3.2.44 \(\int \frac {\sqrt {(a+b x)^3}}{x^2} \, dx\) [144]

Optimal. Leaf size=93 \[ -\frac {\sqrt {(a+b x)^5}}{a x}+\frac {3 b \left (2 \sqrt {a+b x} \left (a+\frac {1}{3} (a+b x)\right )+a^{3/2} \log \left (\frac {-\sqrt {a}+\sqrt {a+b x}}{\sqrt {a}+\sqrt {a+b x}}\right )\right )}{2 a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 80, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1973, 43, 52, 65, 214} \begin {gather*} -\frac {3 b \sqrt {(a+b x)^3} \text {arctanh}\left (\sqrt {\frac {b x}{a}+1}\right )}{a \left (\frac {b x}{a}+1\right )^{3/2}}+\frac {3 b \sqrt {(a+b x)^3}}{a+b x}-\frac {\sqrt {(a+b x)^3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[(a + b*x)^3]/x^2,x]

[Out]

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

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 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]

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 {gather*} \begin {aligned} \text {Integral} &=\frac {\sqrt {(a+b x)^3} \int \frac {\left (1+\frac {b x}{a}\right )^{3/2}}{x^2} \, dx}{\left (1+\frac {b x}{a}\right )^{3/2}}\\ &=-\frac {\sqrt {(a+b x)^3}}{x}+\frac {\left (3 b \sqrt {(a+b x)^3}\right ) \int \frac {\sqrt {1+\frac {b x}{a}}}{x} \, dx}{2 a \left (1+\frac {b x}{a}\right )^{3/2}}\\ &=-\frac {\sqrt {(a+b x)^3}}{x}+\frac {3 b \sqrt {(a+b x)^3}}{a+b x}+\frac {\left (3 b \sqrt {(a+b x)^3}\right ) \int \frac {1}{x \sqrt {1+\frac {b x}{a}}} \, dx}{2 a \left (1+\frac {b x}{a}\right )^{3/2}}\\ &=-\frac {\sqrt {(a+b x)^3}}{x}+\frac {3 b \sqrt {(a+b x)^3}}{a+b x}+\frac {\left (3 \sqrt {(a+b x)^3}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {a x^2}{b}} \, dx,x,\sqrt {1+\frac {b x}{a}}\right )}{\left (1+\frac {b x}{a}\right )^{3/2}}\\ &=-\frac {\sqrt {(a+b x)^3}}{x}+\frac {3 b \sqrt {(a+b x)^3}}{a+b x}-\frac {3 b \sqrt {(a+b x)^3} \text {arctanh}\left (\sqrt {1+\frac {b x}{a}}\right )}{a \left (1+\frac {b x}{a}\right )^{3/2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 67, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {(a+b x)^3} \left ((a-2 b x) \sqrt {a+b x}+3 \sqrt {a} b x \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{x (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(a + b*x)^3]/x^2,x]

[Out]

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

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Maple [A]
time = 0.14, size = 68, normalized size = 0.73

method	result	size
default	\(-\frac {\sqrt {\left (b x +a \right )^{3}}\, \left (-2 b x \sqrt {b x +a}\, \sqrt {a}+3 \,\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b x +\sqrt {b x +a}\, a^{\frac {3}{2}}\right )}{\left (b x +a \right )^{\frac {3}{2}} x \sqrt {a}}\)	\(68\)
risch	\(-\frac {a \sqrt {\left (b x +a \right )^{3}}}{\left (b x +a \right ) x}+\frac {b \left (4 \sqrt {b x +a}-6 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right ) \sqrt {\left (b x +a \right )^{3}}}{2 \left (b x +a \right )^{\frac {3}{2}}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x + a)^3)/x^2, x)

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Fricas [A]
time = 0.62, size = 252, normalized size = 2.71 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} + a b x\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 (2 \, b x - a\right )}}{2 \, {\left (b x^{2} + a x\right )}}, \frac {3 \, {\left (b^{2} x^{2} + a b x\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 (2 \, b x - a\right )}}{b x^{2} + a x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b^2*x^2 + a*b*x)*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)*(2*b*x - a))/(b*x^2 + a*x), (3*
(b^2*x^2 + a*b*x)*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)*(2*b*x - a))/(b*x^2 + a*x)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (a + b x\right )^{3}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((a + b*x)**3)/x**2, x)

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Giac [A]
time = 0.42, size = 56, normalized size = 0.60 \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)^(1/2)/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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {{\left (a+b\,x\right )}^3}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^3)^(1/2)/x^2,x)

[Out]

int(((a + b*x)^3)^(1/2)/x^2, x)

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