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

Optimal. Leaf size=120 \[ -\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{5/2}}+\frac {b \sqrt {a+b x^2} (A b-2 a B)}{16 a^2 x^2}+\frac {\sqrt {a+b x^2} (A b-2 a B)}{8 a x^4}-\frac {A \left (a+b x^2\right )^{3/2}}{6 a x^6} \]

[Out]

-1/6*A*(b*x^2+a)^(3/2)/a/x^6-1/16*b^2*(A*b-2*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+1/8*(A*b-2*B*a)*(b*
x^2+a)^(1/2)/a/x^4+1/16*b*(A*b-2*B*a)*(b*x^2+a)^(1/2)/a^2/x^2

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Rubi [A]  time = 0.09, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{5/2}}+\frac {b \sqrt {a+b x^2} (A b-2 a B)}{16 a^2 x^2}+\frac {\sqrt {a+b x^2} (A b-2 a B)}{8 a x^4}-\frac {A \left (a+b x^2\right )^{3/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.51 \[ -\frac {\left (a+b x^2\right )^{3/2} \left (a^3 A+b^2 x^6 (2 a B-A b) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x^2}{a}+1\right )\right )}{6 a^4 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((a + b*x^2)^(3/2)*(a^3*A + b^2*(-(A*b) + 2*a*B)*x^6*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*x^2)/a]))/(a^4
*x^6)

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fricas [A]  time = 0.86, size = 221, normalized size = 1.84 \[ \left [-\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {a} x^{6} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, a^{3} x^{6}}, -\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a^{3} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="fricas")

[Out]

[-1/96*(3*(2*B*a*b^2 - A*b^3)*sqrt(a)*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*(2*B*a^2*
b - A*a*b^2)*x^4 + 8*A*a^3 + 2*(6*B*a^3 + A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^6), -1/48*(3*(2*B*a*b^2 - A*b^
3)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(2*B*a^2*b - A*a*b^2)*x^4 + 8*A*a^3 + 2*(6*B*a^3 + A*a^2
*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^6)]

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giac [A]  time = 0.31, size = 140, normalized size = 1.17 \[ -\frac {\frac {3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {6 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a b^{3} - 6 \, \sqrt {b x^{2} + a} B a^{3} b^{3} - 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4} + 8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{4} + 3 \, \sqrt {b x^{2} + a} A a^{2} b^{4}}{a^{2} b^{3} x^{6}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="giac")

[Out]

-1/48*(3*(2*B*a*b^3 - A*b^4)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (6*(b*x^2 + a)^(5/2)*B*a*b^3 -
6*sqrt(b*x^2 + a)*B*a^3*b^3 - 3*(b*x^2 + a)^(5/2)*A*b^4 + 8*(b*x^2 + a)^(3/2)*A*a*b^4 + 3*sqrt(b*x^2 + a)*A*a^
2*b^4)/(a^2*b^3*x^6))/b

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maple [A]  time = 0.01, size = 197, normalized size = 1.64 \[ -\frac {A \,b^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}+\frac {B \,b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, A \,b^{3}}{16 a^{3}}-\frac {\sqrt {b \,x^{2}+a}\, B \,b^{2}}{8 a^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2}}{16 a^{3} x^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B b}{8 a^{2} x^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A b}{8 a^{2} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{4 a \,x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 174, normalized size = 1.45 \[ \frac {B b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {A b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a} B b^{2}}{8 \, a^{2}} + \frac {\sqrt {b x^{2} + a} A b^{3}}{16 \, a^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{16 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{4 \, a x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{8 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{6 \, a x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.20, size = 134, normalized size = 1.12 \[ \frac {B\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{3/2}}-\frac {B\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {A\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a\,x^6}+\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^2\,x^6}-\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a\,x^4}+\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*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))/(8*x^4) - (A*(a + b*x^2)^
(1/2))/(16*x^6) + (B*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(3/2)) - (A*(a + b*x^2)^(3/2))/(6*a*x^6) + (A*
(a + b*x^2)^(5/2))/(16*a^2*x^6) - (B*(a + b*x^2)^(3/2))/(8*a*x^4)

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sympy [B]  time = 144.35, size = 226, normalized size = 1.88 \[ - \frac {A a}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {3}{2}}}{48 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {5}{2}}}{16 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {5}{2}}} - \frac {B a}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 B \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B b^{\frac {3}{2}}}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**7,x)

[Out]

-A*a/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 5*A*sqrt(b)/(24*x**5*sqrt(a/(b*x**2) + 1)) + A*b**(3/2)/(48*a*x**
3*sqrt(a/(b*x**2) + 1)) + A*b**(5/2)/(16*a**2*x*sqrt(a/(b*x**2) + 1)) - A*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*
a**(5/2)) - B*a/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*B*sqrt(b)/(8*x**3*sqrt(a/(b*x**2) + 1)) - B*b**(3/2)
/(8*a*x*sqrt(a/(b*x**2) + 1)) + B*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(3/2))

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