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

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

[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)

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Rubi [A]  time = 0.0880568, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.0548632, 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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Maple [A]  time = 0.008, size = 112, normalized size = 1.1 \begin{align*}{\frac{B}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{B}{{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{B\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{A}{ax} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,Abx}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{8\,Abx}{3\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.68784, size = 589, normalized size = 5.66 \begin{align*} \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 ] \end{align*}

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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Sympy [B]  time = 20.0836, size = 910, normalized size = 8.75 \begin{align*} \text{result too large to display} \end{align*}

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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Giac [A]  time = 1.20275, size = 161, normalized size = 1.55 \begin{align*} -\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}} \end{align*}

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)