3.3.91 \(\int \frac {\sqrt {a+b x}}{x^4} \, dx\) [291]

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

[Out]

-1/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(5/2)-1/3*(b*x+a)^(1/2)/x^3-1/12*b*(b*x+a)^(1/2)/a/x^2+1/8*b^2*(b*x+
a)^(1/2)/a^2/x

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Rubi [A]
time = 0.02, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 = {43, 44, 65, 214} \begin {gather*} -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {b^2 \sqrt {a+b x}}{8 a^2 x}-\frac {\sqrt {a+b x}}{3 x^3}-\frac {b \sqrt {a+b x}}{12 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]/x^4,x]

[Out]

-1/3*Sqrt[a + b*x]/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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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 \text {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*}

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Mathematica [A]
time = 0.12, size = 67, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {a+b x} \left (8 a^2+2 a b x-3 b^2 x^2\right )}{24 a^2 x^3}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*(8*a^2 + 2*a*b*x - 3*b^2*x^2))/(a^2*x^3) - (b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2
))

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Maple [A]
time = 0.09, size = 66, normalized size = 0.76

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-3 x^{2} b^{2}+2 a b x +8 a^{2}\right )}{24 x^{3} a^{2}}-\frac {b^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\)	\(56\)
derivativedivides	\(2 b^{3} \left (-\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6 a}+\frac {\sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\)	\(66\)
default	\(2 b^{3} \left (-\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6 a}+\frac {\sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[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))

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Maxima [A]
time = 0.48, size = 121, normalized size = 1.39 \begin {gather*} \frac {b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a^{2} - 3 \, {\left (b x + a\right )}^{2} a^{3} + 3 \, {\left (b x + a\right )} a^{4} - a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(5/2) + 1/24*(3*(b*x + a)^(5/2)*b^3 - 8*(b
*x + a)^(3/2)*a*b^3 - 3*sqrt(b*x + a)*a^2*b^3)/((b*x + a)^3*a^2 - 3*(b*x + a)^2*a^3 + 3*(b*x + a)*a^4 - a^5)

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Fricas [A]
time = 0.42, size = 145, normalized size = 1.67 \begin {gather*} \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 {gather*}

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)]

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Sympy [A]
time = 4.96, size = 122, normalized size = 1.40 \begin {gather*} - \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 {gather*}

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))

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Giac [A]
time = 0.77, size = 84, normalized size = 0.97 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.11, size = 66, normalized size = 0.76 \begin {gather*} \frac {{\left (a+b\,x\right )}^{5/2}}{8\,a^2\,x^3}-\frac {{\left (a+b\,x\right )}^{3/2}}{3\,a\,x^3}-\frac {\sqrt {a+b\,x}}{8\,x^3}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)/x^4,x)

[Out]

(a + b*x)^(5/2)/(8*a^2*x^3) - (a + b*x)^(3/2)/(3*a*x^3) - (a + b*x)^(1/2)/(8*x^3) + (b^3*atan(((a + b*x)^(1/2)
*1i)/a^(1/2))*1i)/(8*a^(5/2))

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