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

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

[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))

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Rubi [A]  time = 0.0860523, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {835, 807, 266, 47, 63, 208} \[ \frac{2 A c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac{B c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 a^{3/2}}-\frac{A \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{B c \sqrt{a+c x^2}}{8 a x^2}-\frac{B \left (a+c x^2\right )^{3/2}}{4 a x^4} \]

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 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 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]

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*}

Mathematica [C]  time = 0.0223373, size = 62, normalized size = 0.51 \[ -\frac{\left (a+c x^2\right )^{3/2} \left (a A \left (3 a-2 c x^2\right )+5 B c^2 x^5 \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x^2}{a}+1\right )\right )}{15 a^3 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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]))/(15*a^3
*x^5)

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Maple [A]  time = 0.012, size = 126, normalized size = 1. \begin{align*} -{\frac{B}{4\,a{x}^{4}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{Bc}{8\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{B{c}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B{c}^{2}}{8\,{a}^{2}}\sqrt{c{x}^{2}+a}}-{\frac{A}{5\,a{x}^{5}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{2\,Ac}{15\,{a}^{2}{x}^{3}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

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]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81095, size = 466, normalized size = 3.82 \begin{align*} \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{align*}

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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Sympy [A]  time = 6.44715, size = 173, normalized size = 1.42 \begin{align*} - \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{align*}

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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Giac [B]  time = 1.14087, size = 360, normalized size = 2.95 \begin{align*} -\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{align*}

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)