\(\int \frac {x^3}{\sqrt {b x+c x^2}} \, dx\) [43]

Optimal result

Integrand size = 17, antiderivative size = 102 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \]

[Out]

-5/8*b^3*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+5/8*b^2*(c*x^2+b*x)^(1/2)/c^3-5/12*b*x*(c*x^2+b*x)^(1/2)
/c^2+1/3*x^2*(c*x^2+b*x)^(1/2)/c

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {684, 654, 634, 212} \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}+\frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c} \]

[In]

Int[x^3/Sqrt[b*x + c*x^2],x]

[Out]

(5*b^2*Sqrt[b*x + c*x^2])/(8*c^3) - (5*b*x*Sqrt[b*x + c*x^2])/(12*c^2) + (x^2*Sqrt[b*x + c*x^2])/(3*c) - (5*b^
3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(7/2))

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 684

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*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{6 c} \\ & = -\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c}+\frac {\left (5 b^2\right ) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{8 c^2} \\ & = \frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {\left (5 b^3\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^3} \\ & = \frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^3} \\ & = \frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x \left (15 b^3+5 b^2 c x-2 b c^2 x^2+8 c^3 x^3\right )+30 b^3 \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x^3/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*x*(15*b^3 + 5*b^2*c*x - 2*b*c^2*x^2 + 8*c^3*x^3) + 30*b^3*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt
[x])/(Sqrt[b] - Sqrt[b + c*x])])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {\left (8 c^{2} x^{2}-10 b c x +15 b^{2}\right ) x \left (c x +b \right )}{24 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}\)	\(73\)
pseudoelliptic	\(\frac {8 c^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}\, x^{2}-10 b \,c^{\frac {3}{2}} x \sqrt {x \left (c x +b \right )}+15 b^{2} \sqrt {c}\, \sqrt {x \left (c x +b \right )}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{3}}{24 c^{\frac {7}{2}}}\)	\(79\)
default	\(\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\)	\(97\)

[In]

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

[Out]

1/24*(8*c^2*x^2-10*b*c*x+15*b^2)*x*(c*x+b)/c^3/(x*(c*x+b))^(1/2)-5/16*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^
2+b*x)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.44 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\left [\frac {15 \, b^{3} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {15 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \]

[In]

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

[Out]

[1/48*(15*b^3*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*x^2 - 10*b*c^2*x + 15*b^2*c)*sqr
t(c*x^2 + b*x))/c^4, 1/24*(15*b^3*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*c^3*x^2 - 10*b*c^2*x
+ 15*b^2*c)*sqrt(c*x^2 + b*x))/c^4]

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\begin {cases} - \frac {5 b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {5 b^{2}}{8 c^{3}} - \frac {5 b x}{12 c^{2}} + \frac {x^{2}}{3 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {7}{2}}}{7 b^{4}} & \text {for}\: b \neq 0 \\\tilde {\infty } x^{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-5*b**3*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c)
+ x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(16*c**3) + sqrt(b*x + c*x**2)*(5*b**2/(8*c**3) - 5*b*
x/(12*c**2) + x**2/(3*c)), Ne(c, 0)), (2*(b*x)**(7/2)/(7*b**4), Ne(b, 0)), (zoo*x**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b x}{12 \, c^{2}} - \frac {5 \, b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{2}}{8 \, c^{3}} \]

[In]

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

[Out]

1/3*sqrt(c*x^2 + b*x)*x^2/c - 5/12*sqrt(c*x^2 + b*x)*b*x/c^2 - 5/16*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq
rt(c))/c^(7/2) + 5/8*sqrt(c*x^2 + b*x)*b^2/c^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, x}{c} - \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2}}{c^{3}}\right )} + \frac {5 \, b^{3} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*x*(4*x/c - 5*b/c^2) + 15*b^2/c^3) + 5/16*b^3*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x
))*sqrt(c) + b))/c^(7/2)

Mupad [F(-1)]

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

[In]

int(x^3/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^3/(b*x + c*x^2)^(1/2), x)