3.616 \(\int x (a+b x^2)^2 (c+d x^2)^{3/2} \, dx\)

Optimal. Leaf size=77 \[ -\frac{2 b \left (c+d x^2\right )^{7/2} (b c-a d)}{7 d^3}+\frac{\left (c+d x^2\right )^{5/2} (b c-a d)^2}{5 d^3}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^3} \]

[Out]

((b*c - a*d)^2*(c + d*x^2)^(5/2))/(5*d^3) - (2*b*(b*c - a*d)*(c + d*x^2)^(7/2))/(7*d^3) + (b^2*(c + d*x^2)^(9/
2))/(9*d^3)

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Rubi [A]  time = 0.0599617, 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} \[ -\frac{2 b \left (c+d x^2\right )^{7/2} (b c-a d)}{7 d^3}+\frac{\left (c+d x^2\right )^{5/2} (b c-a d)^2}{5 d^3}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0463802, size = 67, normalized size = 0.87 \[ \frac{\left (c+d x^2\right )^{5/2} \left (63 a^2 d^2+18 a b d \left (5 d x^2-2 c\right )+b^2 \left (8 c^2-20 c d x^2+35 d^2 x^4\right )\right )}{315 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x^2)^(5/2)*(63*a^2*d^2 + 18*a*b*d*(-2*c + 5*d*x^2) + b^2*(8*c^2 - 20*c*d*x^2 + 35*d^2*x^4)))/(315*d^3)

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Maple [A]  time = 0.005, size = 69, normalized size = 0.9 \begin{align*}{\frac{35\,{b}^{2}{d}^{2}{x}^{4}+90\,ab{d}^{2}{x}^{2}-20\,{b}^{2}cd{x}^{2}+63\,{a}^{2}{d}^{2}-36\,cabd+8\,{b}^{2}{c}^{2}}{315\,{d}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{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)^(3/2),x)

[Out]

1/315*(d*x^2+c)^(5/2)*(35*b^2*d^2*x^4+90*a*b*d^2*x^2-20*b^2*c*d*x^2+63*a^2*d^2-36*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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.39242, size = 305, normalized size = 3.96 \begin{align*} \frac{{\left (35 \, b^{2} d^{4} x^{8} + 10 \,{\left (5 \, b^{2} c d^{3} + 9 \, a b d^{4}\right )} x^{6} + 8 \, b^{2} c^{4} - 36 \, a b c^{3} d + 63 \, a^{2} c^{2} d^{2} + 3 \,{\left (b^{2} c^{2} d^{2} + 48 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} - 2 \,{\left (2 \, b^{2} c^{3} d - 9 \, a b c^{2} d^{2} - 63 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{315 \, 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)^(3/2),x, algorithm="fricas")

[Out]

1/315*(35*b^2*d^4*x^8 + 10*(5*b^2*c*d^3 + 9*a*b*d^4)*x^6 + 8*b^2*c^4 - 36*a*b*c^3*d + 63*a^2*c^2*d^2 + 3*(b^2*
c^2*d^2 + 48*a*b*c*d^3 + 21*a^2*d^4)*x^4 - 2*(2*b^2*c^3*d - 9*a*b*c^2*d^2 - 63*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c)
/d^3

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Sympy [A]  time = 2.38344, size = 303, normalized size = 3.94 \begin{align*} \begin{cases} \frac{a^{2} c^{2} \sqrt{c + d x^{2}}}{5 d} + \frac{2 a^{2} c x^{2} \sqrt{c + d x^{2}}}{5} + \frac{a^{2} d x^{4} \sqrt{c + d x^{2}}}{5} - \frac{4 a b c^{3} \sqrt{c + d x^{2}}}{35 d^{2}} + \frac{2 a b c^{2} x^{2} \sqrt{c + d x^{2}}}{35 d} + \frac{16 a b c x^{4} \sqrt{c + d x^{2}}}{35} + \frac{2 a b d x^{6} \sqrt{c + d x^{2}}}{7} + \frac{8 b^{2} c^{4} \sqrt{c + d x^{2}}}{315 d^{3}} - \frac{4 b^{2} c^{3} x^{2} \sqrt{c + d x^{2}}}{315 d^{2}} + \frac{b^{2} c^{2} x^{4} \sqrt{c + d x^{2}}}{105 d} + \frac{10 b^{2} c x^{6} \sqrt{c + d x^{2}}}{63} + \frac{b^{2} d x^{8} \sqrt{c + d x^{2}}}{9} & \text{for}\: d \neq 0 \\c^{\frac{3}{2}} \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)**(3/2),x)

[Out]

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

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Giac [B]  time = 1.12525, size = 315, normalized size = 4.09 \begin{align*} \frac{105 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} c + 21 \,{\left (3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c\right )} a^{2} + \frac{42 \,{\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 c}{d} + \frac{3 \,{\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} c}{d^{2}} + \frac{6 \,{\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 )} a b}{d} + \frac{{\left (35 \,{\left (d x^{2} + c\right )}^{\frac{9}{2}} - 135 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} c + 189 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} c^{2} - 105 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{3}\right )} b^{2}}{d^{2}}}{315 \, 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)^(3/2),x, algorithm="giac")

[Out]

1/315*(105*(d*x^2 + c)^(3/2)*a^2*c + 21*(3*(d*x^2 + c)^(5/2) - 5*(d*x^2 + c)^(3/2)*c)*a^2 + 42*(3*(d*x^2 + c)^
(5/2) - 5*(d*x^2 + c)^(3/2)*c)*a*b*c/d + 3*(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*c/d^2 + 6*(15*(d*x^2 + c)^(7/2) - 42*(d*x^2 + c)^(5/2)*c + 35*(d*x^2 + c)^(3/2)*c^2)*a*b/d + (35*(
d*x^2 + c)^(9/2) - 135*(d*x^2 + c)^(7/2)*c + 189*(d*x^2 + c)^(5/2)*c^2 - 105*(d*x^2 + c)^(3/2)*c^3)*b^2/d^2)/d