3.342 \(\int \frac {1}{x^4 \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=90 \[ \frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {5 b^2 \sqrt {a+b x}}{8 a^3 x}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}-\frac {\sqrt {a+b x}}{3 a x^3} \]

[Out]

5/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(7/2)-1/3*(b*x+a)^(1/2)/x^3/a+5/12*b*(b*x+a)^(1/2)/a^2/x^2-5/8*b^2*(b
*x+a)^(1/2)/a^3/x

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Rubi [A]  time = 0.03, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 208} \[ -\frac {5 b^2 \sqrt {a+b x}}{8 a^3 x}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}-\frac {\sqrt {a+b x}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a + b*x]),x]

[Out]

-Sqrt[a + b*x]/(3*a*x^3) + (5*b*Sqrt[a + b*x])/(12*a^2*x^2) - (5*b^2*Sqrt[a + b*x])/(8*a^3*x) + (5*b^3*ArcTanh
[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(7/2))

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 {1}{x^4 \sqrt {a+b x}} \, dx &=-\frac {\sqrt {a+b x}}{3 a x^3}-\frac {(5 b) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{6 a}\\ &=-\frac {\sqrt {a+b x}}{3 a x^3}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}+\frac {\left (5 b^2\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{8 a^2}\\ &=-\frac {\sqrt {a+b x}}{3 a x^3}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}-\frac {5 b^2 \sqrt {a+b x}}{8 a^3 x}-\frac {\left (5 b^3\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^3}\\ &=-\frac {\sqrt {a+b x}}{3 a x^3}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}-\frac {5 b^2 \sqrt {a+b x}}{8 a^3 x}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^3}\\ &=-\frac {\sqrt {a+b x}}{3 a x^3}+\frac {5 b \sqrt {a+b x}}{12 a^2 x^2}-\frac {5 b^2 \sqrt {a+b x}}{8 a^3 x}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.37 \[ \frac {2 b^3 \sqrt {a+b x} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};\frac {b x}{a}+1\right )}{a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a + b*x]),x]

[Out]

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

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fricas [A]  time = 0.48, size = 145, normalized size = 1.61 \[ \left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{4} x^{3}}, -\frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{4} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a^4*x^3), -1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 - 10*a^
2*b*x + 8*a^3)*sqrt(b*x + a))/(a^4*x^3)]

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giac [A]  time = 0.91, size = 84, normalized size = 0.93 \[ -\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b
^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^3*x^3))/b

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maple [A]  time = 0.01, size = 90, normalized size = 1.00 \[ 2 \left (-\frac {5 \left (-\frac {3 \left (\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}-\frac {\sqrt {b x +a}}{2 a b x}\right )}{4 a}-\frac {\sqrt {b x +a}}{4 a \,b^{2} x^{2}}\right )}{6 a}-\frac {\sqrt {b x +a}}{6 a \,b^{3} x^{3}}\right ) b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*(-1/6*(b*x+a)^(1/2)/a/x^3/b^3-5/6/a*(-1/4*(b*x+a)^(1/2)/a/b^2/x^2-3/4/a*(-1/2*(b*x+a)^(1/2)/a/b/x+1/2*ar
ctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2))))

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maxima [A]  time = 2.92, size = 121, normalized size = 1.34 \[ -\frac {5 \, b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {7}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 33 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a^{3} - 3 \, {\left (b x + a\right )}^{2} a^{4} + 3 \, {\left (b x + a\right )} a^{5} - a^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(7/2) - 1/24*(15*(b*x + a)^(5/2)*b^3 - 40
*(b*x + a)^(3/2)*a*b^3 + 33*sqrt(b*x + a)*a^2*b^3)/((b*x + a)^3*a^3 - 3*(b*x + a)^2*a^4 + 3*(b*x + a)*a^5 - a^
6)

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mupad [B]  time = 0.05, size = 69, normalized size = 0.77 \[ \frac {5\,{\left (a+b\,x\right )}^{3/2}}{3\,a^2\,x^3}-\frac {11\,\sqrt {a+b\,x}}{8\,a\,x^3}-\frac {5\,{\left (a+b\,x\right )}^{5/2}}{8\,a^3\,x^3}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x)^(1/2)),x)

[Out]

(5*(a + b*x)^(3/2))/(3*a^2*x^3) - (11*(a + b*x)^(1/2))/(8*a*x^3) - (5*(a + b*x)^(5/2))/(8*a^3*x^3) - (b^3*atan
(((a + b*x)^(1/2)*1i)/a^(1/2))*5i)/(8*a^(7/2))

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sympy [A]  time = 7.02, size = 129, normalized size = 1.43 \[ - \frac {1}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {\sqrt {b}}{12 a x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 b^{\frac {3}{2}}}{24 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 b^{\frac {5}{2}}}{8 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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