\(\int x^3 \sqrt {b x+c x^2} \, dx\) [1]

Optimal result

Integrand size = 17, antiderivative size = 131 \[ \int x^3 \sqrt {b x+c x^2} \, dx=-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {7 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}} \]

[Out]

7/48*b^2*(c*x^2+b*x)^(3/2)/c^3-7/40*b*x*(c*x^2+b*x)^(3/2)/c^2+1/5*x^2*(c*x^2+b*x)^(3/2)/c+7/128*b^5*arctanh(x*
c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-7/128*b^3*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {684, 654, 626, 634, 212} \[ \int x^3 \sqrt {b x+c x^2} \, dx=\frac {7 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \]

[In]

Int[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(-7*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (7*b^2*(b*x + c*x^2)^(3/2))/(48*c^3) - (7*b*x*(b*x + c*x^2)
^(3/2))/(40*c^2) + (x^2*(b*x + c*x^2)^(3/2))/(5*c) + (7*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(9/
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^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b) \int x^2 \sqrt {b x+c x^2} \, dx}{10 c} \\ & = -\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 b^2\right ) \int x \sqrt {b x+c x^2} \, dx}{16 c^2} \\ & = \frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {\left (7 b^3\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^3} \\ & = -\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 b^5\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^4} \\ & = -\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 b^5\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^4} \\ & = -\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {7 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.90 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-105 b^4+70 b^3 c x-56 b^2 c^2 x^2+48 b c^3 x^3+384 c^4 x^4\right )+\frac {210 b^5 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{1920 c^{9/2}} \]

[In]

Integrate[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4 + 70*b^3*c*x - 56*b^2*c^2*x^2 + 48*b*c^3*x^3 + 384*c^4*x^4) + (210*b^5*A
rcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(1920*c^(9/2))

Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{5}}{128}-\frac {7 \left (\sqrt {c}\, b^{4}-\frac {2 c^{\frac {3}{2}} b^{3} x}{3}+\frac {8 c^{\frac {5}{2}} b^{2} x^{2}}{15}-\frac {16 c^{\frac {7}{2}} b \,x^{3}}{35}-\frac {128 c^{\frac {9}{2}} x^{4}}{35}\right ) \sqrt {x \left (c x +b \right )}}{128}}{c^{\frac {9}{2}}}\)	\(84\)
risch	\(-\frac {\left (-384 c^{4} x^{4}-48 b \,c^{3} x^{3}+56 b^{2} c^{2} x^{2}-70 b^{3} c x +105 b^{4}\right ) x \left (c x +b \right )}{1920 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {7 b^{5} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}\)	\(95\)
default	\(\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\)	\(129\)

[In]

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

[Out]

7/128*(arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*b^5-(c^(1/2)*b^4-2/3*c^(3/2)*b^3*x+8/15*c^(5/2)*b^2*x^2-16/35*c^(7
/2)*b*x^3-128/35*c^(9/2)*x^4)*(x*(c*x+b))^(1/2))/c^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.47 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\left [\frac {105 \, b^{5} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {105 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \]

[In]

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

[Out]

[1/3840*(105*b^5*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2
*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(105*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x
)*sqrt(-c)/(c*x)) - (384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x)
)/c^5]

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.13 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\begin {cases} \frac {7 b^{5} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{4}} + \sqrt {b x + c x^{2}} \left (- \frac {7 b^{4}}{128 c^{4}} + \frac {7 b^{3} x}{192 c^{3}} - \frac {7 b^{2} x^{2}}{240 c^{2}} + \frac {b x^{3}}{40 c} + \frac {x^{4}}{5}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {9}{2}}}{9 b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((7*b**5*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))/(256*c**4) + sqrt(b*x + c*x**2)*(-7*b**4/(128*c**4) + 7
*b**3*x/(192*c**3) - 7*b**2*x**2/(240*c**2) + b*x**3/(40*c) + x**4/5), Ne(c, 0)), (2*(b*x)**(9/2)/(9*b**4), Ne
(b, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}}{5 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x}{40 \, c^{2}} + \frac {7 \, b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}}{48 \, c^{3}} \]

[In]

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

[Out]

1/5*(c*x^2 + b*x)^(3/2)*x^2/c - 7/64*sqrt(c*x^2 + b*x)*b^3*x/c^3 - 7/40*(c*x^2 + b*x)^(3/2)*b*x/c^2 + 7/256*b^
5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 7/128*sqrt(c*x^2 + b*x)*b^4/c^4 + 7/48*(c*x^2 + b*x)^
(3/2)*b^2/c^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x + \frac {b}{c}\right )} x - \frac {7 \, b^{2}}{c^{2}}\right )} x + \frac {35 \, b^{3}}{c^{3}}\right )} x - \frac {105 \, b^{4}}{c^{4}}\right )} - \frac {7 \, b^{5} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*x + b/c)*x - 7*b^2/c^2)*x + 35*b^3/c^3)*x - 105*b^4/c^4) - 7/256*b^5*log(
abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2)

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.91 \[ \int x^3 \sqrt {b x+c x^2} \, dx=\frac {x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {7\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c} \]

[In]

int(x^3*(b*x + c*x^2)^(1/2),x)

[Out]

(x^2*(b*x + c*x^2)^(3/2))/(5*c) - (7*b*((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)))/(
10*c)