\(\int \frac {1}{x^{3/2} (a+b x)^3} \, dx\) [467]

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=-\frac {15}{4 a^3 \sqrt {x}}+\frac {1}{2 a \sqrt {x} (a+b x)^2}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {44, 53, 65, 211} \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {15}{4 a^3 \sqrt {x}}+\frac {5}{4 a^2 \sqrt {x} (a+b x)}+\frac {1}{2 a \sqrt {x} (a+b x)^2} \]

[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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 211

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

Rubi steps \begin{align*} \text {integral}& = \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) \text {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*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=\frac {-8 a^2-25 a b x-15 b^2 x^2}{4 a^3 \sqrt {x} (a+b x)^2}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[In]

Integrate[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))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(-\frac {2 b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{a^{3} \sqrt {x}}\)	\(56\)
default	\(-\frac {2 b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{a^{3} \sqrt {x}}\)	\(56\)
risch	\(-\frac {2}{a^{3} \sqrt {x}}-\frac {b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{4}+\frac {9 a \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{3}}\)	\(57\)

[In]

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

[Out]

-2/a^3*b*((7/8*b*x^(3/2)+9/8*a*x^(1/2))/(b*x+a)^2+15/8/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))-2/a^3/x^(1/2
)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.61 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (75) = 150\).

Time = 28.04 (sec) , antiderivative size = 779, normalized size of antiderivative = 9.50 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=\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 {15 a^{2} \sqrt {x} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {15 a^{2} \sqrt {x} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {16 a^{2} \sqrt {- \frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {30 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {30 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {50 a b x \sqrt {- \frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {30 b^{2} x^{2} \sqrt {- \frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {- \frac {a}{b}} + 16 a^{4} b x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[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))
, (-15*a**2*sqrt(x)*log(sqrt(x) - sqrt(-a/b))/(8*a**5*sqrt(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a
**3*b**2*x**(5/2)*sqrt(-a/b)) + 15*a**2*sqrt(x)*log(sqrt(x) + sqrt(-a/b))/(8*a**5*sqrt(x)*sqrt(-a/b) + 16*a**4
*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)) - 16*a**2*sqrt(-a/b)/(8*a**5*sqrt(x)*sqrt(-a/b) + 16
*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)) - 30*a*b*x**(3/2)*log(sqrt(x) - sqrt(-a/b))/(8*
a**5*sqrt(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)) + 30*a*b*x**(3/2)*l
og(sqrt(x) + sqrt(-a/b))/(8*a**5*sqrt(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqr
t(-a/b)) - 50*a*b*x*sqrt(-a/b)/(8*a**5*sqrt(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/
2)*sqrt(-a/b)) - 15*b**2*x**(5/2)*log(sqrt(x) - sqrt(-a/b))/(8*a**5*sqrt(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sq
rt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)) + 15*b**2*x**(5/2)*log(sqrt(x) + sqrt(-a/b))/(8*a**5*sqrt(x)*sqrt(
-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)) - 30*b**2*x**2*sqrt(-a/b)/(8*a**5*sqr
t(x)*sqrt(-a/b) + 16*a**4*b*x**(3/2)*sqrt(-a/b) + 8*a**3*b**2*x**(5/2)*sqrt(-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=-\frac {15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} - \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=-\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2}{a^{3} \sqrt {x}} - \frac {7 \, b^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{3/2} (a+b x)^3} \, dx=-\frac {\frac {2}{a}+\frac {15\,b^2\,x^2}{4\,a^3}+\frac {25\,b\,x}{4\,a^2}}{a^2\,\sqrt {x}+b^2\,x^{5/2}+2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]

[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))