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

Optimal. Leaf size=122 \[ \frac {B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac {B c \sqrt {a+c x^2}}{8 a x^2}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4} \]

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Rubi [A]  time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {835, 807, 266, 47, 63, 208} \begin {gather*} \frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac {B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac {B c \sqrt {a+c x^2}}{8 a x^2}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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) \sqrt {a+c x^2}}{x^6} \, dx &=-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {\int \frac {(-5 a B+2 A c x) \sqrt {a+c x^2}}{x^5} \, dx}{5 a}\\ &=-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {\int \frac {(-8 a A c-5 a B c x) \sqrt {a+c x^2}}{x^4} \, dx}{20 a^2}\\ &=-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac {(B c) \int \frac {\sqrt {a+c x^2}}{x^3} \, dx}{4 a}\\ &=-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac {(B c) \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x^2} \, dx,x,x^2\right )}{8 a}\\ &=\frac {B c \sqrt {a+c x^2}}{8 a x^2}-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac {\left (B c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{16 a}\\ &=\frac {B c \sqrt {a+c x^2}}{8 a x^2}-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac {(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}\\ &=\frac {B c \sqrt {a+c x^2}}{8 a x^2}-\frac {A \left (a+c x^2\right )^{3/2}}{5 a x^5}-\frac {B \left (a+c x^2\right )^{3/2}}{4 a x^4}+\frac {2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac {B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.51 \begin {gather*} -\frac {\left (a+c x^2\right )^{3/2} \left (a A \left (3 a-2 c x^2\right )+5 B c^2 x^5 \, _2F_1\left (\frac {3}{2},3;\frac {5}{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)*Sqrt[a + c*x^2])/x^6,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 106, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-24 a^2 A-30 a^2 B x-8 a A c x^2-15 a B c x^3+16 A c^2 x^4\right )}{120 a^2 x^5}-\frac {B c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{4 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-24*a^2*A - 30*a^2*B*x - 8*a*A*c*x^2 - 15*a*B*c*x^3 + 16*A*c^2*x^4))/(120*a^2*x^5) - (B*c^2*
ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(4*a^(3/2))

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fricas [A]  time = 0.48, size = 190, normalized size = 1.56 \begin {gather*} \left [\frac {15 \, 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 (16 \, A c^{2} x^{4} - 15 \, B a c x^{3} - 8 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, a^{2} x^{5}}, -\frac {15 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (16 \, A c^{2} x^{4} - 15 \, B a c x^{3} - 8 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, a^{2} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(15*B*sqrt(a)*c^2*x^5*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(16*A*c^2*x^4 - 15*B*a*c*
x^3 - 8*A*a*c*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))/(a^2*x^5), -1/120*(15*B*sqrt(-a)*c^2*x^5*arctan(sq
rt(-a)/sqrt(c*x^2 + a)) - (16*A*c^2*x^4 - 15*B*a*c*x^3 - 8*A*a*c*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))
/(a^2*x^5)]

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giac [B]  time = 0.19, size = 267, normalized size = 2.19 \begin {gather*} -\frac {B c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a} + \frac {15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B c^{2} + 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a c^{2} + 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} A a c^{\frac {5}{2}} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {5}{2}} - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{3} c^{2} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{3} c^{\frac {5}{2}} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c^{2} - 16 \, A a^{4} c^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*B*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/60*(15*(sqrt(c)*x - sqrt(c*x^2 + a
))^9*B*c^2 + 90*(sqrt(c)*x - sqrt(c*x^2 + a))^7*B*a*c^2 + 240*(sqrt(c)*x - sqrt(c*x^2 + a))^6*A*a*c^(5/2) + 80
*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(5/2) - 90*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^3*c^2 + 80*(sqrt(c)*x
- sqrt(c*x^2 + a))^2*A*a^3*c^(5/2) - 15*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^4*c^2 - 16*A*a^4*c^(5/2))/(((sqrt(c)
*x - sqrt(c*x^2 + a))^2 - a)^5*a)

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maple [A]  time = 0.06, size = 126, normalized size = 1.03 \begin {gather*} \frac {B \,c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+a}\, B \,c^{2}}{8 a^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B c}{8 a^{2} x^{2}}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A c}{15 a^{2} x^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B}{4 a \,x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 114, normalized size = 0.93 \begin {gather*} \frac {B c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {\sqrt {c x^{2} + a} B c^{2}}{8 \, a^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c}{8 \, a^{2} x^{2}} + \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c}{15 \, a^{2} x^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B}{4 \, a x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.37, size = 95, normalized size = 0.78 \begin {gather*} \frac {B\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{3/2}}-\frac {B\,\sqrt {c\,x^2+a}}{8\,x^4}-\frac {B\,{\left (c\,x^2+a\right )}^{3/2}}{8\,a\,x^4}-\frac {A\,\sqrt {c\,x^2+a}\,\left (3\,a^2+a\,c\,x^2-2\,c^2\,x^4\right )}{15\,a^2\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.25, size = 173, normalized size = 1.42 \begin {gather*} - \frac {A \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a x^{2}} + \frac {2 A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{2}} - \frac {B a}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B c^{\frac {3}{2}}}{8 a x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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