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

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

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Rubi [A]  time = 0.12, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {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} \end {gather*}

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 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^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*}

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Mathematica [C]  time = 0.03, size = 64, normalized size = 0.44 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.65, size = 115, normalized size = 0.78 \begin {gather*} \frac {A c^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {\sqrt {a+c x^2} \left (-40 a^2 A-48 a^2 B x-10 a A c x^2-16 a B c x^3+15 A c^2 x^4+32 B c^2 x^5\right )}{240 a^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-40*a^2*A - 48*a^2*B*x - 10*a*A*c*x^2 - 16*a*B*c*x^3 + 15*A*c^2*x^4 + 32*B*c^2*x^5))/(240*a^
2*x^6) + (A*c^3*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(5/2))

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fricas [A]  time = 0.51, size = 219, normalized size = 1.49 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.23, size = 325, normalized size = 2.21 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 147, normalized size = 1.00 \begin {gather*} -\frac {A \,c^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+a}\, A \,c^{3}}{16 a^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{2}}{16 a^{3} x^{2}}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B c}{15 a^{2} x^{3}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A c}{8 a^{2} x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B}{5 a \,x^{5}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A}{6 a \,x^{6}} \end {gather*}

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]

-1/5*B*(c*x^2+a)^(3/2)/a/x^5+2/15*B*c*(c*x^2+a)^(3/2)/a^2/x^3-1/6*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*c^2/a^3/x^2*(c*x^2+a)^(3/2)-1/16*A*c^3/a^(5/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+1/16*A
*c^3/a^3*(c*x^2+a)^(1/2)

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maxima [A]  time = 0.59, size = 135, normalized size = 0.92 \begin {gather*} -\frac {A c^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {\sqrt {c x^{2} + a} A c^{3}}{16 \, a^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{2}}{16 \, a^{3} x^{2}} + \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c}{15 \, a^{2} x^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A c}{8 \, a^{2} x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B}{5 \, a x^{5}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A}{6 \, a x^{6}} \end {gather*}

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]

-1/16*A*c^3*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(5/2) + 1/16*sqrt(c*x^2 + a)*A*c^3/a^3 - 1/16*(c*x^2 + a)^(3/2)*A*
c^2/(a^3*x^2) + 2/15*(c*x^2 + a)^(3/2)*B*c/(a^2*x^3) + 1/8*(c*x^2 + a)^(3/2)*A*c/(a^2*x^4) - 1/5*(c*x^2 + a)^(
3/2)*B/(a*x^5) - 1/6*(c*x^2 + a)^(3/2)*A/(a*x^6)

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mupad [B]  time = 2.70, size = 116, normalized size = 0.79 \begin {gather*} \frac {A\,{\left (c\,x^2+a\right )}^{5/2}}{16\,a^2\,x^6}-\frac {A\,{\left (c\,x^2+a\right )}^{3/2}}{6\,a\,x^6}-\frac {A\,\sqrt {c\,x^2+a}}{16\,x^6}-\frac {B\,\sqrt {c\,x^2+a}\,\left (3\,a^2+a\,c\,x^2-2\,c^2\,x^4\right )}{15\,a^2\,x^5}+\frac {A\,c^3\,\mathrm {atan}\left (\frac {\sqrt {c\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(1/2)*(A + B*x))/x^7,x)

[Out]

(A*c^3*atan(((a + c*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(5/2)) - (A*(a + c*x^2)^(1/2))/(16*x^6) - (A*(a + c*x^2)
^(3/2))/(6*a*x^6) + (A*(a + c*x^2)^(5/2))/(16*a^2*x^6) - (B*(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 = 8.57, size = 201, normalized size = 1.37 \begin {gather*} - \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 {gather*}

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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