3.5.100 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^9} \, dx\)

Optimal. Leaf size=156 \[ \frac {b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}}-\frac {b^2 \sqrt {a+b x^2} (5 A b-8 a B)}{128 a^3 x^2}+\frac {b \sqrt {a+b x^2} (5 A b-8 a B)}{192 a^2 x^4}+\frac {\sqrt {a+b x^2} (5 A b-8 a B)}{48 a x^6}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8} \]

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Rubi [A]  time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 78, 47, 51, 63, 208} \begin {gather*} -\frac {b^2 \sqrt {a+b x^2} (5 A b-8 a B)}{128 a^3 x^2}+\frac {b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {b \sqrt {a+b x^2} (5 A b-8 a B)}{192 a^2 x^4}+\frac {\sqrt {a+b x^2} (5 A b-8 a B)}{48 a x^6}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^2]*(A + B*x^2))/x^9,x]

[Out]

((5*A*b - 8*a*B)*Sqrt[a + b*x^2])/(48*a*x^6) + (b*(5*A*b - 8*a*B)*Sqrt[a + b*x^2])/(192*a^2*x^4) - (b^2*(5*A*b
- 8*a*B)*Sqrt[a + b*x^2])/(128*a^3*x^2) - (A*(a + b*x^2)^(3/2))/(8*a*x^8) + (b^3*(5*A*b - 8*a*B)*ArcTanh[Sqrt
[a + b*x^2]/Sqrt[a]])/(128*a^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}+\frac {\left (-\frac {5 A b}{2}+4 a B\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^4} \, dx,x,x^2\right )}{8 a}\\ &=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}-\frac {(b (5 A b-8 a B)) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right )}{96 a}\\ &=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}+\frac {b (5 A b-8 a B) \sqrt {a+b x^2}}{192 a^2 x^4}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}+\frac {\left (b^2 (5 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{128 a^2}\\ &=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}+\frac {b (5 A b-8 a B) \sqrt {a+b x^2}}{192 a^2 x^4}-\frac {b^2 (5 A b-8 a B) \sqrt {a+b x^2}}{128 a^3 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}-\frac {\left (b^3 (5 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{256 a^3}\\ &=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}+\frac {b (5 A b-8 a B) \sqrt {a+b x^2}}{192 a^2 x^4}-\frac {b^2 (5 A b-8 a B) \sqrt {a+b x^2}}{128 a^3 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}-\frac {\left (b^2 (5 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{128 a^3}\\ &=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}+\frac {b (5 A b-8 a B) \sqrt {a+b x^2}}{192 a^2 x^4}-\frac {b^2 (5 A b-8 a B) \sqrt {a+b x^2}}{128 a^3 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}+\frac {b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.40 \begin {gather*} -\frac {\left (a+b x^2\right )^{3/2} \left (3 a^4 A+b^3 x^8 (5 A b-8 a B) \, _2F_1\left (\frac {3}{2},4;\frac {5}{2};\frac {b x^2}{a}+1\right )\right )}{24 a^5 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/x^9,x]

[Out]

-1/24*((a + b*x^2)^(3/2)*(3*a^4*A + b^3*(5*A*b - 8*a*B)*x^8*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x^2)/a]))/(a
^5*x^8)

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IntegrateAlgebraic [A]  time = 0.19, size = 128, normalized size = 0.82 \begin {gather*} \frac {\left (5 A b^4-8 a b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {\sqrt {a+b x^2} \left (-48 a^3 A-64 a^3 B x^2-8 a^2 A b x^2-16 a^2 b B x^4+10 a A b^2 x^4+24 a b^2 B x^6-15 A b^3 x^6\right )}{384 a^3 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[a + b*x^2]*(A + B*x^2))/x^9,x]

[Out]

(Sqrt[a + b*x^2]*(-48*a^3*A - 8*a^2*A*b*x^2 - 64*a^3*B*x^2 + 10*a*A*b^2*x^4 - 16*a^2*b*B*x^4 - 15*A*b^3*x^6 +
24*a*b^2*B*x^6))/(384*a^3*x^8) + ((5*A*b^4 - 8*a*b^3*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(7/2))

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fricas [A]  time = 1.00, size = 269, normalized size = 1.72 \begin {gather*} \left [-\frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {a} x^{8} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} - 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{768 \, a^{4} x^{8}}, \frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} - 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{384 \, a^{4} x^{8}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^9,x, algorithm="fricas")

[Out]

[-1/768*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(a)*x^8*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*(8*B*a
^2*b^2 - 5*A*a*b^3)*x^6 - 48*A*a^4 - 2*(8*B*a^3*b - 5*A*a^2*b^2)*x^4 - 8*(8*B*a^4 + A*a^3*b)*x^2)*sqrt(b*x^2 +
a))/(a^4*x^8), 1/384*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(-a)*x^8*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(8*B*a^2*b^2
- 5*A*a*b^3)*x^6 - 48*A*a^4 - 2*(8*B*a^3*b - 5*A*a^2*b^2)*x^4 - 8*(8*B*a^4 + A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(
a^4*x^8)]

