\(\int \frac {(a x^2+b x^3)^{3/2}}{x^4} \, dx\) [246]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 a \sqrt {a x^2+b x^3}}{x}+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2046, 2033, 212} \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )+\frac {2 a \sqrt {a x^2+b x^3}}{x}+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3} \]

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^4,x]

[Out]

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

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 2046

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 + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 x \sqrt {a+b x} \left (\sqrt {a+b x} (4 a+b x)-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{3 \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-\sqrt {b x +a}\, \left (\sqrt {a}\, b^{2} x^{2}+\frac {14 a^{\frac {3}{2}} b x}{3}+\frac {8 a^{\frac {5}{2}}}{3}\right )}{8 a^{\frac {3}{2}} x^{3}}\)	\(61\)
default	\(-\frac {2 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\left (b x +a \right )^{\frac {3}{2}}-3 \sqrt {b x +a}\, a \right )}{3 x^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(63\)

[In]

int((b*x^3+a*x^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/8/a^(3/2)*(arctanh((b*x+a)^(1/2)/a^(1/2))*b^3*x^3-(b*x+a)^(1/2)*(a^(1/2)*b^2*x^2+14/3*a^(3/2)*b*x+8/3*a^(5/2
)))/x^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, a^{\frac {3}{2}} x \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (b x + 4 \, a\right )}}{3 \, x}, \frac {2 \, {\left (3 \, \sqrt {-a} a x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (b x + 4 \, a\right )}\right )}}{3 \, x}\right ] \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**4,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**4, x)

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^4, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 \, a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 2 \, \sqrt {b x + a} a \mathrm {sgn}\left (x\right ) - \frac {2 \, {\left (3 \, a^{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 4 \, \sqrt {-a} a^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-a}} \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

2*a^2*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/sqrt(-a) + 2/3*(b*x + a)^(3/2)*sgn(x) + 2*sqrt(b*x + a)*a*sgn(x) -
2/3*(3*a^2*arctan(sqrt(a)/sqrt(-a)) + 4*sqrt(-a)*a^(3/2))*sgn(x)/sqrt(-a)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^4} \,d x \]

[In]

int((a*x^2 + b*x^3)^(3/2)/x^4,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^4, x)