3.525 \(\int x (a+b x^2)^{3/2} (A+B x^2) \, dx\)

Optimal. Leaf size=46 \[ \frac{\left (a+b x^2\right )^{5/2} (A b-a B)}{5 b^2}+\frac{B \left (a+b x^2\right )^{7/2}}{7 b^2} \]

[Out]

((A*b - a*B)*(a + b*x^2)^(5/2))/(5*b^2) + (B*(a + b*x^2)^(7/2))/(7*b^2)

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Rubi [A]  time = 0.0324109, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {444, 43} \[ \frac{\left (a+b x^2\right )^{5/2} (A b-a B)}{5 b^2}+\frac{B \left (a+b x^2\right )^{7/2}}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^(3/2)*(A + B*x^2),x]

[Out]

((A*b - a*B)*(a + b*x^2)^(5/2))/(5*b^2) + (B*(a + b*x^2)^(7/2))/(7*b^2)

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

Mathematica [A]  time = 0.0233831, size = 34, normalized size = 0.74 \[ \frac{\left (a+b x^2\right )^{5/2} \left (-2 a B+7 A b+5 b B x^2\right )}{35 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2)^(3/2)*(A + B*x^2),x]

[Out]

((a + b*x^2)^(5/2)*(7*A*b - 2*a*B + 5*b*B*x^2))/(35*b^2)

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Maple [A]  time = 0.003, size = 31, normalized size = 0.7 \begin{align*}{\frac{5\,bB{x}^{2}+7\,Ab-2\,Ba}{35\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^(3/2)*(B*x^2+A),x)

[Out]

1/35*(b*x^2+a)^(5/2)*(5*B*b*x^2+7*A*b-2*B*a)/b^2

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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)^(3/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62903, size = 161, normalized size = 3.5 \begin{align*} \frac{{\left (5 \, B b^{3} x^{6} +{\left (8 \, B a b^{2} + 7 \, A b^{3}\right )} x^{4} - 2 \, B a^{3} + 7 \, A a^{2} b +{\left (B a^{2} b + 14 \, A a b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{35 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="fricas")

[Out]

1/35*(5*B*b^3*x^6 + (8*B*a*b^2 + 7*A*b^3)*x^4 - 2*B*a^3 + 7*A*a^2*b + (B*a^2*b + 14*A*a*b^2)*x^2)*sqrt(b*x^2 +
a)/b^2

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Sympy [A]  time = 1.26381, size = 158, normalized size = 3.43 \begin{align*} \begin{cases} \frac{A a^{2} \sqrt{a + b x^{2}}}{5 b} + \frac{2 A a x^{2} \sqrt{a + b x^{2}}}{5} + \frac{A b x^{4} \sqrt{a + b x^{2}}}{5} - \frac{2 B a^{3} \sqrt{a + b x^{2}}}{35 b^{2}} + \frac{B a^{2} x^{2} \sqrt{a + b x^{2}}}{35 b} + \frac{8 B a x^{4} \sqrt{a + b x^{2}}}{35} + \frac{B b x^{6} \sqrt{a + b x^{2}}}{7} & \text{for}\: b \neq 0 \\a^{\frac{3}{2}} \left (\frac{A x^{2}}{2} + \frac{B x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)**(3/2)*(B*x**2+A),x)

[Out]

Piecewise((A*a**2*sqrt(a + b*x**2)/(5*b) + 2*A*a*x**2*sqrt(a + b*x**2)/5 + A*b*x**4*sqrt(a + b*x**2)/5 - 2*B*a
**3*sqrt(a + b*x**2)/(35*b**2) + B*a**2*x**2*sqrt(a + b*x**2)/(35*b) + 8*B*a*x**4*sqrt(a + b*x**2)/35 + B*b*x*
*6*sqrt(a + b*x**2)/7, Ne(b, 0)), (a**(3/2)*(A*x**2/2 + B*x**4/4), True))

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Giac [B]  time = 1.09728, size = 162, normalized size = 3.52 \begin{align*} \frac{35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a + 7 \,{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} A + \frac{7 \,{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} B a}{b} + \frac{{\left (15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2}\right )} B}{b}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="giac")

[Out]

1/105*(35*(b*x^2 + a)^(3/2)*A*a + 7*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)*A + 7*(3*(b*x^2 + a)^(5/2) -
5*(b*x^2 + a)^(3/2)*a)*B*a/b + (15*(b*x^2 + a)^(7/2) - 42*(b*x^2 + a)^(5/2)*a + 35*(b*x^2 + a)^(3/2)*a^2)*B/b
)/b