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

Optimal. Leaf size=76 \[ \frac{3 A+2 B x}{3 a^2 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{A+B x}{3 a \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0634318, antiderivative size = 76, 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, 12, 266, 63, 208} \[ \frac{3 A+2 B x}{3 a^2 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{A+B x}{3 a \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*x)/(3*a*(a + b*x^2)^(3/2)) + (3*A + 2*B*x)/(3*a^2*Sqrt[a + b*x^2]) - (A*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0553323, size = 69, normalized size = 0.91 \[ \frac{a (4 A+3 B x)+b x^2 (3 A+2 B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 92, normalized size = 1.2 \begin{align*}{\frac{Bx}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Bx}{3\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{A}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{A}{{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69197, size = 537, normalized size = 7.07 \begin{align*} \left [\frac{3 \,{\left (A b^{2} x^{4} + 2 \, A a b x^{2} + A a^{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 (2 \, B a b x^{3} + 3 \, A a b x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}}, \frac{3 \,{\left (A b^{2} x^{4} + 2 \, A a b x^{2} + A a^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, B a b x^{3} + 3 \, A a b x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 22.9917, size = 840, normalized size = 11.05 \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/(b*x**2+a)**(5/2),x)

[Out]

A*(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**(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) + 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**(13/2)*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 + 18*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**(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)) + B*(3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)) + 2*b*x**3/(3
*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)))

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Giac [A]  time = 1.19623, size = 111, normalized size = 1.46 \begin{align*} \frac{{\left ({\left (\frac{2 \, B b x}{a^{2}} + \frac{3 \, A b}{a^{2}}\right )} x + \frac{3 \, B}{a}\right )} x + \frac{4 \, A}{a}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{2 \, A \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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