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

Optimal result

Integrand size = 17, antiderivative size = 105 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}} \]

[Out]

-5/24*b*(c*x^2+b*x)^(3/2)/c^2+1/4*x*(c*x^2+b*x)^(3/2)/c-5/64*b^4*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+
5/64*b^2*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {684, 654, 626, 634, 212} \[ \int x^2 \sqrt {b x+c x^2} \, dx=-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c} \]

[In]

Int[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(5*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - (5*b*(b*x + c*x^2)^(3/2))/(24*c^2) + (x*(b*x + c*x^2)^(3/2))/
(4*c) - (5*b^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 626

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

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 \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {(5 b) \int x \sqrt {b x+c x^2} \, dx}{8 c} \\ & = -\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (5 b^2\right ) \int \sqrt {b x+c x^2} \, dx}{16 c^2} \\ & = \frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (5 b^4\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^3} \\ & = \frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (5 b^4\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^3} \\ & = \frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^3-10 b^2 c x+8 b c^2 x^2+48 c^3 x^3\right )+\frac {30 b^4 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{192 c^{7/2}} \]

[In]

Integrate[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3 - 10*b^2*c*x + 8*b*c^2*x^2 + 48*c^3*x^3) + (30*b^4*ArcTanh[(Sqrt[c]*Sqrt[x
])/(Sqrt[b] - Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(192*c^(7/2))

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(-\frac {5 \left (\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{4}-\left (\sqrt {c}\, b^{3}-\frac {2 c^{\frac {3}{2}} b^{2} x}{3}+\frac {8 c^{\frac {5}{2}} b \,x^{2}}{15}+\frac {16 c^{\frac {7}{2}} x^{3}}{5}\right ) \sqrt {x \left (c x +b \right )}\right )}{64 c^{\frac {7}{2}}}\)	\(73\)
risch	\(\frac {\left (48 c^{3} x^{3}+8 b \,c^{2} x^{2}-10 b^{2} c x +15 b^{3}\right ) x \left (c x +b \right )}{192 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {5 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}\)	\(84\)
default	\(\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\)	\(103\)

[In]

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

[Out]

-5/64/c^(7/2)*(arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*b^4-(c^(1/2)*b^3-2/3*c^(3/2)*b^2*x+8/15*c^(5/2)*b*x^2+16/5
*c^(7/2)*x^3)*(x*(c*x+b))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.61 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\left [\frac {15 \, b^{4} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, \frac {15 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \]

[In]

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

[Out]

[1/384*(15*b^4*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(48*c^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2
*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4, 1/192*(15*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*c
^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4]

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x}{96 c^{2}} + \frac {b x^{2}}{24 c} + \frac {x^{3}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-5*b**4*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))/(128*c**3) + sqrt(b*x + c*x**2)*(5*b**3/(64*c**3) - 5*
b**2*x/(96*c**2) + b*x**2/(24*c) + x**3/4), Ne(c, 0)), (2*(b*x)**(7/2)/(7*b**3), Ne(b, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\frac {5 \, \sqrt {c x^{2} + b x} b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x}{4 \, c} - \frac {5 \, b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{24 \, c^{2}} \]

[In]

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

[Out]

5/32*sqrt(c*x^2 + b*x)*b^2*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*x/c - 5/128*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)
*sqrt(c))/c^(7/2) + 5/64*sqrt(c*x^2 + b*x)*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3/2)*b/c^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, x + \frac {b}{c}\right )} x - \frac {5 \, b^{2}}{c^{2}}\right )} x + \frac {15 \, b^{3}}{c^{3}}\right )} + \frac {5 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*x + b/c)*x - 5*b^2/c^2)*x + 15*b^3/c^3) + 5/128*b^4*log(abs(2*(sqrt(c)*x - sq
rt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {b x+c x^2} \, dx=\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c} \]

[In]

int(x^2*(b*x + c*x^2)^(1/2),x)

[Out]

(x*(b*x + c*x^2)^(3/2))/(4*c) - (5*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + ((
b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2)))/(8*c)