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

Optimal. Leaf size=97 \[ \frac {B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {835, 807, 266, 63, 208} \begin {gather*} \frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}+\frac {B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[a + c*x^2])/(3*a*x^3) - (B*Sqrt[a + c*x^2])/(2*a*x^2) + (2*A*c*Sqrt[a + c*x^2])/(3*a^2*x) + (B*c*ArcT
anh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2))

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]

Rule 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{x^4 \sqrt {a+c x^2}} \, dx &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {\int \frac {-3 a B+2 A c x}{x^3 \sqrt {a+c x^2}} \, dx}{3 a}\\ &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2}+\frac {\int \frac {-4 a A c-3 a B c x}{x^2 \sqrt {a+c x^2}} \, dx}{6 a^2}\\ &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2}+\frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}-\frac {(B c) \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{2 a}\\ &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2}+\frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}-\frac {(B c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2}+\frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}-\frac {B \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 a}\\ &=-\frac {A \sqrt {a+c x^2}}{3 a x^3}-\frac {B \sqrt {a+c x^2}}{2 a x^2}+\frac {2 A c \sqrt {a+c x^2}}{3 a^2 x}+\frac {B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 73, normalized size = 0.75 \begin {gather*} \frac {\sqrt {a+c x^2} \left (\frac {-2 a A-3 a B x+4 A c x^2}{x^3}+\frac {3 B c \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )}{\sqrt {\frac {c x^2}{a}+1}}\right )}{6 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*((-2*a*A - 3*a*B*x + 4*A*c*x^2)/x^3 + (3*B*c*ArcTanh[Sqrt[1 + (c*x^2)/a]])/Sqrt[1 + (c*x^2)/a
]))/(6*a^2)

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IntegrateAlgebraic [A]  time = 0.37, size = 80, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-2 a A-3 a B x+4 A c x^2\right )}{6 a^2 x^3}-\frac {B c \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-2*a*A - 3*a*B*x + 4*A*c*x^2))/(6*a^2*x^3) - (B*c*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x
^2]/Sqrt[a]])/a^(3/2)

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fricas [A]  time = 0.46, size = 142, normalized size = 1.46 \begin {gather*} \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 (4 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, a^{2} x^{3}}, -\frac {3 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (4 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, a^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.22, size = 151, normalized size = 1.56 \begin {gather*} -\frac {B c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B c + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{2} c - 4 \, A a^{2} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*c
+ 12*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^2*c - 4*A*a^2*c^(3/2))
/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3*a)

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maple [A]  time = 0.06, size = 87, normalized size = 0.90 \begin {gather*} \frac {B c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}+\frac {2 \sqrt {c \,x^{2}+a}\, A c}{3 a^{2} x}-\frac {\sqrt {c \,x^{2}+a}\, B}{2 a \,x^{2}}-\frac {\sqrt {c \,x^{2}+a}\, A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 75, normalized size = 0.77 \begin {gather*} \frac {B c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} + \frac {2 \, \sqrt {c x^{2} + a} A c}{3 \, a^{2} x} - \frac {\sqrt {c x^{2} + a} B}{2 \, a x^{2}} - \frac {\sqrt {c x^{2} + a} A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.64, size = 66, normalized size = 0.68 \begin {gather*} \frac {B\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {B\,\sqrt {c\,x^2+a}}{2\,a\,x^2}-\frac {A\,\sqrt {c\,x^2+a}\,\left (a-2\,c\,x^2\right )}{3\,a^2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.79, size = 97, normalized size = 1.00 \begin {gather*} - \frac {A \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a x^{2}} + \frac {2 A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{2}} - \frac {B \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 a x} + \frac {B c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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