3.4.64 \(\int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac {3 B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{5/2}}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4} \]

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Rubi [A]  time = 0.13, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}-\frac {3 B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^6*Sqrt[a + c*x^2]),x]

[Out]

-(A*Sqrt[a + c*x^2])/(5*a*x^5) - (B*Sqrt[a + c*x^2])/(4*a*x^4) + (4*A*c*Sqrt[a + c*x^2])/(15*a^2*x^3) + (3*B*c
*Sqrt[a + c*x^2])/(8*a^2*x^2) - (8*A*c^2*Sqrt[a + c*x^2])/(15*a^3*x) - (3*B*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^6 \sqrt {a+c x^2}} \, dx &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {\int \frac {-5 a B+4 A c x}{x^5 \sqrt {a+c x^2}} \, dx}{5 a}\\ &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {\int \frac {-16 a A c-15 a B c x}{x^4 \sqrt {a+c x^2}} \, dx}{20 a^2}\\ &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}-\frac {\int \frac {45 a^2 B c-32 a A c^2 x}{x^3 \sqrt {a+c x^2}} \, dx}{60 a^3}\\ &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}+\frac {\int \frac {64 a^2 A c^2+45 a^2 B c^2 x}{x^2 \sqrt {a+c x^2}} \, dx}{120 a^4}\\ &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}+\frac {\left (3 B c^2\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{8 a^2}\\ &=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}+\frac {\left (3 B 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}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}+\frac {(3 B 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}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}-\frac {3 B 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 = 72, normalized size = 0.49 \begin {gather*} -\frac {\sqrt {a+c x^2} \left (A \left (3 a^2-4 a c x^2+8 c^2 x^4\right )+15 B c^2 x^5 \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {c x^2}{a}+1\right )\right )}{15 a^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^6*Sqrt[a + c*x^2]),x]

[Out]

-1/15*(Sqrt[a + c*x^2]*(A*(3*a^2 - 4*a*c*x^2 + 8*c^2*x^4) + 15*B*c^2*x^5*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c
*x^2)/a]))/(a^3*x^5)

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IntegrateAlgebraic [A]  time = 0.52, size = 106, normalized size = 0.72 \begin {gather*} \frac {3 B 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 (-24 a^2 A-30 a^2 B x+32 a A c x^2+45 a B c x^3-64 A c^2 x^4\right )}{120 a^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^6*Sqrt[a + c*x^2]),x]

[Out]

(Sqrt[a + c*x^2]*(-24*a^2*A - 30*a^2*B*x + 32*a*A*c*x^2 + 45*a*B*c*x^3 - 64*A*c^2*x^4))/(120*a^3*x^5) + (3*B*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 = 0.47, size = 190, normalized size = 1.29 \begin {gather*} \left [\frac {45 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (64 \, A c^{2} x^{4} - 45 \, B a c x^{3} - 32 \, A a c x^{2} + 30 \, B a^{2} x + 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, a^{3} x^{5}}, \frac {45 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (64 \, A c^{2} x^{4} - 45 \, B a c x^{3} - 32 \, A a c x^{2} + 30 \, B a^{2} x + 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, a^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^6/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/240*(45*B*sqrt(a)*c^2*x^5*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(64*A*c^2*x^4 - 45*B*a*c*
x^3 - 32*A*a*c*x^2 + 30*B*a^2*x + 24*A*a^2)*sqrt(c*x^2 + a))/(a^3*x^5), 1/120*(45*B*sqrt(-a)*c^2*x^5*arctan(sq
rt(-a)/sqrt(c*x^2 + a)) - (64*A*c^2*x^4 - 45*B*a*c*x^3 - 32*A*a*c*x^2 + 30*B*a^2*x + 24*A*a^2)*sqrt(c*x^2 + a)
)/(a^3*x^5)]

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giac [B]  time = 0.21, size = 241, normalized size = 1.64 \begin {gather*} \frac {3 \, B c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B c^{2} - 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a c^{2} - 640 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {5}{2}} + 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{3} c^{2} + 320 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{3} c^{\frac {5}{2}} - 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c^{2} - 64 \, A a^{4} c^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^6/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

