\(\int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx\) [150]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\frac {\frac {15 b^2}{4 a^3}-\frac {1}{2 a x^2}+\frac {5 b}{4 a^2 x}}{\sqrt {a+b x}}+\frac {15 b^2 \log \left (\frac {-\sqrt {a}+\sqrt {a+b x}}{\sqrt {a}+\sqrt {a+b x}}\right )}{8 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.38, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1973, 44, 53, 65, 214} \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\frac {15 b^2 (a+b x)}{4 a^3 \sqrt {(a+b x)^3}}-\frac {15 b^2 \left (\frac {b x}{a}+1\right )^{3/2} \text {arctanh}\left (\sqrt {\frac {b x}{a}+1}\right )}{4 a^2 \sqrt {(a+b x)^3}}+\frac {5 b (a+b x)}{4 a^2 x \sqrt {(a+b x)^3}}-\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}} \]

[In]

Int[1/(x^3*Sqrt[(a + b*x)^3]),x]

[Out]

(15*b^2*(a + b*x))/(4*a^3*Sqrt[(a + b*x)^3]) - (a + b*x)/(2*a*x^2*Sqrt[(a + b*x)^3]) + (5*b*(a + b*x))/(4*a^2*
x*Sqrt[(a + b*x)^3]) - (15*b^2*(1 + (b*x)/a)^(3/2)*ArcTanh[Sqrt[1 + (b*x)/a]])/(4*a^2*Sqrt[(a + b*x)^3])

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 214

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

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+\frac {b x}{a}\right )^{3/2} \int \frac {1}{x^3 \left (1+\frac {b x}{a}\right )^{3/2}} \, dx}{\sqrt {(a+b x)^3}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}}-\frac {\left (5 b \left (1+\frac {b x}{a}\right )^{3/2}\right ) \int \frac {1}{x^2 \left (1+\frac {b x}{a}\right )^{3/2}} \, dx}{4 a \sqrt {(a+b x)^3}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}}+\frac {5 b (a+b x)}{4 a^2 x \sqrt {(a+b x)^3}}+\frac {\left (15 b^2 \left (1+\frac {b x}{a}\right )^{3/2}\right ) \int \frac {1}{x \left (1+\frac {b x}{a}\right )^{3/2}} \, dx}{8 a^2 \sqrt {(a+b x)^3}} \\ & = \frac {15 b^2 (a+b x)}{4 a^3 \sqrt {(a+b x)^3}}-\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}}+\frac {5 b (a+b x)}{4 a^2 x \sqrt {(a+b x)^3}}+\frac {\left (15 b^2 \left (1+\frac {b x}{a}\right )^{3/2}\right ) \int \frac {1}{x \sqrt {1+\frac {b x}{a}}} \, dx}{8 a^2 \sqrt {(a+b x)^3}} \\ & = \frac {15 b^2 (a+b x)}{4 a^3 \sqrt {(a+b x)^3}}-\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}}+\frac {5 b (a+b x)}{4 a^2 x \sqrt {(a+b x)^3}}+\frac {\left (15 b \left (1+\frac {b x}{a}\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {a x^2}{b}} \, dx,x,\sqrt {1+\frac {b x}{a}}\right )}{4 a \sqrt {(a+b x)^3}} \\ & = \frac {15 b^2 (a+b x)}{4 a^3 \sqrt {(a+b x)^3}}-\frac {a+b x}{2 a x^2 \sqrt {(a+b x)^3}}+\frac {5 b (a+b x)}{4 a^2 x \sqrt {(a+b x)^3}}-\frac {15 b^2 \left (1+\frac {b x}{a}\right )^{3/2} \text {arctanh}\left (\sqrt {1+\frac {b x}{a}}\right )}{4 a^2 \sqrt {(a+b x)^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=-\frac {(a+b x) \left (\sqrt {a} \left (2 a^2-5 a b x-15 b^2 x^2\right )+15 b^2 x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{7/2} x^2 \sqrt {(a+b x)^3}} \]

[In]

Integrate[1/(x^3*Sqrt[(a + b*x)^3]),x]

[Out]

-1/4*((a + b*x)*(Sqrt[a]*(2*a^2 - 5*a*b*x - 15*b^2*x^2) + 15*b^2*x^2*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[
a]]))/(a^(7/2)*x^2*Sqrt[(a + b*x)^3])

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.81

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (15 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-5 a^{\frac {3}{2}} b x -15 b^{2} x^{2} \sqrt {a}+2 a^{\frac {5}{2}}\right )}{4 \sqrt {\left (b x +a \right )^{3}}\, a^{\frac {7}{2}} x^{2}}\)	\(74\)
risch	\(-\frac {\left (b x +a \right )^{2} \left (-7 b x +2 a \right )}{4 a^{3} x^{2} \sqrt {\left (b x +a \right )^{3}}}+\frac {b^{2} \left (\frac {16}{\sqrt {b x +a}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a^{3} \sqrt {\left (b x +a \right )^{3}}}\)	\(85\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (72) = 144\).

Time = 0.26 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.81 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\left [\frac {15 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b^{2} x^{2} + 3 \, a b x + 2 \, a^{2} - 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {a}}{b x^{2} + a x}\right ) + 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )}}{8 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {15 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {-a}}{a b x + a^{2}}\right ) + \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )}}{4 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/8*(15*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*log((b^2*x^2 + 3*a*b*x + 2*a^2 - 2*sqrt(b^3*x^3 + 3*a*b
^2*x^2 + 3*a^2*b*x + a^3)*sqrt(a))/(b*x^2 + a*x)) + 2*sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(15*a*b^2*
x^2 + 5*a^2*b*x - 2*a^3))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), 1/4*(15*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)
*sqrt(-a)*arctan(sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(-a)/(a*b*x + a^2)) + sqrt(b^3*x^3 + 3*a*b^
2*x^2 + 3*a^2*b*x + a^3)*(15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^{3} \sqrt {\left (a + b x\right )^{3}}}\, dx \]

[In]

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

[Out]

Integral(1/(x**3*sqrt((a + b*x)**3)), x)

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{3}} x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^3\,\sqrt {{\left (a+b\,x\right )}^3}} \,d x \]

[In]

int(1/(x^3*((a + b*x)^3)^(1/2)),x)

[Out]

int(1/(x^3*((a + b*x)^3)^(1/2)), x)