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

Optimal. Leaf size=74 \[ \frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {5 b}{a^3 \sqrt {a+b x}}-\frac {5 b}{3 a^2 (a+b x)^{3/2}}-\frac {1}{a x (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.08, 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} \begin {gather*} -\frac {5 \sqrt {a+b x}}{a^3 x}+\frac {10}{3 a^2 x \sqrt {a+b x}}+\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {2}{3 a x (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(5/2)),x]

[Out]

2/(3*a*x*(a + b*x)^(3/2)) + 10/(3*a^2*x*Sqrt[a + b*x]) - (5*Sqrt[a + b*x])/(a^3*x) + (5*b*ArcTanh[Sqrt[a + b*x
]/Sqrt[a]])/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^2 (a+b x)^{5/2}} \, dx &=\frac {2}{3 a x (a+b x)^{3/2}}+\frac {5 \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx}{3 a}\\ &=\frac {2}{3 a x (a+b x)^{3/2}}+\frac {10}{3 a^2 x \sqrt {a+b x}}+\frac {5 \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{a^2}\\ &=\frac {2}{3 a x (a+b x)^{3/2}}+\frac {10}{3 a^2 x \sqrt {a+b x}}-\frac {5 \sqrt {a+b x}}{a^3 x}-\frac {(5 b) \int \frac {1}{x \sqrt {a+b x}} \, dx}{2 a^3}\\ &=\frac {2}{3 a x (a+b x)^{3/2}}+\frac {10}{3 a^2 x \sqrt {a+b x}}-\frac {5 \sqrt {a+b x}}{a^3 x}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{a^3}\\ &=\frac {2}{3 a x (a+b x)^{3/2}}+\frac {10}{3 a^2 x \sqrt {a+b x}}-\frac {5 \sqrt {a+b x}}{a^3 x}+\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.45 \begin {gather*} -\frac {2 b \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {b x}{a}+1\right )}{3 a^2 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 67, normalized size = 0.91 \begin {gather*} \frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {2 a^2+10 a (a+b x)-15 (a+b x)^2}{3 a^3 x (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2 + 10*a*(a + b*x) - 15*(a + b*x)^2)/(3*a^3*x*(a + b*x)^(3/2)) + (5*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(
7/2)

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fricas [A]  time = 1.13, size = 221, normalized size = 2.99 \begin {gather*} \left [\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{6 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*
x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), -1/3*(15*(b^3*x^3 + 2*a*b^2*x^2
+ a^2*b*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4
*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]

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giac [A]  time = 1.03, size = 65, normalized size = 0.88 \begin {gather*} -\frac {5 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} b + a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x + a}}{a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(6*(b*x + a)*b + a*b)/((b*x + a)^(3/2)*a^3) - sqrt(b*
x + a)/(a^3*x)

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maple [A]  time = 0.01, size = 67, normalized size = 0.91 \begin {gather*} 2 \left (-\frac {1}{3 \left (b x +a \right )^{\frac {3}{2}} a^{2}}-\frac {2}{\sqrt {b x +a}\, a^{3}}-\frac {-\frac {5 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\frac {\sqrt {b x +a}}{2 b x}}{a^{3}}\right ) b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.03, size = 89, normalized size = 1.20 \begin {gather*} -\frac {15 \, {\left (b x + a\right )}^{2} b - 10 \, {\left (b x + a\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} a^{3} - {\left (b x + a\right )}^{\frac {3}{2}} a^{4}\right )}} - \frac {5 \, b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(15*(b*x + a)^2*b - 10*(b*x + a)*a*b - 2*a^2*b)/((b*x + a)^(5/2)*a^3 - (b*x + a)^(3/2)*a^4) - 5/2*b*log((
sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(7/2)

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mupad [B]  time = 0.11, size = 73, normalized size = 0.99 \begin {gather*} \frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {\frac {2\,b}{3\,a}+\frac {10\,b\,\left (a+b\,x\right )}{3\,a^2}-\frac {5\,b\,{\left (a+b\,x\right )}^2}{a^3}}{a\,{\left (a+b\,x\right )}^{3/2}-{\left (a+b\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)^(5/2)),x)

[Out]

(5*b*atanh((a + b*x)^(1/2)/a^(1/2)))/a^(7/2) - ((2*b)/(3*a) + (10*b*(a + b*x))/(3*a^2) - (5*b*(a + b*x)^2)/a^3
)/(a*(a + b*x)^(3/2) - (a + b*x)^(5/2))

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sympy [B]  time = 5.60, size = 818, normalized size = 11.05 \begin {gather*} - \frac {6 a^{17} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {46 a^{16} b x \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {15 a^{16} b x \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {30 a^{16} b x \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {70 a^{15} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {45 a^{15} b^{2} x^{2} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {90 a^{15} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {30 a^{14} b^{3} x^{3} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {45 a^{14} b^{3} x^{3} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {90 a^{14} b^{3} x^{3} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {15 a^{13} b^{4} x^{4} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {30 a^{13} b^{4} x^{4} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4
) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b
**3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2
)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*
x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b**2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a
**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x*
*2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**15*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2
)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b*x/
a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**14*b**3*x**3
*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**14*
b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)
*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*
a**(33/2)*b**3*x**4) + 30*a**13*b**4*x**4*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a
**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)

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