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

Optimal. Leaf size=84 \[ \frac {x \sqrt {a+b x^2} (a B+2 A b)}{2 a}+\frac {(a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}-\frac {A \left (a+b x^2\right )^{3/2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {453, 195, 217, 206} \[ \frac {x \sqrt {a+b x^2} (a B+2 A b)}{2 a}+\frac {(a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}-\frac {A \left (a+b x^2\right )^{3/2}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 453

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 71, normalized size = 0.85 \[ \frac {1}{2} \sqrt {a+b x^2} \left (\frac {(a B+2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \sqrt {\frac {b x^2}{a}+1}}-\frac {2 A}{x}+B x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 134, normalized size = 1.60 \[ \left [\frac {{\left (B a + 2 \, A b\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B b x^{2} - 2 \, A b\right )} \sqrt {b x^{2} + a}}{4 \, b x}, -\frac {{\left (B a + 2 \, A b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (B b x^{2} - 2 \, A b\right )} \sqrt {b x^{2} + a}}{2 \, b x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 84, normalized size = 1.00 \[ \frac {1}{2} \, \sqrt {b x^{2} + a} B x + \frac {2 \, A a \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} - \frac {{\left (B a \sqrt {b} + 2 \, A b^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 93, normalized size = 1.11 \[ A \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {B a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}+\frac {\sqrt {b \,x^{2}+a}\, A b x}{a}+\frac {\sqrt {b \,x^{2}+a}\, B x}{2}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.04, size = 59, normalized size = 0.70 \[ \frac {1}{2} \, \sqrt {b x^{2} + a} B x + \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + A \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {\sqrt {b x^{2} + a} A}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.26, size = 94, normalized size = 1.12 \[ \frac {B\,x\,\sqrt {b\,x^2+a}}{2}-\frac {A\,\sqrt {b\,x^2+a}}{x}+\frac {B\,a\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{2\,\sqrt {b}}-\frac {A\,\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}\,\sqrt {\frac {b\,x^2}{a}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x*(a + b*x^2)^(1/2))/2 - (A*(a + b*x^2)^(1/2))/x + (B*a*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/(2*b^(1/2)) - (
A*b^(1/2)*asin((b^(1/2)*x*1i)/a^(1/2))*(a + b*x^2)^(1/2)*1i)/(a^(1/2)*((b*x^2)/a + 1)^(1/2))

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sympy [A]  time = 4.41, size = 107, normalized size = 1.27 \[ - \frac {A \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B \sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*sqrt(a)/(x*sqrt(1 + b*x**2/a)) + A*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) - A*b*x/(sqrt(a)*sqrt(1 + b*x**2/a)) +
B*sqrt(a)*x*sqrt(1 + b*x**2/a)/2 + B*a*asinh(sqrt(b)*x/sqrt(a))/(2*sqrt(b))

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