\(\int \frac {(a+b x)^{3/2}}{x^3} \, dx\) [298]

Optimal result

Integrand size = 13, antiderivative size = 62 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=-\frac {3 b \sqrt {a+b x}}{4 x}-\frac {(a+b x)^{3/2}}{2 x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {43, 65, 214} \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {(a+b x)^{3/2}}{2 x^2}-\frac {3 b \sqrt {a+b x}}{4 x} \]

[In]

Int[(a + b*x)^(3/2)/x^3,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=-\frac {\sqrt {a+b x} (2 a+5 b x)}{4 x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[In]

Integrate[(a + b*x)^(3/2)/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (5 b x +2 a \right )}{4 x^{2}}-\frac {3 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\)	\(42\)
derivativedivides	\(2 b^{2} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\)	\(52\)
default	\(2 b^{2} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\)	\(52\)
pseudoelliptic	\(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-2 \sqrt {b x +a}\, a^{\frac {3}{2}}-5 b x \sqrt {b x +a}\, \sqrt {a}}{4 x^{2} \sqrt {a}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.00 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a x^{2}}, \frac {3 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=- \frac {a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{2 x^{\frac {3}{2}}} - \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{4 \sqrt {x}} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 \sqrt {a}} \]

[In]

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

[Out]

-a*sqrt(b)*sqrt(a/(b*x) + 1)/(2*x**(3/2)) - 5*b**(3/2)*sqrt(a/(b*x) + 1)/(4*sqrt(x)) - 3*b**2*asinh(sqrt(a)/(s
qrt(b)*sqrt(x)))/(4*sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=\frac {3 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, \sqrt {a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {b x + a} a b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + a^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x + a} a b^{3}}{b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^{3/2}}{x^3} \, dx=\frac {3\,a\,\sqrt {a+b\,x}}{4\,x^2}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,\sqrt {a}}-\frac {5\,{\left (a+b\,x\right )}^{3/2}}{4\,x^2} \]

[In]

int((a + b*x)^(3/2)/x^3,x)

[Out]

(3*a*(a + b*x)^(1/2))/(4*x^2) - (3*b^2*atanh((a + b*x)^(1/2)/a^(1/2)))/(4*a^(1/2)) - (5*(a + b*x)^(3/2))/(4*x^
2)