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

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

[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))

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Rubi [A]  time = 0.0730243, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {835, 807, 266, 63, 208} \[ \frac{2 A c \sqrt{a+c x^2}}{3 a^2 x}+\frac{B c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+c x^2}}{3 a x^3}-\frac{B \sqrt{a+c x^2}}{2 a x^2} \]

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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.152729, size = 73, normalized size = 0.75 \[ \frac{\sqrt{a+c x^2} \left (\frac{-2 a A-3 a B x+4 A c x^2}{x^3}+\frac{3 B c \tanh ^{-1}\left (\sqrt{\frac{c x^2}{a}+1}\right )}{\sqrt{\frac{c x^2}{a}+1}}\right )}{6 a^2} \]

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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Maple [A]  time = 0.01, size = 87, normalized size = 0.9 \begin{align*} -{\frac{A}{3\,a{x}^{3}}\sqrt{c{x}^{2}+a}}+{\frac{2\,Ac}{3\,{a}^{2}x}\sqrt{c{x}^{2}+a}}-{\frac{B}{2\,a{x}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{Bc}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

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*a^(1/2)*(c*x^2+a)^(1/2))/x)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61304, size = 351, normalized size = 3.62 \begin{align*} \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{align*}

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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Sympy [A]  time = 4.09461, size = 97, normalized size = 1. \begin{align*} - \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{align*}

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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Giac [A]  time = 1.19154, size = 204, normalized size = 2.1 \begin{align*} -\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{align*}

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)