\(\int \frac {(a+b x)^{3/2}}{x} \, dx\) [296]

Optimal result

Integrand size = 13, antiderivative size = 49 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=2 a \sqrt {a+b x}+\frac {2}{3} (a+b x)^{3/2}-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, 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 = {52, 65, 214} \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 a \sqrt {a+b x}+\frac {2}{3} (a+b x)^{3/2} \]

[In]

Int[(a + b*x)^(3/2)/x,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=\frac {2}{3} \sqrt {a+b x} (4 a+b x)-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

Integrate[(a + b*x)^(3/2)/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (b x +4 a \right )}{3}\)	\(35\)
derivativedivides	\(\frac {2 \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a \sqrt {b x +a}\)	\(38\)
default	\(\frac {2 \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a \sqrt {b x +a}\)	\(38\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.80 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=\left [a^{\frac {3}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{3} \, {\left (b x + 4 \, a\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \frac {2}{3} \, {\left (b x + 4 \, a\right )} \sqrt {b x + a}\right ] \]

[In]

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

[Out]

[a^(3/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/3*(b*x + 4*a)*sqrt(b*x + a), 2*sqrt(-a)*a*arctan(sqr
t(b*x + a)*sqrt(-a)/a) + 2/3*(b*x + 4*a)*sqrt(b*x + a)]

Sympy [A] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.45 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=\frac {8 a^{\frac {3}{2}} \sqrt {1 + \frac {b x}{a}}}{3} + a^{\frac {3}{2}} \log {\left (\frac {b x}{a} \right )} - 2 a^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {2 \sqrt {a} b x \sqrt {1 + \frac {b x}{a}}}{3} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} + 2 \, \sqrt {b x + a} a \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=\frac {2 \, a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} + 2 \, \sqrt {b x + a} a \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x)^{3/2}}{x} \, dx=2\,a\,\sqrt {a+b\,x}+\frac {2\,{\left (a+b\,x\right )}^{3/2}}{3}-2\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right ) \]

[In]

int((a + b*x)^(3/2)/x,x)

[Out]

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