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

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

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Rubi [A]  time = 0.10, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {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} \end {gather*}

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 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^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*}

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.49 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(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]))/(a^3*x^
3)

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IntegrateAlgebraic [A]  time = 0.44, size = 91, normalized size = 0.75 \begin {gather*} \frac {3 A c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {\sqrt {a+c x^2} \left (-6 a A-8 a B x+9 A c x^2+16 B c x^3\right )}{24 a^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-6*a*A - 8*a*B*x + 9*A*c*x^2 + 16*B*c*x^3))/(24*a^2*x^4) + (3*A*c^2*ArcTanh[(Sqrt[c]*x)/Sqrt
[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(4*a^(5/2))

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fricas [A]  time = 0.48, size = 171, normalized size = 1.40 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.21, size = 241, normalized size = 1.98 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 108, normalized size = 0.89 \begin {gather*} -\frac {3 A \,c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}+\frac {2 \sqrt {c \,x^{2}+a}\, B c}{3 a^{2} x}+\frac {3 \sqrt {c \,x^{2}+a}\, A c}{8 a^{2} x^{2}}-\frac {\sqrt {c \,x^{2}+a}\, B}{3 a \,x^{3}}-\frac {\sqrt {c \,x^{2}+a}\, A}{4 a \,x^{4}} \end {gather*}

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]

-1/4*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*c^2/a^(5/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/
2))/x)-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

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maxima [A]  time = 0.64, size = 96, normalized size = 0.79 \begin {gather*} -\frac {3 \, A c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {2 \, \sqrt {c x^{2} + a} B c}{3 \, a^{2} x} + \frac {3 \, \sqrt {c x^{2} + a} A c}{8 \, a^{2} x^{2}} - \frac {\sqrt {c x^{2} + a} B}{3 \, a x^{3}} - \frac {\sqrt {c x^{2} + a} A}{4 \, a x^{4}} \end {gather*}

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]

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

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mupad [B]  time = 1.72, size = 86, normalized size = 0.70 \begin {gather*} \frac {3\,A\,{\left (c\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {5\,A\,\sqrt {c\,x^2+a}}{8\,a\,x^4}-\frac {3\,A\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {B\,\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^5*(a + c*x^2)^(1/2)),x)

[Out]

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

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sympy [A]  time = 5.71, size = 153, normalized size = 1.25 \begin {gather*} - \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 {gather*}

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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