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

Optimal. Leaf size=51 \[ -\frac {(a+b x)^{3/2}}{x}+3 b \sqrt {a+b x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 50, 63, 208} \begin {gather*} -\frac {(a+b x)^{3/2}}{x}+3 b \sqrt {a+b x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^2,x]

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 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 208

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 33, normalized size = 0.65 \begin {gather*} \frac {2 b (a+b x)^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b x}{a}+1\right )}{5 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^2,x]

[Out]

(2*b*(a + b*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x)/a])/(5*a^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.06, size = 49, normalized size = 0.96 \begin {gather*} \frac {\sqrt {a+b x} (2 (a+b x)-3 a)}{x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.04, size = 102, normalized size = 2.00 \begin {gather*} \left [\frac {3 \, \sqrt {a} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b x - a\right )} \sqrt {b x + a}}{2 \, x}, \frac {3 \, \sqrt {-a} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b x - a\right )} \sqrt {b x + a}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.04, size = 56, normalized size = 1.10 \begin {gather*} \frac {\frac {3 \, a b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x + a} b^{2} - \frac {\sqrt {b x + a} a b}{x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 47, normalized size = 0.92 \begin {gather*} 2 \left (\left (-\frac {3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b x +a}}{2 b x}\right ) a +\sqrt {b x +a}\right ) b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.87, size = 58, normalized size = 1.14 \begin {gather*} \frac {3}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x + a} b - \frac {\sqrt {b x + a} a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 42, normalized size = 0.82 \begin {gather*} 2\,b\,\sqrt {a+b\,x}-3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )-\frac {a\,\sqrt {a+b\,x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/x^2,x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 2.66, size = 92, normalized size = 1.80 \begin {gather*} - 3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {a^{2}}{\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {a \sqrt {b}}{\sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {3}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________