3.325 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{x^7} \, dx\)

Optimal. Leaf size=147 \[ -\frac{A c^2 \sqrt{a+c x^2}}{16 a^2 x^2}-\frac{A c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{5/2}}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5} \]

[Out]

-(A*c^2*Sqrt[a + c*x^2])/(16*a^2*x^2) - (A*(a + c*x^2)^(3/2))/(6*a*x^6) - (B*(a + c*x^2)^(3/2))/(5*a*x^5) + (A
*c*(a + c*x^2)^(3/2))/(8*a^2*x^4) + (2*B*c*(a + c*x^2)^(3/2))/(15*a^2*x^3) - (A*c^3*ArcTanh[Sqrt[a + c*x^2]/Sq
rt[a]])/(16*a^(5/2))

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Rubi [A]  time = 0.116914, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{A c^2 \sqrt{a+c x^2}}{16 a^2 x^2}-\frac{A c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{5/2}}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + c*x^2])/x^7,x]

[Out]

-(A*c^2*Sqrt[a + c*x^2])/(16*a^2*x^2) - (A*(a + c*x^2)^(3/2))/(6*a*x^6) - (B*(a + c*x^2)^(3/2))/(5*a*x^5) + (A
*c*(a + c*x^2)^(3/2))/(8*a^2*x^4) + (2*B*c*(a + c*x^2)^(3/2))/(15*a^2*x^3) - (A*c^3*ArcTanh[Sqrt[a + c*x^2]/Sq
rt[a]])/(16*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 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^7} \, dx &=-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{\int \frac{(-6 a B+3 A c x) \sqrt{a+c x^2}}{x^6} \, dx}{6 a}\\ &=-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{\int \frac{(-15 a A c-12 a B c x) \sqrt{a+c x^2}}{x^5} \, dx}{30 a^2}\\ &=-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}-\frac{\int \frac{\left (48 a^2 B c-15 a A c^2 x\right ) \sqrt{a+c x^2}}{x^4} \, dx}{120 a^3}\\ &=-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac{\left (A c^2\right ) \int \frac{\sqrt{a+c x^2}}{x^3} \, dx}{8 a^2}\\ &=-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac{\left (A c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+c x}}{x^2} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac{A c^2 \sqrt{a+c x^2}}{16 a^2 x^2}-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac{\left (A c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{32 a^2}\\ &=-\frac{A c^2 \sqrt{a+c x^2}}{16 a^2 x^2}-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}+\frac{\left (A c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{16 a^2}\\ &=-\frac{A c^2 \sqrt{a+c x^2}}{16 a^2 x^2}-\frac{A \left (a+c x^2\right )^{3/2}}{6 a x^6}-\frac{B \left (a+c x^2\right )^{3/2}}{5 a x^5}+\frac{A c \left (a+c x^2\right )^{3/2}}{8 a^2 x^4}+\frac{2 B c \left (a+c x^2\right )^{3/2}}{15 a^2 x^3}-\frac{A c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0304825, size = 64, normalized size = 0.44 \[ \frac{\left (a+c x^2\right )^{3/2} \left (a^2 B \left (2 c x^2-3 a\right )+5 A c^3 x^5 \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{c x^2}{a}+1\right )\right )}{15 a^4 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + c*x^2])/x^7,x]

[Out]

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

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Maple [A]  time = 0.013, size = 147, normalized size = 1. \begin{align*} -{\frac{A}{6\,a{x}^{6}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ac}{8\,{a}^{2}{x}^{4}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{A{c}^{2}}{16\,{a}^{3}{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{A{c}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{A{c}^{3}}{16\,{a}^{3}}\sqrt{c{x}^{2}+a}}-{\frac{B}{5\,a{x}^{5}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{2\,Bc}{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^7,x)

[Out]

