\(\int \sqrt {a+b x^2} (A+B x^2) \, dx\) [509]

Optimal result

Integrand size = 19, antiderivative size = 87 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {(4 A b-a B) x \sqrt {a+b x^2}}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {a (4 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \]

[Out]

1/4*B*x*(b*x^2+a)^(3/2)/b+1/8*a*(4*A*b-B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)+1/8*(4*A*b-B*a)*x*(b*x^
2+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {396, 201, 223, 212} \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {a (4 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 A b-a B)}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b} \]

[In]

Int[Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

((4*A*b - a*B)*x*Sqrt[a + b*x^2])/(8*b) + (B*x*(a + b*x^2)^(3/2))/(4*b) + (a*(4*A*b - a*B)*ArcTanh[(Sqrt[b]*x)
/Sqrt[a + b*x^2]])/(8*b^(3/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 396

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

Rubi steps \begin{align*} \text {integral}& = \frac {B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {(-4 A b+a B) \int \sqrt {a+b x^2} \, dx}{4 b} \\ & = \frac {(4 A b-a B) x \sqrt {a+b x^2}}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {(a (4 A b-a B)) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b} \\ & = \frac {(4 A b-a B) x \sqrt {a+b x^2}}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {(a (4 A b-a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b} \\ & = \frac {(4 A b-a B) x \sqrt {a+b x^2}}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {a (4 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (4 A b+a B+2 b B x^2\right )}{8 b}+\frac {a (-4 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{3/2}} \]

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(4*A*b + a*B + 2*b*B*x^2))/(8*b) + (a*(-4*A*b + a*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(
8*b^(3/2))

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {x \left (2 b B \,x^{2}+4 A b +B a \right ) \sqrt {b \,x^{2}+a}}{8 b}+\frac {a \left (4 A b -B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}\)	\(63\)
pseudoelliptic	\(\frac {a \left (A b -\frac {B a}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \left (\left (\frac {x^{2} B}{2}+A \right ) b^{\frac {3}{2}}+\frac {B \sqrt {b}\, a}{4}\right ) \sqrt {b \,x^{2}+a}}{2 b^{\frac {3}{2}}}\)	\(65\)
default	\(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(98\)

[In]

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

[Out]

1/8*x*(2*B*b*x^2+4*A*b+B*a)*(b*x^2+a)^(1/2)/b+1/8*a*(4*A*b-B*a)/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.78 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\left [-\frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, B b^{2} x^{3} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{2}}, \frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B b^{2} x^{3} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{2}}\right ] \]

[In]

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

[Out]

[-1/16*((B*a^2 - 4*A*a*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2*B*b^2*x^3 + (B*a*b +
4*A*b^2)*x)*sqrt(b*x^2 + a))/b^2, 1/8*((B*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (2*B*b^
2*x^3 + (B*a*b + 4*A*b^2)*x)*sqrt(b*x^2 + a))/b^2]

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B x^{3}}{4} + \frac {x \left (A b + \frac {B a}{4}\right )}{2 b}\right ) + \left (A a - \frac {a \left (A b + \frac {B a}{4}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} \]

[In]

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

[Out]

1/2*sqrt(b*x^2 + a)*A*x + 1/4*(b*x^2 + a)^(3/2)*B*x/b - 1/8*sqrt(b*x^2 + a)*B*a*x/b - 1/8*B*a^2*arcsinh(b*x/sq
rt(a*b))/b^(3/2) + 1/2*A*a*arcsinh(b*x/sqrt(a*b))/sqrt(b)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {1}{8} \, {\left (2 \, B x^{2} + \frac {B a b + 4 \, A b^{2}}{b^{2}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \]

[In]

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

[Out]

1/8*(2*B*x^2 + (B*a*b + 4*A*b^2)/b^2)*sqrt(b*x^2 + a)*x + 1/8*(B*a^2 - 4*A*a*b)*log(abs(-sqrt(b)*x + sqrt(b*x^
2 + a)))/b^(3/2)

Mupad [F(-1)]

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

[In]

int((A + B*x^2)*(a + b*x^2)^(1/2),x)

[Out]

int((A + B*x^2)*(a + b*x^2)^(1/2), x)