3.4.56 \(\int \frac {1}{x (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=54 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2}{a^2 \sqrt {a+b x}}+\frac {2}{3 a (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 208} \begin {gather*} \frac {2}{a^2 \sqrt {a+b x}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2}{3 a (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x)^(5/2)),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.59 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b x}{a}+1\right )}{3 a (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 49, normalized size = 0.91 \begin {gather*} \frac {2 (3 (a+b x)+a)}{3 a^2 (a+b x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(a + b*x)^(5/2)),x]

[Out]

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

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fricas [B]  time = 1.02, size = 177, normalized size = 3.28 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x + 4 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}}, \frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x + 4 \, a^{2}\right )} \sqrt {b x + a}\right )}}{3 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.00, size = 45, normalized size = 0.83 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {2 \, {\left (3 \, b x + 4 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 43, normalized size = 0.80 \begin {gather*} \frac {2}{3 \left (b x +a \right )^{\frac {3}{2}} a}-\frac {2 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {2}{\sqrt {b x +a}\, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x+a)^(5/2),x)

[Out]

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

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maxima [A]  time = 2.91, size = 53, normalized size = 0.98 \begin {gather*} \frac {\log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, b x + 4 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 42, normalized size = 0.78 \begin {gather*} \frac {\frac {2\,\left (a+b\,x\right )}{a^2}+\frac {2}{3\,a}}{{\left (a+b\,x\right )}^{3/2}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x)^(5/2)),x)

[Out]

((2*(a + b*x))/a^2 + 2/(3*a))/(a + b*x)^(3/2) - (2*atanh((a + b*x)^(1/2)/a^(1/2)))/a^(5/2)

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sympy [B]  time = 2.99, size = 697, normalized size = 12.91 \begin {gather*} \frac {8 a^{7} \sqrt {1 + \frac {b x}{a}}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {3 a^{7} \log {\left (\frac {b x}{a} \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} - \frac {6 a^{7} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {14 a^{6} b x \sqrt {1 + \frac {b x}{a}}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {9 a^{6} b x \log {\left (\frac {b x}{a} \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} - \frac {18 a^{6} b x \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {6 a^{5} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {9 a^{5} b^{2} x^{2} \log {\left (\frac {b x}{a} \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} - \frac {18 a^{5} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} + \frac {3 a^{4} b^{3} x^{3} \log {\left (\frac {b x}{a} \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} - \frac {6 a^{4} b^{3} x^{3} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} + 9 a^{\frac {17}{2}} b x + 9 a^{\frac {15}{2}} b^{2} x^{2} + 3 a^{\frac {13}{2}} b^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)**(5/2),x)

[Out]

8*a**7*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 3*a**
7*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 6*a**7*log(sqrt
(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 14*a**6*b*x
*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a**6*b*x*
log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*a**6*b*x*log(s
qrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 6*a**5*b
**2*x**2*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a
**5*b**2*x**2*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*
a**5*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b
**3*x**3) + 3*a**4*b**3*x**3*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b
**3*x**3) - 6*a**4*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 +
3*a**(13/2)*b**3*x**3)

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