\(\int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx\) [271]

Optimal result

Integrand size = 21, antiderivative size = 60 \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {a x^2+b x^3}}{b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2049, 2054, 212} \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {a x^2+b x^3}}{b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{b^{3/2}} \]

[In]

Int[x^(3/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

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 2049

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

Rule 2054

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.33 \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {b} x^{3/2} (a+b x)+2 a x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{b^{3/2} \sqrt {x^2 (a+b x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30

method	result	size
risch	\(\frac {x^{\frac {3}{2}} \left (b x +a \right )}{b \sqrt {x^{2} \left (b x +a \right )}}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x}\, \sqrt {x \left (b x +a \right )}}{2 b^{\frac {3}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(78\)
default	\(\frac {\sqrt {x}\, \left (2 b^{\frac {5}{2}} x^{2}+2 b^{\frac {3}{2}} a x -a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b \right )}{2 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{\frac {5}{2}}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.18 \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\left [\frac {a \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b} \sqrt {x}}{x}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} b \sqrt {x}}{2 \, b^{2} x}, \frac {a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + \sqrt {b x^{3} + a x^{2}} b \sqrt {x}}{b^{2} x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{\frac {3}{2}}}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\int { \frac {x^{\frac {3}{2}}}{\sqrt {b x^{3} + a x^{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=-\frac {a \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, b^{\frac {3}{2}}} + \frac {\frac {a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b x + a} \sqrt {x}}{b}}{\mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/2*a*log(abs(a))*sgn(x)/b^(3/2) + (a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(3/2) + sqrt(b*x + a)*sqrt
(x)/b)/sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{3/2}}{\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

int(x^(3/2)/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(x^(3/2)/(a*x^2 + b*x^3)^(1/2), x)