3.5.60 \(\int \frac {1}{x^{3/2} (a+b x)^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {3}{a^2 \sqrt {x}}+\frac {1}{a \sqrt {x} (a+b x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 56, 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, 205} \begin {gather*} -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {3}{a^2 \sqrt {x}}+\frac {1}{a \sqrt {x} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x)^2),x]

[Out]

-3/(a^2*Sqrt[x]) + 1/(a*Sqrt[x]*(a + b*x)) - (3*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

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Mathematica [C]  time = 0.00, size = 25, normalized size = 0.45 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b x}{a}\right )}{a^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x)^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 54, normalized size = 0.96 \begin {gather*} \frac {-2 a-3 b x}{a^2 \sqrt {x} (a+b x)}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 147, normalized size = 2.62 \begin {gather*} \left [\frac {3 \, {\left (b x^{2} + a x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (3 \, b x + 2 \, a\right )} \sqrt {x}}{2 \, {\left (a^{2} b x^{2} + a^{3} x\right )}}, \frac {3 \, {\left (b x^{2} + a x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (3 \, b x + 2 \, a\right )} \sqrt {x}}{a^{2} b x^{2} + a^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.01, size = 49, normalized size = 0.88 \begin {gather*} -\frac {3 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {3 \, b x + 2 \, a}{{\left (b x^{\frac {3}{2}} + a \sqrt {x}\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 48, normalized size = 0.86 \begin {gather*} -\frac {3 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}-\frac {b \sqrt {x}}{\left (b x +a \right ) a^{2}}-\frac {2}{a^{2} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(b*x+a)^2,x)

[Out]

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

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maxima [A]  time = 2.94, size = 51, normalized size = 0.91 \begin {gather*} -\frac {3 \, b x + 2 \, a}{a^{2} b x^{\frac {3}{2}} + a^{3} \sqrt {x}} - \frac {3 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.12, size = 48, normalized size = 0.86 \begin {gather*} -\frac {\frac {2}{a}+\frac {3\,b\,x}{a^2}}{a\,\sqrt {x}+b\,x^{3/2}}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(a + b*x)^2),x)

[Out]

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

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sympy [A]  time = 17.73, size = 434, normalized size = 7.75 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b^{2} x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{a^{2} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {4 i a^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {6 i \sqrt {a} b x \sqrt {\frac {1}{b}}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {3 a \sqrt {x} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} + \frac {3 a \sqrt {x} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {3 b x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} + \frac {3 b x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 2 i a^{\frac {5}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x+a)**2,x)

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b**2*x**(5/2)), Eq(a, 0)), (-2/(a**2*sqrt(x)), Eq(b, 0))
, (-4*I*a**(3/2)*sqrt(1/b)/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sqrt(1/b)) - 6*I*sqrt(a)*
b*x*sqrt(1/b)/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sqrt(1/b)) - 3*a*sqrt(x)*log(-I*sqrt(a
)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sqrt(1/b)) + 3*a*sqrt(x)*log(
I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sqrt(1/b)) - 3*b*x**(
3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sqrt(1/b))
+ 3*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b*x**(3/2)*sq
rt(1/b)), True))

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