3.6.25 \(\int x (a+b x^2)^{3/2} (A+B x^2) \, dx\) [525]

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

[Out]

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

________________________________________________________________________________________

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 )^{5/2} (A b-a B)}{5 b^2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b^2} \end {gather*}

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 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 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int (a+b x)^{3/2} (A+B x) \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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 = 0.03, size = 34, normalized size = 0.74 \begin {gather*} \frac {\left (a+b x^2\right )^{5/2} \left (7 A b-2 a B+5 b B x^2\right )}{35 b^2} \end {gather*}

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)

________________________________________________________________________________________

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

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (5 b B \,x^{2}+7 A b -2 B a \right )}{35 b^{2}}\)	\(31\)
default	\(B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}\)	\(52\)
trager	\(\frac {\left (5 B \,x^{6} b^{3}+7 A \,b^{3} x^{4}+8 B a \,b^{2} x^{4}+14 A a \,b^{2} x^{2}+B \,a^{2} b \,x^{2}+7 A \,a^{2} b -2 B \,a^{3}\right ) \sqrt {b \,x^{2}+a}}{35 b^{2}}\)	\(76\)
risch	\(\frac {\left (5 B \,x^{6} b^{3}+7 A \,b^{3} x^{4}+8 B a \,b^{2} x^{4}+14 A a \,b^{2} x^{2}+B \,a^{2} b \,x^{2}+7 A \,a^{2} b -2 B \,a^{3}\right ) \sqrt {b \,x^{2}+a}}{35 b^{2}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 50, normalized size = 1.09 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{2}}{7 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a}{35 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, b} \end {gather*}

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]

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

________________________________________________________________________________________

Fricas [A]
time = 1.38, size = 73, normalized size = 1.59 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (39) = 78\).
time = 0.19, size = 158, normalized size = 3.43 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________

Giac [A]
time = 2.27, size = 44, normalized size = 0.96 \begin {gather*} \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B - 7 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a + 7 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{35 \, b^{2}} \end {gather*}

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]

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

________________________________________________________________________________________

Mupad [B]
time = 0.29, size = 76, normalized size = 1.65 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {x^4\,\left (7\,A\,b^3+8\,B\,a\,b^2\right )}{35\,b^2}-\frac {2\,B\,a^3-7\,A\,a^2\,b}{35\,b^2}+\frac {B\,b\,x^6}{7}+\frac {a\,x^2\,\left (14\,A\,b+B\,a\right )}{35\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________