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

Optimal. Leaf size=82 \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {15}{4 a^3 \sqrt {x}}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}+\frac {1}{2 a \sqrt {x} (a+b x)^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 82, 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, 205} \begin {gather*} \frac {5}{4 a^2 \sqrt {x} (a+b x)}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {15}{4 a^3 \sqrt {x}}+\frac {1}{2 a \sqrt {x} (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x)^3),x]

[Out]

-15/(4*a^3*Sqrt[x]) + 1/(2*a*Sqrt[x]*(a + b*x)^2) + 5/(4*a^2*Sqrt[x]*(a + b*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*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 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)^3} \, dx &=\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5 \int \frac {1}{x^{3/2} (a+b x)^2} \, dx}{4 a}\\ &=\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}+\frac {15 \int \frac {1}{x^{3/2} (a+b x)} \, dx}{8 a^2}\\ &=-\frac {15}{4 a^3 \sqrt {x}}+\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}-\frac {(15 b) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 a^3}\\ &=-\frac {15}{4 a^3 \sqrt {x}}+\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=-\frac {15}{4 a^3 \sqrt {x}}+\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 25, normalized size = 0.30 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b x}{a}\right )}{a^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x)^3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 70, normalized size = 0.85 \begin {gather*} \frac {-8 a^2-25 a b x-15 b^2 x^2}{4 a^3 \sqrt {x} (a+b x)^2}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2 - 25*a*b*x - 15*b^2*x^2)/(4*a^3*Sqrt[x]*(a + b*x)^2) - (15*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/
(4*a^(7/2))

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fricas [A]  time = 0.72, size = 214, normalized size = 2.61 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} 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 (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^
2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*
sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^
2 + a^5*x)]

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giac [A]  time = 1.14, size = 59, normalized size = 0.72 \begin {gather*} -\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2}{a^{3} \sqrt {x}} - \frac {7 \, b^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*b*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/(a^3*sqrt(x)) - 1/4*(7*b^2*x^(3/2) + 9*a*b*sqrt(x))/((
b*x + a)^2*a^3)

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maple [A]  time = 0.02, size = 66, normalized size = 0.80 \begin {gather*} -\frac {7 b^{2} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} a^{3}}-\frac {9 b \sqrt {x}}{4 \left (b x +a \right )^{2} a^{2}}-\frac {15 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{3}}-\frac {2}{a^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a^3/x^(1/2)-7/4/a^3*b^2/(b*x+a)^2*x^(3/2)-9/4/a^2*b/(b*x+a)^2*x^(1/2)-15/4/a^3*b/(a*b)^(1/2)*arctan(1/(a*b)
^(1/2)*b*x^(1/2))

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maxima [A]  time = 2.99, size = 73, normalized size = 0.89 \begin {gather*} -\frac {15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} - \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*b^2*x^2 + 25*a*b*x + 8*a^2)/(a^3*b^2*x^(5/2) + 2*a^4*b*x^(3/2) + a^5*sqrt(x)) - 15/4*b*arctan(b*sqrt(
x)/sqrt(a*b))/(sqrt(a*b)*a^3)

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mupad [B]  time = 0.15, size = 70, normalized size = 0.85 \begin {gather*} -\frac {\frac {2}{a}+\frac {15\,b^2\,x^2}{4\,a^3}+\frac {25\,b\,x}{4\,a^2}}{a^2\,\sqrt {x}+b^2\,x^{5/2}+2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(a + b*x)^3),x)

[Out]

- (2/a + (15*b^2*x^2)/(4*a^3) + (25*b*x)/(4*a^2))/(a^2*x^(1/2) + b^2*x^(5/2) + 2*a*b*x^(3/2)) - (15*b^(1/2)*at
an((b^(1/2)*x^(1/2))/a^(1/2)))/(4*a^(7/2))

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sympy [A]  time = 54.35, size = 865, normalized size = 10.55 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{a^{3} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{7 b^{3} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {16 i a^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {50 i a^{\frac {3}{2}} b x \sqrt {\frac {1}{b}}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {30 i \sqrt {a} b^{2} x^{2} \sqrt {\frac {1}{b}}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 a^{2} \sqrt {x} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 a^{2} \sqrt {x} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {30 a b x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 a b x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} + 16 i a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 i a^{\frac {7}{2}} b^{2} x^{\frac {5}{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)**3,x)

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(a**3*sqrt(x)), Eq(b, 0)), (-2/(7*b**3*x**(7/2)), Eq(a, 0))
, (-16*I*a**(5/2)*sqrt(1/b)/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/
2)*b**2*x**(5/2)*sqrt(1/b)) - 50*I*a**(3/2)*b*x*sqrt(1/b)/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x
**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*I*sqrt(a)*b**2*x**2*sqrt(1/b)/(8*I*a**(11/2)*sq
rt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*a**2*sqrt(x)
*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8
*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*a**2*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(
x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*a*b*x**(3/2)*lo
g(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*
a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*a*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*
sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*log(
-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a*
*(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*s
qrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)), True))

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