3/4*B*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/60*(45*(sqrt(c)*x - sqrt(c*x^2 +
a))^9*B*c^2 - 210*(sqrt(c)*x - sqrt(c*x^2 + a))^7*B*a*c^2 - 640*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(5/2)
+ 210*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^3*c^2 + 320*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^3*c^(5/2) - 45*(sqrt
(c)*x - sqrt(c*x^2 + a))*B*a^4*c^2 - 64*A*a^4*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^5*a^2)

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maple [A]  time = 0.06, size = 129, normalized size = 0.88 \begin {gather*} -\frac {3 B \,c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}-\frac {8 \sqrt {c \,x^{2}+a}\, A \,c^{2}}{15 a^{3} x}+\frac {3 \sqrt {c \,x^{2}+a}\, B c}{8 a^{2} x^{2}}+\frac {4 \sqrt {c \,x^{2}+a}\, A c}{15 a^{2} x^{3}}-\frac {\sqrt {c \,x^{2}+a}\, B}{4 a \,x^{4}}-\frac {\sqrt {c \,x^{2}+a}\, A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^6/(c*x^2+a)^(1/2),x)

[Out]

-1/5*A*(c*x^2+a)^(1/2)/a/x^5+4/15*A*c*(c*x^2+a)^(1/2)/a^2/x^3-8/15*A*c^2*(c*x^2+a)^(1/2)/a^3/x-1/4*B*(c*x^2+a)
^(1/2)/a/x^4+3/8*B*c*(c*x^2+a)^(1/2)/a^2/x^2-3/8*B*c^2/a^(5/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 0.48, size = 117, normalized size = 0.80 \begin {gather*} -\frac {3 \, B c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {8 \, \sqrt {c x^{2} + a} A c^{2}}{15 \, a^{3} x} + \frac {3 \, \sqrt {c x^{2} + a} B c}{8 \, a^{2} x^{2}} + \frac {4 \, \sqrt {c x^{2} + a} A c}{15 \, a^{2} x^{3}} - \frac {\sqrt {c x^{2} + a} B}{4 \, a x^{4}} - \frac {\sqrt {c x^{2} + a} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^6/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-3/8*B*c^2*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(5/2) - 8/15*sqrt(c*x^2 + a)*A*c^2/(a^3*x) + 3/8*sqrt(c*x^2 + a)*B*
c/(a^2*x^2) + 4/15*sqrt(c*x^2 + a)*A*c/(a^2*x^3) - 1/4*sqrt(c*x^2 + a)*B/(a*x^4) - 1/5*sqrt(c*x^2 + a)*A/(a*x^
5)

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mupad [B]  time = 1.79, size = 99, normalized size = 0.67 \begin {gather*} \frac {3\,B\,{\left (c\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {5\,B\,\sqrt {c\,x^2+a}}{8\,a\,x^4}-\frac {3\,B\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {A\,\sqrt {c\,x^2+a}\,\left (3\,a^2-4\,a\,c\,x^2+8\,c^2\,x^4\right )}{15\,a^3\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^6*(a + c*x^2)^(1/2)),x)

[Out]

(3*B*(a + c*x^2)^(3/2))/(8*a^2*x^4) - (5*B*(a + c*x^2)^(1/2))/(8*a*x^4) - (3*B*c^2*atanh((a + c*x^2)^(1/2)/a^(
1/2)))/(8*a^(5/2)) - (A*(a + c*x^2)^(1/2)*(3*a^2 + 8*c^2*x^4 - 4*a*c*x^2))/(15*a^3*x^5)

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sympy [B]  time = 6.46, size = 408, normalized size = 2.78 \begin {gather*} - \frac {3 A a^{4} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {2 A a^{3} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {3 A a^{2} c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {12 A a c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {8 A c^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {B}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B \sqrt {c}}{8 a x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {3 B c^{\frac {3}{2}}}{8 a^{2} x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**6/(c*x**2+a)**(1/2),x)

[Out]

-3*A*a**4*c**(9/2)*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 2*A*a**3
*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 3*A*a**2*c*
*(13/2)*x**4*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 12*A*a*c**(15/
2)*x**6*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 8*A*c**(17/2)*x**8*
sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - B/(4*sqrt(c)*x**5*sqrt(a/(c
*x**2) + 1)) + B*sqrt(c)/(8*a*x**3*sqrt(a/(c*x**2) + 1)) + 3*B*c**(3/2)/(8*a**2*x*sqrt(a/(c*x**2) + 1)) - 3*B*
c**2*asinh(sqrt(a)/(sqrt(c)*x))/(8*a**(5/2))

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