3.41 \(\int \frac {A+B x}{x^2 (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac {B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {8 A \sqrt {a+b x^2}}{3 a^3 x}+\frac {4 A+3 B x}{3 a^2 x \sqrt {a+b x^2}}+\frac {A+B x}{3 a x \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {823, 807, 266, 63, 208} \[ \frac {4 A+3 B x}{3 a^2 x \sqrt {a+b x^2}}-\frac {8 A \sqrt {a+b x^2}}{3 a^3 x}-\frac {B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {A+B x}{3 a x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 266

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 807

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 823

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] + Di
st[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])

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 95, normalized size = 0.91 \[ \frac {a^2 (4 B x-3 A)+3 a b x^2 (B x-4 A)-3 \sqrt {a} B x \left (a+b x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-8 A b^2 x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 264, normalized size = 2.54 \[ \left [\frac {3 \, {\left (B b^{2} x^{5} + 2 \, B a b x^{3} + B a^{2} x\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (8 \, A b^{2} x^{4} - 3 \, B a b x^{3} + 12 \, A a b x^{2} - 4 \, B a^{2} x + 3 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, \frac {3 \, {\left (B b^{2} x^{5} + 2 \, B a b x^{3} + B a^{2} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, A b^{2} x^{4} - 3 \, B a b x^{3} + 12 \, A a b x^{2} - 4 \, B a^{2} x + 3 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(B*b^2*x^5 + 2*B*a*b*x^3 + B*a^2*x)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(8
*A*b^2*x^4 - 3*B*a*b*x^3 + 12*A*a*b*x^2 - 4*B*a^2*x + 3*A*a^2)*sqrt(b*x^2 + a))/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a
^5*x), 1/3*(3*(B*b^2*x^5 + 2*B*a*b*x^3 + B*a^2*x)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (8*A*b^2*x^4 - 3
*B*a*b*x^3 + 12*A*a*b*x^2 - 4*B*a^2*x + 3*A*a^2)*sqrt(b*x^2 + a))/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x)]

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giac [A]  time = 0.55, size = 119, normalized size = 1.14 \[ -\frac {{\left ({\left (\frac {5 \, A b^{2} x}{a^{3}} - \frac {3 \, B b}{a^{2}}\right )} x + \frac {6 \, A b}{a^{2}}\right )} x - \frac {4 \, B}{a}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {2 \, B \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 112, normalized size = 1.08 \[ -\frac {4 A b x}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}-\frac {8 A b x}{3 \sqrt {b \,x^{2}+a}\, a^{3}}+\frac {B}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a}-\frac {B \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {5}{2}}}-\frac {A}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} a x}+\frac {B}{\sqrt {b \,x^{2}+a}\, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 100, normalized size = 0.96 \[ -\frac {8 \, A b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, A b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a^{2}} + \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.58, size = 96, normalized size = 0.92 \[ \frac {\frac {B}{3\,a}+\frac {B\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {A\,a^2-8\,A\,{\left (b\,x^2+a\right )}^2+4\,A\,a\,\left (b\,x^2+a\right )}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 24.18, size = 910, normalized size = 8.75 \[ A \left (- \frac {3 a^{2} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {12 a b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {8 b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}}\right ) + B \left (\frac {8 a^{7} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {3 a^{7} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {6 a^{7} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {14 a^{6} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {9 a^{6} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {18 a^{6} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {6 a^{5} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {9 a^{5} b^{2} x^{4} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {18 a^{5} b^{2} x^{4} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {3 a^{4} b^{3} x^{6} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {6 a^{4} b^{3} x^{6} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-3*a**2*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 12*a*b**(11/2)*
x**2*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 8*b**(13/2)*x**4*sqrt(a/(b*x**
2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4)) + B*(8*a**7*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*
a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**7*log(b*x**2/a)/(6*a**(19/2) + 18*a*
*(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**7*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(1
9/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 14*a**6*b*x**2*sqrt(1 + b*x**2/
a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 9*a**6*b*x**2*log(b*
x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 18*a**6*b*x**2*
log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**
6) + 6*a**5*b**2*x**4*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(1
3/2)*b**3*x**6) + 9*a**5*b**2*x**4*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 +
6*a**(13/2)*b**3*x**6) - 18*a**5*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 1
8*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**4*b**3*x**6*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*
x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**4*b**3*x**6*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(1
9/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6))

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