\(\int \frac {x^3}{(b x+c x^2)^{3/2}} \, dx\) [53]

Optimal result

Integrand size = 17, antiderivative size = 69 \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {682, 654, 634, 212} \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {2 x^2}{c \sqrt {b x+c x^2}} \]

[In]

Int[x^3/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*x^2)/(c*Sqrt[b*x + c*x^2]) + (3*Sqrt[b*x + c*x^2])/c^2 - (3*b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Dist[e^2*((m + p)/(c*(p + 1))), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{c} \\ & = -\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {(3 b) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{2 c^2} \\ & = -\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{c^2} \\ & = -\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.16 \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c} x (3 b+c x)+6 b \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{c^{5/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x^3/(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[c]*x*(3*b + c*x) + 6*b*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])])/(c^(5
/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {c^{\frac {3}{2}} x^{2}+3 x b \sqrt {c}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) \sqrt {x \left (c x +b \right )}\, b}{c^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\)	\(58\)
risch	\(\frac {x \left (c x +b \right )}{c^{2} \sqrt {x \left (c x +b \right )}}-\frac {3 b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}}+\frac {2 b \sqrt {c \left (\frac {b}{c}+x \right )^{2}-b \left (\frac {b}{c}+x \right )}}{c^{3} \left (\frac {b}{c}+x \right )}\)	\(90\)
default	\(\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\)	\(119\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.20 \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b c x + b^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (c^{2} x + 3 \, b c\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{4} x + b c^{3}\right )}}, \frac {3 \, {\left (b c x + b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (c^{2} x + 3 \, b c\right )} \sqrt {c x^{2} + b x}}{c^{4} x + b c^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(x**3/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**3/(x*(b + c*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {3 \, b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

x^2/(sqrt(c*x^2 + b*x)*c) + 3*b*x/(sqrt(c*x^2 + b*x)*c^2) - 3/2*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))
/c^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {3 \, b \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} + \frac {2 \, b^{2}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{\frac {5}{2}}} + \frac {\sqrt {c x^{2} + b x}}{c^{2}} \]

[In]

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

[Out]

3/2*b*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) + 2*b^2/(((sqrt(c)*x - sqrt(c*x^2 + b*x
))*sqrt(c) + b)*c^(5/2)) + sqrt(c*x^2 + b*x)/c^2

Mupad [F(-1)]

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

[In]

int(x^3/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^3/(b*x + c*x^2)^(3/2), x)