3.430 \(\int \frac{A+B x}{x^3 \sqrt{a+b x}} \, dx\)

Optimal. Leaf size=84 \[ \frac{\sqrt{a+b x} (3 A b-4 a B)}{4 a^2 x}-\frac{b (3 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{5/2}}-\frac{A \sqrt{a+b x}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0347658, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \[ \frac{\sqrt{a+b x} (3 A b-4 a B)}{4 a^2 x}-\frac{b (3 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{5/2}}-\frac{A \sqrt{a+b x}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^3*Sqrt[a + b*x]),x]

[Out]

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

Rule 78

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] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.145176, size = 73, normalized size = 0.87 \[ \frac{\sqrt{a+b x} \left (\frac{a (3 A b x-2 a (A+2 B x))}{x^2}+\frac{b (4 a B-3 A b) \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*((a*(3*A*b*x - 2*a*(A + 2*B*x)))/x^2 + (b*(-3*A*b + 4*a*B)*ArcTanh[Sqrt[1 + (b*x)/a]])/Sqrt[1 +
(b*x)/a]))/(4*a^3)

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Maple [A]  time = 0.01, size = 81, normalized size = 1. \begin{align*} 2\,b \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( 1/8\,{\frac{ \left ( 3\,Ab-4\,Ba \right ) \left ( bx+a \right ) ^{3/2}}{{a}^{2}}}-1/8\,{\frac{ \left ( 5\,Ab-4\,Ba \right ) \sqrt{bx+a}}{a}} \right ) }-1/8\,{\frac{3\,Ab-4\,Ba}{{a}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(b*x+a)^(1/2),x)

[Out]

2*b*((1/8*(3*A*b-4*B*a)/a^2*(b*x+a)^(3/2)-1/8*(5*A*b-4*B*a)/a*(b*x+a)^(1/2))/b^2/x^2-1/8*(3*A*b-4*B*a)/a^(5/2)
*arctanh((b*x+a)^(1/2)/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^3/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64296, size = 385, normalized size = 4.58 \begin{align*} \left [-\frac{{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt{a} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{b x + a}}{8 \, a^{3} x^{2}}, -\frac{{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{b x + a}}{4 \, a^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 38.4763, size = 156, normalized size = 1.86 \begin{align*} - \frac{A}{2 \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{A \sqrt{b}}{4 a x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{3 A b^{\frac{3}{2}}}{4 a^{2} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} - \frac{3 A b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{5}{2}}} - \frac{B \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{a \sqrt{x}} + \frac{B b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**3/(b*x+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.22662, size = 150, normalized size = 1.79 \begin{align*} -\frac{\frac{{\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{4 \,{\left (b x + a\right )}^{\frac{3}{2}} B a b^{2} - 4 \, \sqrt{b x + a} B a^{2} b^{2} - 3 \,{\left (b x + a\right )}^{\frac{3}{2}} A b^{3} + 5 \, \sqrt{b x + a} A a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

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