3.6.10 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x} \, dx\) [510]

Optimal. Leaf size=59 \[ A \sqrt {a+b x^2}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b}-\sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 81, 52, 65, 214} \begin {gather*} A \sqrt {a+b x^2}-\sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {B \left (a+b x^2\right )^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 59, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x^2} \left (3 A b+a B+b B x^2\right )}{3 b}-\sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 57, normalized size = 0.97

method	result	size
default	\(\frac {B \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+A \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 45, normalized size = 0.76 \begin {gather*} -A \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \sqrt {b x^{2} + a} A + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.78, size = 123, normalized size = 2.08 \begin {gather*} \left [\frac {3 \, A \sqrt {a} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B b x^{2} + B a + 3 \, A b\right )} \sqrt {b x^{2} + a}}{6 \, b}, \frac {3 \, A \sqrt {-a} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (B b x^{2} + B a + 3 \, A b\right )} \sqrt {b x^{2} + a}}{3 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*A*sqrt(a)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(B*b*x^2 + B*a + 3*A*b)*sqrt(b*x^2
+ a))/b, 1/3*(3*A*sqrt(-a)*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (B*b*x^2 + B*a + 3*A*b)*sqrt(b*x^2 + a))/b]

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Sympy [A]
time = 10.93, size = 76, normalized size = 1.29 \begin {gather*} - \frac {A \left (- \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - 2 \sqrt {a + b x^{2}}\right )}{2} - \frac {B \left (\begin {cases} - \sqrt {a} x^{2} & \text {for}\: b = 0 \\- \frac {2 \left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.61, size = 60, normalized size = 1.02 \begin {gather*} \frac {A a \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2} + 3 \, \sqrt {b x^{2} + a} A b^{3}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/3*((b*x^2 + a)^(3/2)*B*b^2 + 3*sqrt(b*x^2 + a)*A*b^3)/b^3

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Mupad [B]
time = 0.42, size = 47, normalized size = 0.80 \begin {gather*} A\,\sqrt {b\,x^2+a}+\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3\,b}-A\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(a + b*x^2)^(1/2) + (B*(a + b*x^2)^(3/2))/(3*b) - A*a^(1/2)*atanh((a + b*x^2)^(1/2)/a^(1/2))

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