\(\int x^2 \sqrt {a+b x^2} (A+B x^2) \, dx\) [507]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {a (2 A b-a B) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 A b-a B) x^3 \sqrt {a+b x^2}}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a^2 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \]

[Out]

1/6*B*x^3*(b*x^2+a)^(3/2)/b-1/16*a^2*(2*A*b-B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)+1/16*a*(2*A*b-B*a)
*x*(b*x^2+a)^(1/2)/b^2+1/8*(2*A*b-B*a)*x^3*(b*x^2+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {470, 285, 327, 223, 212} \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=-\frac {a^2 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}+\frac {a x \sqrt {a+b x^2} (2 A b-a B)}{16 b^2}+\frac {x^3 \sqrt {a+b x^2} (2 A b-a B)}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]

[In]

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

[Out]

(a*(2*A*b - a*B)*x*Sqrt[a + b*x^2])/(16*b^2) + ((2*A*b - a*B)*x^3*Sqrt[a + b*x^2])/(8*b) + (B*x^3*(a + b*x^2)^
(3/2))/(6*b) - (a^2*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {(-6 A b+3 a B) \int x^2 \sqrt {a+b x^2} \, dx}{6 b} \\ & = \frac {(2 A b-a B) x^3 \sqrt {a+b x^2}}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {(a (2 A b-a B)) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{8 b} \\ & = \frac {a (2 A b-a B) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 A b-a B) x^3 \sqrt {a+b x^2}}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\left (a^2 (2 A b-a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^2} \\ & = \frac {a (2 A b-a B) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 A b-a B) x^3 \sqrt {a+b x^2}}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\left (a^2 (2 A b-a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b^2} \\ & = \frac {a (2 A b-a B) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 A b-a B) x^3 \sqrt {a+b x^2}}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a^2 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {\sqrt {a+b x^2} \left (6 a A b x-3 a^2 B x+12 A b^2 x^3+2 a b B x^3+8 b^2 B x^5\right )}{48 b^2}+\frac {a^2 (-2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(6*a*A*b*x - 3*a^2*B*x + 12*A*b^2*x^3 + 2*a*b*B*x^3 + 8*b^2*B*x^5))/(48*b^2) + (a^2*(-2*A*b +
a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(5/2))

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {x \left (8 b^{2} B \,x^{4}+12 A \,b^{2} x^{2}+2 B a b \,x^{2}+6 a b A -3 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b^{2}}-\frac {a^{2} \left (2 A b -B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\)	\(88\)
pseudoelliptic	\(\frac {\left (-\frac {1}{2} a^{2} b A +\frac {1}{4} a^{3} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \left (\frac {\left (\frac {x^{2} B}{3}+A \right ) a \,b^{\frac {3}{2}}}{2}+x^{2} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {5}{2}}-\frac {B \,a^{2} \sqrt {b}}{4}\right ) \sqrt {b \,x^{2}+a}}{4 b^{\frac {5}{2}}}\)	\(89\)
default	\(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(144\)

[In]

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

[Out]

1/48*x*(8*B*b^2*x^4+12*A*b^2*x^2+2*B*a*b*x^2+6*A*a*b-3*B*a^2)*(b*x^2+a)^(1/2)/b^2-1/16*a^2*(2*A*b-B*a)/b^(5/2)
*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.69 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x^{3} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, B b^{3} x^{5} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x^{3} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \]

[In]

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

[Out]

[-1/96*(3*(B*a^3 - 2*A*a^2*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*B*b^3*x^5 + 2*(B*
a*b^2 + 6*A*b^3)*x^3 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/48*(3*(B*a^3 - 2*A*a^2*b)*sqrt(-b)*
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*B*b^3*x^5 + 2*(B*a*b^2 + 6*A*b^3)*x^3 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqr
t(b*x^2 + a))/b^3]

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\begin {cases} - \frac {a \left (A a - \frac {3 a \left (A b + \frac {B a}{6}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {B x^{5}}{6} + \frac {x^{3} \left (A b + \frac {B a}{6}\right )}{4 b} + \frac {x \left (A a - \frac {3 a \left (A b + \frac {B a}{6}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x}{4 \, b} - \frac {\sqrt {b x^{2} + a} A a x}{8 \, b} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} \]

[In]

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

[Out]

1/6*(b*x^2 + a)^(3/2)*B*x^3/b - 1/8*(b*x^2 + a)^(3/2)*B*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*B*a^2*x/b^2 + 1/4*(b*x^
2 + a)^(3/2)*A*x/b - 1/8*sqrt(b*x^2 + a)*A*a*x/b + 1/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/8*A*a^2*arcsi
nh(b*x/sqrt(a*b))/b^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.82 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, B x^{2} + \frac {B a b^{3} + 6 \, A b^{4}}{b^{4}}\right )} x^{2} - \frac {3 \, {\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (B a^{3} - 2 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

1/48*(2*(4*B*x^2 + (B*a*b^3 + 6*A*b^4)/b^4)*x^2 - 3*(B*a^2*b^2 - 2*A*a*b^3)/b^4)*sqrt(b*x^2 + a)*x - 1/16*(B*a
^3 - 2*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\int x^2\,\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a} \,d x \]

[In]

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

[Out]

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