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

Optimal. Leaf size=66 \[ -\frac{A \left (a+b x^2\right )^{3/2}}{3 a x^3}-\frac{B \sqrt{a+b x^2}}{x}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

[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]]

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Rubi [A]  time = 0.0249793, antiderivative size = 66, normalized size of antiderivative = 1., 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 = {451, 277, 217, 206} \[ -\frac{A \left (a+b x^2\right )^{3/2}}{3 a x^3}-\frac{B \sqrt{a+b x^2}}{x}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

Antiderivative was successfully verified.

[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 451

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]))

Rule 277

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 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 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])

Rubi steps

\begin{align*} \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}+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) \operatorname{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]  time = 0.165591, size = 81, normalized size = 1.23 \[ \frac{\sqrt{a+b x^2} \left (\frac{3 \sqrt{a} \sqrt{b} B \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}-\frac{a A+3 a B x^2+A b x^2}{x^3}\right )}{3 a} \]

Antiderivative was successfully verified.

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

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Maple [A]  time = 0.009, size = 75, normalized size = 1.1 \begin{align*} -{\frac{A}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{B}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{bBx}{a}\sqrt{b{x}^{2}+a}}+B\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5914, size = 331, normalized size = 5.02 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [A]  time = 2.55674, size = 107, normalized size = 1.62 \begin{align*} - \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [B]  time = 1.20675, size = 204, normalized size = 3.09 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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