3.6.8 \(\int x \sqrt {a+b x^2} (A+B x^2) \, dx\) [508]

Optimal. Leaf size=46 \[ \frac {(A b-a B) \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^2} \]

[Out]

1/3*(A*b-B*a)*(b*x^2+a)^(3/2)/b^2+1/5*B*(b*x^2+a)^(5/2)/b^2

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Rubi [A]
time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \begin {gather*} \frac {\left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^2}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

((A*b - a*B)*(a + b*x^2)^(3/2))/(3*b^2) + (B*(a + b*x^2)^(5/2))/(5*b^2)

Rule 45

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])

Rule 455

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 34, normalized size = 0.74 \begin {gather*} \frac {\left (a+b x^2\right )^{3/2} \left (5 A b-2 a B+3 b B x^2\right )}{15 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(3/2)*(5*A*b - 2*a*B + 3*b*B*x^2))/(15*b^2)

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Maple [A]
time = 0.08, size = 52, normalized size = 1.13

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (3 b B \,x^{2}+5 A b -2 B a \right )}{15 b^{2}}\)	\(31\)
default	\(B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}\)	\(52\)
trager	\(\frac {\left (3 b^{2} B \,x^{4}+5 A \,b^{2} x^{2}+B a b \,x^{2}+5 a b A -2 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{15 b^{2}}\)	\(52\)
risch	\(\frac {\left (3 b^{2} B \,x^{4}+5 A \,b^{2} x^{2}+B a b \,x^{2}+5 a b A -2 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{15 b^{2}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x^2+A)*(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 50, normalized size = 1.09 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a}{15 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.77, size = 50, normalized size = 1.09 \begin {gather*} \frac {{\left (3 \, B b^{2} x^{4} - 2 \, B a^{2} + 5 \, A a b + {\left (B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*B*b^2*x^4 - 2*B*a^2 + 5*A*a*b + (B*a*b + 5*A*b^2)*x^2)*sqrt(b*x^2 + a)/b^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (39) = 78\).
time = 0.10, size = 110, normalized size = 2.39 \begin {gather*} \begin {cases} \frac {A a \sqrt {a + b x^{2}}}{3 b} + \frac {A x^{2} \sqrt {a + b x^{2}}}{3} - \frac {2 B a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {B a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {B x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{2}}{2} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((A*a*sqrt(a + b*x**2)/(3*b) + A*x**2*sqrt(a + b*x**2)/3 - 2*B*a**2*sqrt(a + b*x**2)/(15*b**2) + B*a*
x**2*sqrt(a + b*x**2)/(15*b) + B*x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*(A*x**2/2 + B*x**4/4), True))

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Giac [A]
time = 1.44, size = 44, normalized size = 0.96 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B - 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{15 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.26, size = 53, normalized size = 1.15 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {B\,x^4}{5}-\frac {2\,B\,a^2-5\,A\,a\,b}{15\,b^2}+\frac {x^2\,\left (5\,A\,b^2+B\,a\,b\right )}{15\,b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(A + B*x^2)*(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)*((B*x^4)/5 - (2*B*a^2 - 5*A*a*b)/(15*b^2) + (x^2*(5*A*b^2 + B*a*b))/(15*b^2))

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