\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^5} \, dx\) [531]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {3 b (A b+4 a B) \sqrt {a+b x^2}}{8 a}-\frac {(A b+4 a B) \left (a+b x^2\right )^{3/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}-\frac {3 b (A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 79, 43, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=-\frac {3 b (4 a B+A b) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {\left (a+b x^2\right )^{3/2} (4 a B+A b)}{8 a x^2}+\frac {3 b \sqrt {a+b x^2} (4 a B+A b)}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4} \]

[In]

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

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{3/2} (A+B x)}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}+\frac {(A b+4 a B) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )}{8 a} \\ & = -\frac {(A b+4 a B) \left (a+b x^2\right )^{3/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}+\frac {(3 b (A b+4 a B)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )}{16 a} \\ & = \frac {3 b (A b+4 a B) \sqrt {a+b x^2}}{8 a}-\frac {(A b+4 a B) \left (a+b x^2\right )^{3/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}+\frac {1}{16} (3 b (A b+4 a B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {3 b (A b+4 a B) \sqrt {a+b x^2}}{8 a}-\frac {(A b+4 a B) \left (a+b x^2\right )^{3/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}+\frac {1}{8} (3 (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 ) \\ & = \frac {3 b (A b+4 a B) \sqrt {a+b x^2}}{8 a}-\frac {(A b+4 a B) \left (a+b x^2\right )^{3/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^4}-\frac {3 b (A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a A-5 A b x^2-4 a B x^2+8 b B x^4\right )}{8 x^4}-\frac {3 b (A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {3 \left (b \,x^{4} \left (A b +4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\frac {5 \sqrt {b \,x^{2}+a}\, \left (\frac {2 \left (2 x^{2} B +A \right ) a^{\frac {3}{2}}}{5}+b \,x^{2} \sqrt {a}\, \left (-\frac {8 x^{2} B}{5}+A \right )\right )}{3}\right )}{8 \sqrt {a}\, x^{4}}\)	\(78\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (5 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 x^{4}}+\frac {b \left (8 \sqrt {b \,x^{2}+a}\, B -\frac {\left (3 A b +12 B a \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}\right )}{8}\)	\(88\)
default	\(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )\)	\(182\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=\left [\frac {3 \, {\left (4 \, B a b + A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, B a b x^{4} - 2 \, A a^{2} - {\left (4 \, B a^{2} + 5 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a x^{4}}, \frac {3 \, {\left (4 \, B a b + A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B a b x^{4} - 2 \, A a^{2} - {\left (4 \, B a^{2} + 5 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a x^{4}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (105) = 210\).

Time = 39.62 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.88 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=- \frac {A a^{2}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A a \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {A b^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 \sqrt {a}} - \frac {3 B \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {B a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {B a \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=-\frac {3}{2} \, B \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {3 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {3}{2} \, \sqrt {b x^{2} + a} B b + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{2 \, a} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{2 \, a x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{4 \, a x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {8 \, \sqrt {b x^{2} + a} B b^{2} + \frac {3 \, {\left (4 \, B a b^{2} + A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{2} b^{2} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} - 3 \, \sqrt {b x^{2} + a} A a b^{3}}{b^{2} x^{4}}}{8 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^5} \, dx=B\,b\,\sqrt {b\,x^2+a}-\frac {5\,A\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}-\frac {3\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}+\frac {3\,A\,a\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {B\,a\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {3\,B\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2} \]

[In]

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

[Out]

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