\(\int x (a+b x^2)^{5/2} (A+B x^2) \, dx\) [542]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {(A b-a B) \left (a+b x^2\right )^{7/2}}{7 b^2}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^2} \]

[Out]

1/7*(A*b-B*a)*(b*x^2+a)^(7/2)/b^2+1/9*B*(b*x^2+a)^(9/2)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {\left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^2}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^2} \]

[In]

Int[x*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

((A*b - a*B)*(a + b*x^2)^(7/2))/(7*b^2) + (B*(a + b*x^2)^(9/2))/(9*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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (a+b x)^{5/2} (A+B x) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {(A b-a B) (a+b x)^{5/2}}{b}+\frac {B (a+b x)^{7/2}}{b}\right ) \, dx,x,x^2\right ) \\ & = \frac {(A b-a B) \left (a+b x^2\right )^{7/2}}{7 b^2}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {\left (a+b x^2\right )^{7/2} \left (9 A b-2 a B+7 b B x^2\right )}{63 b^2} \]

[In]

Integrate[x*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

((a + b*x^2)^(7/2)*(9*A*b - 2*a*B + 7*b*B*x^2))/(63*b^2)

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (7 b B \,x^{2}+9 A b -2 B a \right )}{63 b^{2}}\)	\(31\)
pseudoelliptic	\(\frac {\left (\left (7 x^{2} B +9 A \right ) b -2 B a \right ) \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{2}}\)	\(32\)
default	\(B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{9 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{2}}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7 b}\)	\(52\)
trager	\(\frac {\left (7 B \,x^{8} b^{4}+9 A \,x^{6} b^{4}+19 B \,x^{6} a \,b^{3}+27 A a \,b^{3} x^{4}+15 B \,a^{2} b^{2} x^{4}+27 A \,a^{2} b^{2} x^{2}+B \,a^{3} b \,x^{2}+9 A \,a^{3} b -2 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{63 b^{2}}\)	\(100\)
risch	\(\frac {\left (7 B \,x^{8} b^{4}+9 A \,x^{6} b^{4}+19 B \,x^{6} a \,b^{3}+27 A a \,b^{3} x^{4}+15 B \,a^{2} b^{2} x^{4}+27 A \,a^{2} b^{2} x^{2}+B \,a^{3} b \,x^{2}+9 A \,a^{3} b -2 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{63 b^{2}}\)	\(100\)

[In]

int(x*(b*x^2+a)^(5/2)*(B*x^2+A),x,method=_RETURNVERBOSE)

[Out]

1/63*(b*x^2+a)^(7/2)*(7*B*b*x^2+9*A*b-2*B*a)/b^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (38) = 76\).

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.11 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {{\left (7 \, B b^{4} x^{8} + {\left (19 \, B a b^{3} + 9 \, A b^{4}\right )} x^{6} - 2 \, B a^{4} + 9 \, A a^{3} b + 3 \, {\left (5 \, B a^{2} b^{2} + 9 \, A a b^{3}\right )} x^{4} + {\left (B a^{3} b + 27 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{63 \, b^{2}} \]

[In]

integrate(x*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="fricas")

[Out]

1/63*(7*B*b^4*x^8 + (19*B*a*b^3 + 9*A*b^4)*x^6 - 2*B*a^4 + 9*A*a^3*b + 3*(5*B*a^2*b^2 + 9*A*a*b^3)*x^4 + (B*a^
3*b + 27*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (39) = 78\).

Time = 0.43 (sec) , antiderivative size = 209, normalized size of antiderivative = 4.54 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\begin {cases} \frac {A a^{3} \sqrt {a + b x^{2}}}{7 b} + \frac {3 A a^{2} x^{2} \sqrt {a + b x^{2}}}{7} + \frac {3 A a b x^{4} \sqrt {a + b x^{2}}}{7} + \frac {A b^{2} x^{6} \sqrt {a + b x^{2}}}{7} - \frac {2 B a^{4} \sqrt {a + b x^{2}}}{63 b^{2}} + \frac {B a^{3} x^{2} \sqrt {a + b x^{2}}}{63 b} + \frac {5 B a^{2} x^{4} \sqrt {a + b x^{2}}}{21} + \frac {19 B a b x^{6} \sqrt {a + b x^{2}}}{63} + \frac {B b^{2} x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{2}}{2} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x*(b*x**2+a)**(5/2)*(B*x**2+A),x)

[Out]

Piecewise((A*a**3*sqrt(a + b*x**2)/(7*b) + 3*A*a**2*x**2*sqrt(a + b*x**2)/7 + 3*A*a*b*x**4*sqrt(a + b*x**2)/7
+ A*b**2*x**6*sqrt(a + b*x**2)/7 - 2*B*a**4*sqrt(a + b*x**2)/(63*b**2) + B*a**3*x**2*sqrt(a + b*x**2)/(63*b) +
5*B*a**2*x**4*sqrt(a + b*x**2)/21 + 19*B*a*b*x**6*sqrt(a + b*x**2)/63 + B*b**2*x**8*sqrt(a + b*x**2)/9, Ne(b,
0)), (a**(5/2)*(A*x**2/2 + B*x**4/4), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{2}}{9 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a}{63 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{7 \, b} \]

[In]

integrate(x*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

1/9*(b*x^2 + a)^(7/2)*B*x^2/b - 2/63*(b*x^2 + a)^(7/2)*B*a/b^2 + 1/7*(b*x^2 + a)^(7/2)*A/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {7 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B - 9 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a + 9 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{63 \, b^{2}} \]

[In]

integrate(x*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="giac")

[Out]

1/63*(7*(b*x^2 + a)^(9/2)*B - 9*(b*x^2 + a)^(7/2)*B*a + 9*(b*x^2 + a)^(7/2)*A*b)/b^2

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int x \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {7\,B\,{\left (b\,x^2+a\right )}^{9/2}+9\,A\,b\,{\left (b\,x^2+a\right )}^{7/2}-9\,B\,a\,{\left (b\,x^2+a\right )}^{7/2}}{63\,b^2} \]

[In]

int(x*(A + B*x^2)*(a + b*x^2)^(5/2),x)

[Out]

(7*B*(a + b*x^2)^(9/2) + 9*A*b*(a + b*x^2)^(7/2) - 9*B*a*(a + b*x^2)^(7/2))/(63*b^2)