\(\int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx\) [270]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=-\frac {3 a \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {x}}+\frac {\sqrt {x} \sqrt {a x^2+b x^3}}{2 b}+\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{4 b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2049, 2054, 212} \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{4 b^{5/2}}-\frac {3 a \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {x}}+\frac {\sqrt {x} \sqrt {a x^2+b x^3}}{2 b} \]

[In]

Int[x^(5/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(-3*a*Sqrt[a*x^2 + b*x^3])/(4*b^2*Sqrt[x]) + (Sqrt[x]*Sqrt[a*x^2 + b*x^3])/(2*b) + (3*a^2*ArcTanh[(Sqrt[b]*x^(
3/2))/Sqrt[a*x^2 + b*x^3]])/(4*b^(5/2))

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 {x} \sqrt {a x^2+b x^3}}{2 b}-\frac {(3 a) \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx}{4 b} \\ & = -\frac {3 a \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {x}}+\frac {\sqrt {x} \sqrt {a x^2+b x^3}}{2 b}+\frac {\left (3 a^2\right ) \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^3}} \, dx}{8 b^2} \\ & = -\frac {3 a \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {x}}+\frac {\sqrt {x} \sqrt {a x^2+b x^3}}{2 b}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{4 b^2} \\ & = -\frac {3 a \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {x}}+\frac {\sqrt {x} \sqrt {a x^2+b x^3}}{2 b}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{4 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {b} x^{3/2} \left (-3 a^2-a b x+2 b^2 x^2\right )+6 a^2 x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{5/2} \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[x^(5/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(Sqrt[b]*x^(3/2)*(-3*a^2 - a*b*x + 2*b^2*x^2) + 6*a^2*x*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sq
rt[a + b*x])])/(4*b^(5/2)*Sqrt[x^2*(a + b*x)])

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\left (-2 b x +3 a \right ) x^{\frac {3}{2}} \left (b x +a \right )}{4 b^{2} \sqrt {x^{2} \left (b x +a \right )}}+\frac {3 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x}\, \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(89\)
default	\(\frac {\sqrt {x}\, \left (4 b^{\frac {7}{2}} x^{3}-2 b^{\frac {5}{2}} a \,x^{2}-6 a^{2} b^{\frac {3}{2}} x +3 \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b \right )}{8 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{\frac {7}{2}}}\)	\(92\)

[In]

int(x^(5/2)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(-2*b*x+3*a)*x^(3/2)*(b*x+a)/b^2/(x^2*(b*x+a))^(1/2)+3/8*a^2/b^(5/2)*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 = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.67 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b} \sqrt {x}}{x}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt {x}}{8 \, b^{3} x}, -\frac {3 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) - \sqrt {b x^{3} + a x^{2}} {\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt {x}}{4 \, b^{3} x}\right ] \]

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*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)*(
2*b^2*x - 3*a*b)*sqrt(x))/(b^3*x), -1/4*(3*a^2*sqrt(-b)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-b)/(b*x^(3/2))) - s
qrt(b*x^3 + a*x^2)*(2*b^2*x - 3*a*b)*sqrt(x))/(b^3*x)]

Sympy [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{\frac {5}{2}}}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \]

[In]

integrate(x**(5/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(x**(5/2)/sqrt(x**2*(a + b*x)), x)

Maxima [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {b x^{3} + a x^{2}}} \,d x } \]

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/sqrt(b*x^3 + a*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\frac {3 \, a^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x + a} \sqrt {x} {\left (\frac {2 \, x}{b} - \frac {3 \, a}{b^{2}}\right )} - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}}}{4 \, \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

3/8*a^2*log(abs(a))*sgn(x)/b^(5/2) + 1/4*(sqrt(b*x + a)*sqrt(x)*(2*x/b - 3*a/b^2) - 3*a^2*log(abs(-sqrt(b)*sqr
t(x) + sqrt(b*x + a)))/b^(5/2))/sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{5/2}}{\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

int(x^(5/2)/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(x^(5/2)/(a*x^2 + b*x^3)^(1/2), x)