3}d{x}^{2}+105\,{b}^{4}c{x}^{2}+128\,{a}^{4}f-144\,{a}^{3}be+168\,{a}^{2}{b}^{2}d-210\,a{b}^{3}c}{315\,{b}^{5}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x)

[Out]

1/315*(b*x^2+a)^(1/2)*(35*b^4*f*x^8-40*a*b^3*f*x^6+45*b^4*e*x^6+48*a^2*b^2*f*x^4-54*a*b^3*e*x^4+63*b^4*d*x^4-6
4*a^3*b*f*x^2+72*a^2*b^2*e*x^2-84*a*b^3*d*x^2+105*b^4*c*x^2+128*a^4*f-144*a^3*b*e+168*a^2*b^2*d-210*a*b^3*c)/b
^5

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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^3*(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 = 1.35754, size = 316, normalized size = 1.89 \begin{align*} \frac{{\left (35 \, b^{4} f x^{8} + 5 \,{\left (9 \, b^{4} e - 8 \, a b^{3} f\right )} x^{6} - 210 \, a b^{3} c + 168 \, a^{2} b^{2} d - 144 \, a^{3} b e + 128 \, a^{4} f + 3 \,{\left (21 \, b^{4} d - 18 \, a b^{3} e + 16 \, a^{2} b^{2} f\right )} x^{4} +{\left (105 \, b^{4} c - 84 \, a b^{3} d + 72 \, a^{2} b^{2} e - 64 \, a^{3} b f\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{315 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/315*(35*b^4*f*x^8 + 5*(9*b^4*e - 8*a*b^3*f)*x^6 - 210*a*b^3*c + 168*a^2*b^2*d - 144*a^3*b*e + 128*a^4*f + 3*
(21*b^4*d - 18*a*b^3*e + 16*a^2*b^2*f)*x^4 + (105*b^4*c - 84*a*b^3*d + 72*a^2*b^2*e - 64*a^3*b*f)*x^2)*sqrt(b*
x^2 + a)/b^5

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Sympy [A]  time = 2.7518, size = 340, normalized size = 2.04 \begin{align*} \begin{cases} \frac{128 a^{4} f \sqrt{a + b x^{2}}}{315 b^{5}} - \frac{16 a^{3} e \sqrt{a + b x^{2}}}{35 b^{4}} - \frac{64 a^{3} f x^{2} \sqrt{a + b x^{2}}}{315 b^{4}} + \frac{8 a^{2} d \sqrt{a + b x^{2}}}{15 b^{3}} + \frac{8 a^{2} e x^{2} \sqrt{a + b x^{2}}}{35 b^{3}} + \frac{16 a^{2} f x^{4} \sqrt{a + b x^{2}}}{105 b^{3}} - \frac{2 a c \sqrt{a + b x^{2}}}{3 b^{2}} - \frac{4 a d x^{2} \sqrt{a + b x^{2}}}{15 b^{2}} - \frac{6 a e x^{4} \sqrt{a + b x^{2}}}{35 b^{2}} - \frac{8 a f x^{6} \sqrt{a + b x^{2}}}{63 b^{2}} + \frac{c x^{2} \sqrt{a + b x^{2}}}{3 b} + \frac{d x^{4} \sqrt{a + b x^{2}}}{5 b} + \frac{e x^{6} \sqrt{a + b x^{2}}}{7 b} + \frac{f x^{8} \sqrt{a + b x^{2}}}{9 b} & \text{for}\: b \neq 0 \\\frac{\frac{c x^{4}}{4} + \frac{d x^{6}}{6} + \frac{e x^{8}}{8} + \frac{f x^{10}}{10}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**(1/2),x)

[Out]

Piecewise((128*a**4*f*sqrt(a + b*x**2)/(315*b**5) - 16*a**3*e*sqrt(a + b*x**2)/(35*b**4) - 64*a**3*f*x**2*sqrt
(a + b*x**2)/(315*b**4) + 8*a**2*d*sqrt(a + b*x**2)/(15*b**3) + 8*a**2*e*x**2*sqrt(a + b*x**2)/(35*b**3) + 16*
a**2*f*x**4*sqrt(a + b*x**2)/(105*b**3) - 2*a*c*sqrt(a + b*x**2)/(3*b**2) - 4*a*d*x**2*sqrt(a + b*x**2)/(15*b*
*2) - 6*a*e*x**4*sqrt(a + b*x**2)/(35*b**2) - 8*a*f*x**6*sqrt(a + b*x**2)/(63*b**2) + c*x**2*sqrt(a + b*x**2)/
(3*b) + d*x**4*sqrt(a + b*x**2)/(5*b) + e*x**6*sqrt(a + b*x**2)/(7*b) + f*x**8*sqrt(a + b*x**2)/(9*b), Ne(b, 0
)), ((c*x**4/4 + d*x**6/6 + e*x**8/8 + f*x**10/10)/sqrt(a), True))

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Giac [A]  time = 1.2444, size = 296, normalized size = 1.77 \begin{align*} \frac{105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{3} c - 315 \, \sqrt{b x^{2} + a} a b^{3} c + 63 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} b^{2} d - 210 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a b^{2} d + 315 \, \sqrt{b x^{2} + a} a^{2} b^{2} d + 35 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} f - 180 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a f + 378 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} f - 420 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} f + 315 \, \sqrt{b x^{2} + a} a^{4} f + 45 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} b e - 189 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a b e + 315 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2} b e - 315 \, \sqrt{b x^{2} + a} a^{3} b e}{315 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/315*(105*(b*x^2 + a)^(3/2)*b^3*c - 315*sqrt(b*x^2 + a)*a*b^3*c + 63*(b*x^2 + a)^(5/2)*b^2*d - 210*(b*x^2 + a
)^(3/2)*a*b^2*d + 315*sqrt(b*x^2 + a)*a^2*b^2*d + 35*(b*x^2 + a)^(9/2)*f - 180*(b*x^2 + a)^(7/2)*a*f + 378*(b*
x^2 + a)^(5/2)*a^2*f - 420*(b*x^2 + a)^(3/2)*a^3*f + 315*sqrt(b*x^2 + a)*a^4*f + 45*(b*x^2 + a)^(7/2)*b*e - 18
9*(b*x^2 + a)^(5/2)*a*b*e + 315*(b*x^2 + a)^(3/2)*a^2*b*e - 315*sqrt(b*x^2 + a)*a^3*b*e)/b^5