3.1.35 \(\int \frac {A+B x}{x^3 (a+b x^2)^{3/2}} \, dx\) [35]

Optimal. Leaf size=95 \[ \frac {A+B x}{a x^2 \sqrt {a+b x^2}}-\frac {3 A \sqrt {a+b x^2}}{2 a^2 x^2}-\frac {2 B \sqrt {a+b x^2}}{a^2 x}+\frac {3 A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {837, 849, 821, 272, 65, 214} \begin {gather*} \frac {3 A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3 A \sqrt {a+b x^2}}{2 a^2 x^2}-\frac {2 B \sqrt {a+b x^2}}{a^2 x}+\frac {A+B x}{a x^2 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 272

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 821

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 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 849

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

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Mathematica [A]
time = 0.34, size = 83, normalized size = 0.87 \begin {gather*} \frac {-a (A+2 B x)-b x^2 (3 A+4 B x)}{2 a^2 x^2 \sqrt {a+b x^2}}-\frac {3 A b \tanh ^{-1}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 106, normalized size = 1.12

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 a^{2} x^{2}}-\frac {b A}{a^{2} \sqrt {b \,x^{2}+a}}-\frac {b B x}{a^{2} \sqrt {b \,x^{2}+a}}+\frac {3 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A b}{2 a^{\frac {5}{2}}}\)	\(88\)
default	\(A \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )+B \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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/a/x/(b*x^2+a)^(1/2)-2*b/a^2*x/(b*x^2+a)^(1/2))

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Maxima [A]
time = 0.32, size = 89, normalized size = 0.94 \begin {gather*} -\frac {2 \, B b x}{\sqrt {b x^{2} + a} a^{2}} + \frac {3 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {3 \, A b}{2 \, \sqrt {b x^{2} + a} a^{2}} - \frac {B}{\sqrt {b x^{2} + a} a x} - \frac {A}{2 \, \sqrt {b x^{2} + a} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.40, size = 211, normalized size = 2.22 \begin {gather*} \left [\frac {3 \, {\left (A b^{2} x^{4} + A a b x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (4 \, B a b x^{3} + 3 \, A a b x^{2} + 2 \, B a^{2} x + A a^{2}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, -\frac {3 \, {\left (A b^{2} x^{4} + A a b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (4 \, B a b x^{3} + 3 \, A a b x^{2} + 2 \, B a^{2} x + A a^{2}\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 4.25, size = 124, normalized size = 1.31 \begin {gather*} A \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 ) + B \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-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))) + B*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1
)))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (79) = 158\).
time = 0.78, size = 171, normalized size = 1.80 \begin {gather*} -\frac {\frac {B b x}{a^{2}} + \frac {A b}{a^{2}}}{\sqrt {b x^{2} + a}} - \frac {3 \, A b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.59, size = 94, normalized size = 0.99 \begin {gather*} \frac {3\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}}-\frac {3\,A\,b}{2\,a^2\,\sqrt {b\,x^2+a}}-\frac {A}{2\,a\,x^2\,\sqrt {b\,x^2+a}}-\frac {\sqrt {b\,x^2+a}\,\left (\frac {B}{a}+\frac {2\,B\,b\,x^2}{a^2}\right )}{b\,x^3+a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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