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

Optimal. Leaf size=71 \[ -\frac {A \left (a+c x^2\right )^{3/2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 x^2}-\frac {B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {807, 266, 47, 63, 208} \[ -\frac {A \left (a+c x^2\right )^{3/2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 x^2}-\frac {B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 85, normalized size = 1.20 \[ \frac {-\left (a+c x^2\right ) \left (2 a A+3 a B x+2 A c x^2\right )-3 a B c x^3 \sqrt {\frac {c x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )}{6 a x^3 \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + c*x^2)*(2*a*A + 3*a*B*x + 2*A*c*x^2)) - 3*a*B*c*x^3*Sqrt[1 + (c*x^2)/a]*ArcTanh[Sqrt[1 + (c*x^2)/a]])/
(6*a*x^3*Sqrt[a + c*x^2])

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fricas [A]  time = 0.95, size = 142, normalized size = 2.00 \[ \left [\frac {3 \, B \sqrt {a} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, A c x^{2} + 3 \, B a x + 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, a x^{3}}, \frac {3 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, A c x^{2} + 3 \, B a x + 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, a x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*B*sqrt(a)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(2*A*c*x^2 + 3*B*a*x + 2*A*a)
*sqrt(c*x^2 + a))/(a*x^3), 1/6*(3*B*sqrt(-a)*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (2*A*c*x^2 + 3*B*a*x + 2
*A*a)*sqrt(c*x^2 + a))/(a*x^3)]

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giac [B]  time = 0.22, size = 143, normalized size = 2.01 \[ \frac {B c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B c + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{2} c + 2 \, A a^{2} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*c + 6*
(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^2*c + 2*A*a^2*c^(3/2))/((sqrt(
c)*x - sqrt(c*x^2 + a))^2 - a)^3

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maple [A]  time = 0.07, size = 84, normalized size = 1.18 \[ -\frac {B c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}}+\frac {\sqrt {c \,x^{2}+a}\, B c}{2 a}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B}{2 a \,x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 72, normalized size = 1.01 \[ -\frac {B c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {\sqrt {c x^{2} + a} B c}{2 \, a} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B}{2 \, a x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.89, size = 55, normalized size = 0.77 \[ -\frac {B\,\sqrt {c\,x^2+a}}{2\,x^2}-\frac {A\,{\left (c\,x^2+a\right )}^{3/2}}{3\,a\,x^3}-\frac {B\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.21, size = 92, normalized size = 1.30 \[ - \frac {A \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a} - \frac {B \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {B c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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