3.600 \(\int x (a+b x^2)^2 \sqrt{c+d x^2} \, dx\)

Optimal. Leaf size=77 \[ -\frac{2 b \left (c+d x^2\right )^{5/2} (b c-a d)}{5 d^3}+\frac{\left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^3}+\frac{b^2 \left (c+d x^2\right )^{7/2}}{7 d^3} \]

[Out]

((b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^3) - (2*b*(b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^3) + (b^2*(c + d*x^2)^(7/
2))/(7*d^3)

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Rubi [A]  time = 0.0595577, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {444, 43} \[ -\frac{2 b \left (c+d x^2\right )^{5/2} (b c-a d)}{5 d^3}+\frac{\left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^3}+\frac{b^2 \left (c+d x^2\right )^{7/2}}{7 d^3} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

((b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^3) - (2*b*(b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^3) + (b^2*(c + d*x^2)^(7/
2))/(7*d^3)

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0419449, size = 67, normalized size = 0.87 \[ \frac{\left (c+d x^2\right )^{3/2} \left (35 a^2 d^2+14 a b d \left (3 d x^2-2 c\right )+b^2 \left (8 c^2-12 c d x^2+15 d^2 x^4\right )\right )}{105 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x^2)^(3/2)*(35*a^2*d^2 + 14*a*b*d*(-2*c + 3*d*x^2) + b^2*(8*c^2 - 12*c*d*x^2 + 15*d^2*x^4)))/(105*d^3)

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Maple [A]  time = 0.008, size = 69, normalized size = 0.9 \begin{align*}{\frac{15\,{b}^{2}{d}^{2}{x}^{4}+42\,ab{d}^{2}{x}^{2}-12\,{b}^{2}cd{x}^{2}+35\,{a}^{2}{d}^{2}-28\,cabd+8\,{b}^{2}{c}^{2}}{105\,{d}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^2*(d*x^2+c)^(1/2),x)

[Out]

1/105*(d*x^2+c)^(3/2)*(15*b^2*d^2*x^4+42*a*b*d^2*x^2-12*b^2*c*d*x^2+35*a^2*d^2-28*a*b*c*d+8*b^2*c^2)/d^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59331, size = 225, normalized size = 2.92 \begin{align*} \frac{{\left (15 \, b^{2} d^{3} x^{6} + 8 \, b^{2} c^{3} - 28 \, a b c^{2} d + 35 \, a^{2} c d^{2} + 3 \,{\left (b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{4} -{\left (4 \, b^{2} c^{2} d - 14 \, a b c d^{2} - 35 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{105 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*b^2*d^3*x^6 + 8*b^2*c^3 - 28*a*b*c^2*d + 35*a^2*c*d^2 + 3*(b^2*c*d^2 + 14*a*b*d^3)*x^4 - (4*b^2*c^2*
d - 14*a*b*c*d^2 - 35*a^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^3

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Sympy [A]  time = 0.790295, size = 226, normalized size = 2.94 \begin{align*} \begin{cases} \frac{a^{2} c \sqrt{c + d x^{2}}}{3 d} + \frac{a^{2} x^{2} \sqrt{c + d x^{2}}}{3} - \frac{4 a b c^{2} \sqrt{c + d x^{2}}}{15 d^{2}} + \frac{2 a b c x^{2} \sqrt{c + d x^{2}}}{15 d} + \frac{2 a b x^{4} \sqrt{c + d x^{2}}}{5} + \frac{8 b^{2} c^{3} \sqrt{c + d x^{2}}}{105 d^{3}} - \frac{4 b^{2} c^{2} x^{2} \sqrt{c + d x^{2}}}{105 d^{2}} + \frac{b^{2} c x^{4} \sqrt{c + d x^{2}}}{35 d} + \frac{b^{2} x^{6} \sqrt{c + d x^{2}}}{7} & \text{for}\: d \neq 0 \\\sqrt{c} \left (\frac{a^{2} x^{2}}{2} + \frac{a b x^{4}}{2} + \frac{b^{2} x^{6}}{6}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c*sqrt(c + d*x**2)/(3*d) + a**2*x**2*sqrt(c + d*x**2)/3 - 4*a*b*c**2*sqrt(c + d*x**2)/(15*d**2
) + 2*a*b*c*x**2*sqrt(c + d*x**2)/(15*d) + 2*a*b*x**4*sqrt(c + d*x**2)/5 + 8*b**2*c**3*sqrt(c + d*x**2)/(105*d
**3) - 4*b**2*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + b**2*c*x**4*sqrt(c + d*x**2)/(35*d) + b**2*x**6*sqrt(c +
d*x**2)/7, Ne(d, 0)), (sqrt(c)*(a**2*x**2/2 + a*b*x**4/2 + b**2*x**6/6), True))

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Giac [A]  time = 1.54361, size = 130, normalized size = 1.69 \begin{align*} \frac{35 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} + \frac{14 \,{\left (3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c\right )} a b}{d} + \frac{{\left (15 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} - 42 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} c + 35 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{2}\right )} b^{2}}{d^{2}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(35*(d*x^2 + c)^(3/2)*a^2 + 14*(3*(d*x^2 + c)^(5/2) - 5*(d*x^2 + c)^(3/2)*c)*a*b/d + (15*(d*x^2 + c)^(7/
2) - 42*(d*x^2 + c)^(5/2)*c + 35*(d*x^2 + c)^(3/2)*c^2)*b^2/d^2)/d