3.291 \(\int \frac{\sqrt{a+b x}}{x^4} \, dx\)

Optimal. Leaf size=87 \[ \frac{b^2 \sqrt{a+b x}}{8 a^2 x}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{b \sqrt{a+b x}}{12 a x^2}-\frac{\sqrt{a+b x}}{3 x^3} \]

[Out]

-Sqrt[a + b*x]/(3*x^3) - (b*Sqrt[a + b*x])/(12*a*x^2) + (b^2*Sqrt[a + b*x])/(8*a^2*x) - (b^3*ArcTanh[Sqrt[a +
b*x]/Sqrt[a]])/(8*a^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0269141, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 51, 63, 208} \[ \frac{b^2 \sqrt{a+b x}}{8 a^2 x}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{b \sqrt{a+b x}}{12 a x^2}-\frac{\sqrt{a+b x}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]/x^4,x]

[Out]

-Sqrt[a + b*x]/(3*x^3) - (b*Sqrt[a + b*x])/(12*a*x^2) + (b^2*Sqrt[a + b*x])/(8*a^2*x) - (b^3*ArcTanh[Sqrt[a +
b*x]/Sqrt[a]])/(8*a^(5/2))

Rule 47

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

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{\sqrt{a+b x}}{x^4} \, dx &=-\frac{\sqrt{a+b x}}{3 x^3}+\frac{1}{6} b \int \frac{1}{x^3 \sqrt{a+b x}} \, dx\\ &=-\frac{\sqrt{a+b x}}{3 x^3}-\frac{b \sqrt{a+b x}}{12 a x^2}-\frac{b^2 \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{8 a}\\ &=-\frac{\sqrt{a+b x}}{3 x^3}-\frac{b \sqrt{a+b x}}{12 a x^2}+\frac{b^2 \sqrt{a+b x}}{8 a^2 x}+\frac{b^3 \int \frac{1}{x \sqrt{a+b x}} \, dx}{16 a^2}\\ &=-\frac{\sqrt{a+b x}}{3 x^3}-\frac{b \sqrt{a+b x}}{12 a x^2}+\frac{b^2 \sqrt{a+b x}}{8 a^2 x}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{8 a^2}\\ &=-\frac{\sqrt{a+b x}}{3 x^3}-\frac{b \sqrt{a+b x}}{12 a x^2}+\frac{b^2 \sqrt{a+b x}}{8 a^2 x}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0104371, size = 35, normalized size = 0.4 \[ \frac{2 b^3 (a+b x)^{3/2} \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{b x}{a}+1\right )}{3 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]/x^4,x]

[Out]

(2*b^3*(a + b*x)^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x)/a])/(3*a^4)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 65, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( 1/16\,{\frac{ \left ( bx+a \right ) ^{5/2}}{{a}^{2}}}-1/6\,{\frac{ \left ( bx+a \right ) ^{3/2}}{a}}-1/16\,\sqrt{bx+a} \right ) }-1/16\,{\frac{1}{{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)^(1/2)/x^4,x)

[Out]

2*b^3*((1/16/a^2*(b*x+a)^(5/2)-1/6/a*(b*x+a)^(3/2)-1/16*(b*x+a)^(1/2))/b^3/x^3-1/16*arctanh((b*x+a)^(1/2)/a^(1
/2))/a^(5/2))

________________________________________________________________________________________

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)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.45495, size = 347, normalized size = 3.99 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a^{3} x^{3}}, \frac{3 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sq
rt(b*x + a))/(a^3*x^3), 1/24*(3*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b^2*x^2 - 2*a^2*b*x -
8*a^3)*sqrt(b*x + a))/(a^3*x^3)]

________________________________________________________________________________________

Sympy [A]  time = 7.4923, size = 122, normalized size = 1.4 \begin{align*} - \frac{a}{3 \sqrt{b} x^{\frac{7}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{5 \sqrt{b}}{12 x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{b^{\frac{3}{2}}}{24 a x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{b^{\frac{5}{2}}}{8 a^{2} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} - \frac{b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{8 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/x**4,x)

[Out]

-a/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 5*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) + b**(3/2)/(24*a*x**(3/2
)*sqrt(a/(b*x) + 1)) + b**(5/2)/(8*a**2*sqrt(x)*sqrt(a/(b*x) + 1)) - b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*
a**(5/2))

________________________________________________________________________________________

Giac [A]  time = 1.22449, size = 113, normalized size = 1.3 \begin{align*} \frac{\frac{3 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 8 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} - 3 \, \sqrt{b x + a} a^{2} b^{4}}{a^{2} b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*b^4 - 8*(b*x + a)^(3/2)*a*b^4 -
3*sqrt(b*x + a)*a^2*b^4)/(a^2*b^3*x^3))/b