3.363 \(\int \frac{A+B x}{x^5 \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3} \]

[Out]

-(A*Sqrt[a + c*x^2])/(4*a*x^4) - (B*Sqrt[a + c*x^2])/(3*a*x^3) + (3*A*c*Sqrt[a + c*x^2])/(8*a^2*x^2) + (2*B*c*
Sqrt[a + c*x^2])/(3*a^2*x) - (3*A*c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(5/2))

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Rubi [A]  time = 0.0995186, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^5*Sqrt[a + c*x^2]),x]

[Out]

-(A*Sqrt[a + c*x^2])/(4*a*x^4) - (B*Sqrt[a + c*x^2])/(3*a*x^3) + (3*A*c*Sqrt[a + c*x^2])/(8*a^2*x^2) + (2*B*c*
Sqrt[a + c*x^2])/(3*a^2*x) - (3*A*c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(5/2))

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^5 \sqrt{a+c x^2}} \, dx &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{\int \frac{-4 a B+3 A c x}{x^4 \sqrt{a+c x^2}} \, dx}{4 a}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{\int \frac{-9 a A c-8 a B c x}{x^3 \sqrt{a+c x^2}} \, dx}{12 a^2}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}-\frac{\int \frac{16 a^2 B c-9 a A c^2 x}{x^2 \sqrt{a+c x^2}} \, dx}{24 a^3}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}+\frac{\left (3 A c^2\right ) \int \frac{1}{x \sqrt{a+c x^2}} \, dx}{8 a^2}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}+\frac{\left (3 A c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}+\frac{(3 A c) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{8 a^2}\\ &=-\frac{A \sqrt{a+c x^2}}{4 a x^4}-\frac{B \sqrt{a+c x^2}}{3 a x^3}+\frac{3 A c \sqrt{a+c x^2}}{8 a^2 x^2}+\frac{2 B c \sqrt{a+c x^2}}{3 a^2 x}-\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0206227, size = 60, normalized size = 0.49 \[ -\frac{\sqrt{a+c x^2} \left (3 A c^2 x^3 \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{c x^2}{a}+1\right )+a B \left (a-2 c x^2\right )\right )}{3 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^5*Sqrt[a + c*x^2]),x]

[Out]

-(Sqrt[a + c*x^2]*(a*B*(a - 2*c*x^2) + 3*A*c^2*x^3*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c*x^2)/a]))/(3*a^3*x^3)

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Maple [A]  time = 0.01, size = 108, normalized size = 0.9 \begin{align*} -{\frac{B}{3\,a{x}^{3}}\sqrt{c{x}^{2}+a}}+{\frac{2\,Bc}{3\,{a}^{2}x}\sqrt{c{x}^{2}+a}}-{\frac{A}{4\,a{x}^{4}}\sqrt{c{x}^{2}+a}}+{\frac{3\,Ac}{8\,{a}^{2}{x}^{2}}\sqrt{c{x}^{2}+a}}-{\frac{3\,A{c}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^5/(c*x^2+a)^(1/2),x)

[Out]

-1/3*B*(c*x^2+a)^(1/2)/a/x^3+2/3*B*c*(c*x^2+a)^(1/2)/a^2/x-1/4*A*(c*x^2+a)^(1/2)/a/x^4+3/8*A*c*(c*x^2+a)^(1/2)
/a^2/x^2-3/8*A/a^(5/2)*c^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^5/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70864, size = 413, normalized size = 3.39 \begin{align*} \left [\frac{9 \, A \sqrt{a} c^{2} x^{4} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (16 \, B a c x^{3} + 9 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{48 \, a^{3} x^{4}}, \frac{9 \, A \sqrt{-a} c^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) +{\left (16 \, B a c x^{3} + 9 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{24 \, a^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(9*A*sqrt(a)*c^2*x^4*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(16*B*a*c*x^3 + 9*A*a*c*x^2
- 8*B*a^2*x - 6*A*a^2)*sqrt(c*x^2 + a))/(a^3*x^4), 1/24*(9*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x^2 + a)
) + (16*B*a*c*x^3 + 9*A*a*c*x^2 - 8*B*a^2*x - 6*A*a^2)*sqrt(c*x^2 + a))/(a^3*x^4)]

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Sympy [A]  time = 6.29403, size = 153, normalized size = 1.25 \begin{align*} - \frac{A}{4 \sqrt{c} x^{5} \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{A \sqrt{c}}{8 a x^{3} \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{3 A c^{\frac{3}{2}}}{8 a^{2} x \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{3 A c^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{8 a^{\frac{5}{2}}} - \frac{B \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{3 a x^{2}} + \frac{2 B c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{3 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**5/(c*x**2+a)**(1/2),x)

[Out]

-A/(4*sqrt(c)*x**5*sqrt(a/(c*x**2) + 1)) + A*sqrt(c)/(8*a*x**3*sqrt(a/(c*x**2) + 1)) + 3*A*c**(3/2)/(8*a**2*x*
sqrt(a/(c*x**2) + 1)) - 3*A*c**2*asinh(sqrt(a)/(sqrt(c)*x))/(8*a**(5/2)) - B*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*a
*x**2) + 2*B*c**(3/2)*sqrt(a/(c*x**2) + 1)/(3*a**2)

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Giac [B]  time = 1.16287, size = 325, normalized size = 2.66 \begin{align*} \frac{3 \, A c^{2} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{2}} - \frac{9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{7} A c^{2} - 33 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} A a c^{2} - 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} B a^{2} c^{\frac{3}{2}} - 33 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} A a^{2} c^{2} + 64 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} B a^{3} c^{\frac{3}{2}} + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} A a^{3} c^{2} - 16 \, B a^{4} c^{\frac{3}{2}}}{12 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{4} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

3/4*A*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/12*(9*(sqrt(c)*x - sqrt(c*x^2 + a
))^7*A*c^2 - 33*(sqrt(c)*x - sqrt(c*x^2 + a))^5*A*a*c^2 - 48*(sqrt(c)*x - sqrt(c*x^2 + a))^4*B*a^2*c^(3/2) - 3
3*(sqrt(c)*x - sqrt(c*x^2 + a))^3*A*a^2*c^2 + 64*(sqrt(c)*x - sqrt(c*x^2 + a))^2*B*a^3*c^(3/2) + 9*(sqrt(c)*x
- sqrt(c*x^2 + a))*A*a^3*c^2 - 16*B*a^4*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^4*a^2)