3.3.37 \(\int \frac {\sqrt {a x^2+b x^3}}{x^3} \, dx\) [237]

Optimal. Leaf size=52 \[ -\frac {\sqrt {a x^2+b x^3}}{x^2}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2045, 2033, 212} \begin {gather*} -\frac {\sqrt {a x^2+b x^3}}{x^2}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^3,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2045

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*
x^n)^p/(c*(m + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 64, normalized size = 1.23 \begin {gather*} -\frac {\sqrt {a+b x} \left (\sqrt {a} \sqrt {a+b x}+b x \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{\sqrt {a} \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^3,x]

[Out]

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

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Maple [A]
time = 0.51, size = 56, normalized size = 1.08

method	result	size
default	\(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x +\sqrt {b x +a}\, \sqrt {a}\right )}{x^{2} \sqrt {b x +a}\, \sqrt {a}}\)	\(56\)
risch	\(-\frac {\sqrt {x^{2} \left (b x +a \right )}}{x^{2}}-\frac {b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{\sqrt {a}\, x \sqrt {b x +a}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^3, x)

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Fricas [A]
time = 1.42, size = 127, normalized size = 2.44 \begin {gather*} \left [\frac {\sqrt {a} b x^{2} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} a}{2 \, a x^{2}}, \frac {\sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) - \sqrt {b x^{3} + a x^{2}} a}{a x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2*(a + b*x))/x**3, x)

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Giac [A]
time = 0.96, size = 45, normalized size = 0.87 \begin {gather*} \frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} - \frac {\sqrt {b x + a} b \mathrm {sgn}\left (x\right )}{x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {b\,x^3+a\,x^2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2)/x^3,x)

[Out]

int((a*x^2 + b*x^3)^(1/2)/x^3, x)

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