\(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^4} \, dx\) [513]

Optimal result

Integrand size = 22, antiderivative size = 66 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=-\frac {B \sqrt {a+b x^2}}{x}-\frac {A \left (a+b x^2\right )^{3/2}}{3 a x^3}+\sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {462, 283, 223, 212} \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=-\frac {A \left (a+b x^2\right )^{3/2}}{3 a x^3}+\sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {B \sqrt {a+b x^2}}{x} \]

[In]

Int[(Sqrt[a + b*x^2]*(A + B*x^2))/x^4,x]

[Out]

-((B*Sqrt[a + b*x^2])/x) - (A*(a + b*x^2)^(3/2))/(3*a*x^3) + Sqrt[b]*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^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 283

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

Rule 462

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=\frac {\sqrt {a+b x^2} \left (-a A-A b x^2-3 a B x^2\right )}{3 a x^3}-\sqrt {b} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]

[In]

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 2.90 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (A b \,x^{2}+3 B a \,x^{2}+A a \right )}{3 x^{3} a}+B \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )\)	\(57\)
pseudoelliptic	\(\frac {3 a \sqrt {b}\, B \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) x^{3}-\left (\left (3 x^{2} B +A \right ) a +A b \,x^{2}\right ) \sqrt {b \,x^{2}+a}}{3 a \,x^{3}}\)	\(65\)
default	\(-\frac {A \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 a \,x^{3}}+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )\)	\(81\)

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.08 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=\left [\frac {3 \, B a \sqrt {b} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (3 \, B a + A b\right )} x^{2} + A a\right )} \sqrt {b x^{2} + a}}{6 \, a x^{3}}, -\frac {3 \, B a \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (3 \, B a + A b\right )} x^{2} + A a\right )} \sqrt {b x^{2} + a}}{3 \, a x^{3}}\right ] \]

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} - \frac {B \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + B \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {B b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=B \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {\sqrt {b x^{2} + a} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, a x^{3}} \]

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (54) = 108\).

Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.29 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=-\frac {1}{2} \, B \sqrt {b} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a \sqrt {b} + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + 3 \, B a^{3} \sqrt {b} + A a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \]

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.84 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^4} \, dx=-\frac {B\,\sqrt {b\,x^2+a}}{x}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{3\,a\,x^3}-\frac {B\,\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}\,\sqrt {\frac {b\,x^2}{a}+1}} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/x^4,x)

[Out]

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