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

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

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Rubi [A]  time = 0.02, antiderivative size = 66, 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 = {451, 277, 217, 206} \begin {gather*} -\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 ) \end {gather*}

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 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 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 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]))

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*}

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Mathematica [A]  time = 0.17, size = 81, normalized size = 1.23 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.12, size = 70, normalized size = 1.06 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-a A-3 a B x^2-A b x^2\right )}{3 a x^3}-\sqrt {b} B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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]]

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fricas [A]  time = 0.88, size = 137, normalized size = 2.08 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.52, size = 151, normalized size = 2.29 \begin {gather*} -\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 {gather*}

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

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maple [A]  time = 0.01, size = 75, normalized size = 1.14 \begin {gather*} B \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {\sqrt {b \,x^{2}+a}\, B b x}{a}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{a x}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{3 a \,x^{3}} \end {gather*}

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]

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

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maxima [A]  time = 1.08, size = 48, normalized size = 0.73 \begin {gather*} 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}} \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^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)

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mupad [B]  time = 1.47, size = 76, normalized size = 1.15 \begin {gather*} -\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}} \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^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))

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sympy [A]  time = 3.35, size = 107, normalized size = 1.62 \begin {gather*} - \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 {gather*}

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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