-1/6*A*(c*x^2+a)^(3/2)/a/x^6+1/8*A*c*(c*x^2+a)^(3/2)/a^2/x^4-1/16*A/a^3*c^2/x^2*(c*x^2+a)^(3/2)-1/16*A/a^(5/2)
*c^3*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)+1/16*A/a^3*c^3*(c*x^2+a)^(1/2)-1/5*B*(c*x^2+a)^(3/2)/a/x^5+2/15*B*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^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82498, size = 529, normalized size = 3.6 \begin{align*} \left [\frac{15 \, A \sqrt{a} c^{3} x^{6} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (32 \, B a c^{2} x^{5} + 15 \, A a c^{2} x^{4} - 16 \, B a^{2} c x^{3} - 10 \, A a^{2} c x^{2} - 48 \, B a^{3} x - 40 \, A a^{3}\right )} \sqrt{c x^{2} + a}}{480 \, a^{3} x^{6}}, \frac{15 \, A \sqrt{-a} c^{3} x^{6} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) +{\left (32 \, B a c^{2} x^{5} + 15 \, A a c^{2} x^{4} - 16 \, B a^{2} c x^{3} - 10 \, A a^{2} c x^{2} - 48 \, B a^{3} x - 40 \, A a^{3}\right )} \sqrt{c x^{2} + a}}{240 \, a^{3} x^{6}}\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^7,x, algorithm="fricas")

[Out]

[1/480*(15*A*sqrt(a)*c^3*x^6*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(32*B*a*c^2*x^5 + 15*A*a*
c^2*x^4 - 16*B*a^2*c*x^3 - 10*A*a^2*c*x^2 - 48*B*a^3*x - 40*A*a^3)*sqrt(c*x^2 + a))/(a^3*x^6), 1/240*(15*A*sqr
t(-a)*c^3*x^6*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (32*B*a*c^2*x^5 + 15*A*a*c^2*x^4 - 16*B*a^2*c*x^3 - 10*A*a^2*
c*x^2 - 48*B*a^3*x - 40*A*a^3)*sqrt(c*x^2 + a))/(a^3*x^6)]

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Sympy [A]  time = 9.0688, size = 201, normalized size = 1.37 \begin{align*} - \frac{A a}{6 \sqrt{c} x^{7} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{5 A \sqrt{c}}{24 x^{5} \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{A c^{\frac{3}{2}}}{48 a x^{3} \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{A c^{\frac{5}{2}}}{16 a^{2} x \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{A c^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{16 a^{\frac{5}{2}}} - \frac{B \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{5 x^{4}} - \frac{B c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{15 a x^{2}} + \frac{2 B c^{\frac{5}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{15 a^{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**7,x)

[Out]

-A*a/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 5*A*sqrt(c)/(24*x**5*sqrt(a/(c*x**2) + 1)) + A*c**(3/2)/(48*a*x**
3*sqrt(a/(c*x**2) + 1)) + A*c**(5/2)/(16*a**2*x*sqrt(a/(c*x**2) + 1)) - A*c**3*asinh(sqrt(a)/(sqrt(c)*x))/(16*
a**(5/2)) - B*sqrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - B*c**(3/2)*sqrt(a/(c*x**2) + 1)/(15*a*x**2) + 2*B*c**(5/
2)*sqrt(a/(c*x**2) + 1)/(15*a**2)

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Giac [B]  time = 1.16027, size = 439, normalized size = 2.99 \begin{align*} \frac{A c^{3} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{2}} - \frac{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{11} A c^{3} - 85 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{9} A a c^{3} - 480 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{8} B a^{2} c^{\frac{5}{2}} - 570 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{7} A a^{2} c^{3} + 320 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{6} B a^{3} c^{\frac{5}{2}} - 570 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} A a^{3} c^{3} - 85 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} A a^{4} c^{3} + 192 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} B a^{5} c^{\frac{5}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} A a^{5} c^{3} - 32 \, B a^{6} c^{\frac{5}{2}}}{120 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{6} a^{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^7,x, algorithm="giac")

[Out]

1/8*A*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/120*(15*(sqrt(c)*x - sqrt(c*x^2 +
a))^11*A*c^3 - 85*(sqrt(c)*x - sqrt(c*x^2 + a))^9*A*a*c^3 - 480*(sqrt(c)*x - sqrt(c*x^2 + a))^8*B*a^2*c^(5/2)
- 570*(sqrt(c)*x - sqrt(c*x^2 + a))^7*A*a^2*c^3 + 320*(sqrt(c)*x - sqrt(c*x^2 + a))^6*B*a^3*c^(5/2) - 570*(sq
rt(c)*x - sqrt(c*x^2 + a))^5*A*a^3*c^3 - 85*(sqrt(c)*x - sqrt(c*x^2 + a))^3*A*a^4*c^3 + 192*(sqrt(c)*x - sqrt(
c*x^2 + a))^2*B*a^5*c^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + a))*A*a^5*c^3 - 32*B*a^6*c^(5/2))/(((sqrt(c)*x - sq
rt(c*x^2 + a))^2 - a)^6*a^2)