3.6.80 \(\int \frac {A+B x^2}{x^5 (a+b x^2)^{3/2}} \, dx\) [580]

Optimal. Leaf size=118 \[ \frac {3 b (5 A b-4 a B)}{8 a^3 \sqrt {a+b x^2}}-\frac {A}{4 a x^4 \sqrt {a+b x^2}}+\frac {5 A b-4 a B}{8 a^2 x^2 \sqrt {a+b x^2}}-\frac {3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{7/2}} \]

[Out]

-3/8*b*(5*A*b-4*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(7/2)+3/8*b*(5*A*b-4*B*a)/a^3/(b*x^2+a)^(1/2)-1/4*A/a/
x^4/(b*x^2+a)^(1/2)+1/8*(5*A*b-4*B*a)/a^2/x^2/(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 118, 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 = {457, 79, 44, 53, 65, 214} \begin {gather*} -\frac {3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {3 b (5 A b-4 a B)}{8 a^3 \sqrt {a+b x^2}}+\frac {5 A b-4 a B}{8 a^2 x^2 \sqrt {a+b x^2}}-\frac {A}{4 a x^4 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]

[Out]

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

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

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Mathematica [A]
time = 0.18, size = 100, normalized size = 0.85 \begin {gather*} \frac {-2 a^2 A+5 a A b x^2-4 a^2 B x^2+15 A b^2 x^4-12 a b B x^4}{8 a^3 x^4 \sqrt {a+b x^2}}+\frac {3 b (-5 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]

[Out]

(-2*a^2*A + 5*a*A*b*x^2 - 4*a^2*B*x^2 + 15*A*b^2*x^4 - 12*a*b*B*x^4)/(8*a^3*x^4*Sqrt[a + b*x^2]) + (3*b*(-5*A*
b + 4*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*a^(7/2))

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Maple [A]
time = 0.10, size = 162, normalized size = 1.37

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-7 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 a^{3} x^{4}}+\frac {b^{2} A}{a^{3} \sqrt {b \,x^{2}+a}}-\frac {b B}{a^{2} \sqrt {b \,x^{2}+a}}-\frac {15 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A}{8 a^{\frac {7}{2}}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{2 a^{\frac {5}{2}}}\)	\(133\)
default	\(A \left (-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{2}+a}}-\frac {5 b \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )}{4 a}\right )+B \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 130, normalized size = 1.10 \begin {gather*} \frac {3 \, B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {15 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {7}{2}}} - \frac {3 \, B b}{2 \, \sqrt {b x^{2} + a} a^{2}} + \frac {15 \, A b^{2}}{8 \, \sqrt {b x^{2} + a} a^{3}} - \frac {B}{2 \, \sqrt {b x^{2} + a} a x^{2}} + \frac {5 \, A b}{8 \, \sqrt {b x^{2} + a} a^{2} x^{2}} - \frac {A}{4 \, \sqrt {b x^{2} + a} a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

3/2*B*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) - 15/8*A*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 3/2*B*b/(sq
rt(b*x^2 + a)*a^2) + 15/8*A*b^2/(sqrt(b*x^2 + a)*a^3) - 1/2*B/(sqrt(b*x^2 + a)*a*x^2) + 5/8*A*b/(sqrt(b*x^2 +
a)*a^2*x^2) - 1/4*A/(sqrt(b*x^2 + a)*a*x^4)

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Fricas [A]
time = 1.27, size = 287, normalized size = 2.43 \begin {gather*} \left [-\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}, -\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A*a*b^2)*x^4)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sq
rt(a) + 2*a)/x^2) + 2*(3*(4*B*a^2*b - 5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(
a^4*b*x^6 + a^5*x^4), -1/8*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A*a*b^2)*x^4)*sqrt(-a)*arctan(sqrt(-
a)/sqrt(b*x^2 + a)) + (3*(4*B*a^2*b - 5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(
a^4*b*x^6 + a^5*x^4)]

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Sympy [A]
time = 32.94, size = 180, normalized size = 1.53 \begin {gather*} A \left (- \frac {1}{4 a \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 \sqrt {b}}{8 a^{2} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {15 b^{\frac {3}{2}}}{8 a^{3} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {7}{2}}}\right ) + B \left (- \frac {1}{2 a \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{2 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {5}{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**(3/2),x)

[Out]

A*(-1/(4*a*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + 5*sqrt(b)/(8*a**2*x**3*sqrt(a/(b*x**2) + 1)) + 15*b**(3/2)/(8*
a**3*x*sqrt(a/(b*x**2) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(7/2))) + B*(-1/(2*a*sqrt(b)*x**3*sqrt
(a/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(5/2)))

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Giac [A]
time = 1.06, size = 137, normalized size = 1.16 \begin {gather*} -\frac {3 \, {\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{3}} - \frac {B a b - A b^{2}}{\sqrt {b x^{2} + a} a^{3}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x^{2} + a} B a^{2} b - 7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} + 9 \, \sqrt {b x^{2} + a} A a b^{2}}{8 \, a^{3} b^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-3/8*(4*B*a*b - 5*A*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) - (B*a*b - A*b^2)/(sqrt(b*x^2 + a)*a^
3) - 1/8*(4*(b*x^2 + a)^(3/2)*B*a*b - 4*sqrt(b*x^2 + a)*B*a^2*b - 7*(b*x^2 + a)^(3/2)*A*b^2 + 9*sqrt(b*x^2 + a
)*A*a*b^2)/(a^3*b^2*x^4)

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Mupad [B]
time = 0.91, size = 134, normalized size = 1.14 \begin {gather*} \frac {15\,A\,b^2}{8\,a^3\,\sqrt {b\,x^2+a}}-\frac {3\,B\,b}{2\,a^2\,\sqrt {b\,x^2+a}}-\frac {15\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{7/2}}-\frac {A}{4\,a\,x^4\,\sqrt {b\,x^2+a}}-\frac {B}{2\,a\,x^2\,\sqrt {b\,x^2+a}}+\frac {3\,B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}}+\frac {5\,A\,b}{8\,a^2\,x^2\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x)

[Out]

(15*A*b^2)/(8*a^3*(a + b*x^2)^(1/2)) - (3*B*b)/(2*a^2*(a + b*x^2)^(1/2)) - (15*A*b^2*atanh((a + b*x^2)^(1/2)/a
^(1/2)))/(8*a^(7/2)) - A/(4*a*x^4*(a + b*x^2)^(1/2)) - B/(2*a*x^2*(a + b*x^2)^(1/2)) + (3*B*b*atanh((a + b*x^2
)^(1/2)/a^(1/2)))/(2*a^(5/2)) + (5*A*b)/(8*a^2*x^2*(a + b*x^2)^(1/2))

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