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

Optimal. Leaf size=87 \[ \frac {a (4 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 A b-a B)}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b} \]

[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

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Rubi [A]  time = 0.03, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {388, 195, 217, 206} \[ \frac {a (4 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 A b-a B)}{8 b}+\frac {B x \left (a+b x^2\right )^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[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 195

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 206

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

Rule 217

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 388

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*} \int \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx &=\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)) \operatorname {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*}

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Mathematica [A]  time = 0.16, size = 85, normalized size = 0.98 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (B \left (a+2 b x^2\right )+4 A b\right )-\frac {\sqrt {a} (a B-4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{8 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(4*A*b + B*(a + 2*b*x^2)) - (Sqrt[a]*(-4*A*b + a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/
Sqrt[1 + (b*x^2)/a]))/(8*b^(3/2))

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fricas [A]  time = 0.78, size = 155, normalized size = 1.78 \[ \left [-\frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, B b^{2} x^{3} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{2}}, \frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B b^{2} x^{3} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.49, size = 69, normalized size = 0.79 \[ \frac {1}{8} \, {\left (2 \, B x^{2} + \frac {B a b + 4 \, A b^{2}}{b^{2}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 96, normalized size = 1.10 \[ \frac {A a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}-\frac {B \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, A x}{2}-\frac {\sqrt {b \,x^{2}+a}\, B a x}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B x}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.12, size = 81, normalized size = 0.93 \[ \frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (B\,x^2+A\right )\,\sqrt {b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 6.30, size = 144, normalized size = 1.66 \[ \frac {A \sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {A a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {B a^{\frac {3}{2}} x}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {B b x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*sqrt(a)*x*sqrt(1 + b*x**2/a)/2 + A*a*asinh(sqrt(b)*x/sqrt(a))/(2*sqrt(b)) + B*a**(3/2)*x/(8*b*sqrt(1 + b*x**
2/a)) + 3*B*sqrt(a)*x**3/(8*sqrt(1 + b*x**2/a)) - B*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*b**(3/2)) + B*b*x**5/(4*s
qrt(a)*sqrt(1 + b*x**2/a))

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