3.1.7 \(\int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx\)

Optimal. Leaf size=80 \[ -\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}-\frac {A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {811, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}-\frac {A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 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 208

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx &=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}-\frac {\int \frac {-2 a A b-4 a b B x}{x \sqrt {a+b x^2}} \, dx}{4 a}\\ &=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\frac {1}{2} (A b) \int \frac {1}{x \sqrt {a+b x^2}} \, dx+(b B) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\frac {1}{4} (A b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )+(b B) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {1}{2} A \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 108, normalized size = 1.35 \begin {gather*} -\frac {\sqrt {a+b x^2} \left (a \sqrt {\frac {b x^2}{a}+1} (A+2 B x)+A b x^2 \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )-2 \sqrt {a} \sqrt {b} B x^2 \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{2 a x^2 \sqrt {\frac {b x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 96, normalized size = 1.20 \begin {gather*} \frac {\sqrt {a+b x^2} (-A-2 B x)}{2 x^2}+\frac {A b \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\sqrt {b} B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 377, normalized size = 4.71 \begin {gather*} \left [\frac {2 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{4 \, a x^{2}}, -\frac {4 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{4 \, a x^{2}}, \frac {A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}}, -\frac {2 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.48, size = 163, normalized size = 2.04 \begin {gather*} \frac {A b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - B \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 121, normalized size = 1.51 \begin {gather*} -\frac {A b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}}+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 {\sqrt {b \,x^{2}+a}\, A b}{2 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}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 83, normalized size = 1.04 \begin {gather*} B \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {\sqrt {b x^{2} + a} A b}{2 \, a} - \frac {\sqrt {b x^{2} + a} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.79, size = 94, normalized size = 1.18 \begin {gather*} -\frac {A\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {B\,\sqrt {b\,x^2+a}}{x}-\frac {A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}}-\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)^(1/2)*(A + B*x))/x^3,x)

[Out]

- (A*(a + b*x^2)^(1/2))/(2*x^2) - (B*(a + b*x^2)^(1/2))/x - (A*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(1/2))
- (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 = 5.50, size = 107, normalized size = 1.34 \begin {gather*} - \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {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+A)*(b*x**2+a)**(1/2)/x**3,x)

[Out]

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