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giac [A]  time = 0.45, size = 194, normalized size = 1.24 \begin {gather*} \frac {\frac {3 \, {\left (8 \, B a b^{4} - 5 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {24 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a b^{4} - 88 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x^{2} + a} B a^{4} b^{4} - 15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{5} + 55 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b^{5} - 73 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 15 \, \sqrt {b x^{2} + a} A a^{3} b^{5}}{a^{3} b^{4} x^{8}}}{384 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^9,x, algorithm="giac")

[Out]

1/384*(3*(8*B*a*b^4 - 5*A*b^5)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + (24*(b*x^2 + a)^(7/2)*B*a*b^4
- 88*(b*x^2 + a)^(5/2)*B*a^2*b^4 + 40*(b*x^2 + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x^2 + a)*B*a^4*b^4 - 15*(b*x^2
+ a)^(7/2)*A*b^5 + 55*(b*x^2 + a)^(5/2)*A*a*b^5 - 73*(b*x^2 + a)^(3/2)*A*a^2*b^5 - 15*sqrt(b*x^2 + a)*A*a^3*b^
5)/(a^3*b^4*x^8))/b

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maple [A]  time = 0.02, size = 239, normalized size = 1.53 \begin {gather*} \frac {5 A \,b^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {7}{2}}}-\frac {B \,b^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}-\frac {5 \sqrt {b \,x^{2}+a}\, A \,b^{4}}{128 a^{4}}+\frac {\sqrt {b \,x^{2}+a}\, B \,b^{3}}{16 a^{3}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{3}}{128 a^{4} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{2}}{16 a^{3} x^{2}}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2}}{64 a^{3} x^{4}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B b}{8 a^{2} x^{4}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A b}{48 a^{2} x^{6}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{6 a \,x^{6}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{8 a \,x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/x^9,x)

[Out]

-1/8*A*(b*x^2+a)^(3/2)/a/x^8+5/48*A*b/a^2/x^6*(b*x^2+a)^(3/2)-5/64*A*b^2/a^3/x^4*(b*x^2+a)^(3/2)+5/128*A*b^3/a
^4/x^2*(b*x^2+a)^(3/2)+5/128*A*b^4/a^(7/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-5/128*A*b^4/a^4*(b*x^2+a)^(1/
2)-1/6*B/a/x^6*(b*x^2+a)^(3/2)+1/8*B*b/a^2/x^4*(b*x^2+a)^(3/2)-1/16*B*b^2/a^3/x^2*(b*x^2+a)^(3/2)-1/16*B*b^3/a
^(5/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+1/16*B*b^3/a^3*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.17, size = 216, normalized size = 1.38 \begin {gather*} -\frac {B b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {5 \, A b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {7}{2}}} + \frac {\sqrt {b x^{2} + a} B b^{3}}{16 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} A b^{4}}{128 \, a^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{16 \, a^{3} x^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{8 \, a^{2} x^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{64 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{6 \, a x^{6}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{8 \, a x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^9,x, algorithm="maxima")

[Out]

-1/16*B*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 5/128*A*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) + 1/16*s
qrt(b*x^2 + a)*B*b^3/a^3 - 5/128*sqrt(b*x^2 + a)*A*b^4/a^4 - 1/16*(b*x^2 + a)^(3/2)*B*b^2/(a^3*x^2) + 5/128*(b
*x^2 + a)^(3/2)*A*b^3/(a^4*x^2) + 1/8*(b*x^2 + a)^(3/2)*B*b/(a^2*x^4) - 5/64*(b*x^2 + a)^(3/2)*A*b^2/(a^3*x^4)
- 1/6*(b*x^2 + a)^(3/2)*B/(a*x^6) + 5/48*(b*x^2 + a)^(3/2)*A*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(3/2)*A/(a*x^8)

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mupad [B]  time = 2.68, size = 173, normalized size = 1.11 \begin {gather*} \frac {55\,A\,{\left (b\,x^2+a\right )}^{5/2}}{384\,a^2\,x^8}-\frac {B\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {73\,A\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^8}-\frac {5\,A\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^3\,x^8}-\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a\,x^6}+\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^2\,x^6}-\frac {A\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{7/2}}+\frac {B\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,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 + B*x^2)*(a + b*x^2)^(1/2))/x^9,x)

[Out]

(B*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(5/2)) - (B*(a + b*x^2)^(1/2))/(16*x^6) - (A*b^4*atan(((
a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(7/2)) - (5*A*(a + b*x^2)^(1/2))/(128*x^8) - (73*A*(a + b*x^2)^(3/2))
/(384*a*x^8) + (55*A*(a + b*x^2)^(5/2))/(384*a^2*x^8) - (5*A*(a + b*x^2)^(7/2))/(128*a^3*x^8) - (B*(a + b*x^2)
^(3/2))/(6*a*x^6) + (B*(a + b*x^2)^(5/2))/(16*a^2*x^6)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**9,x)

[Out]

Timed out

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