\(\int x (b x+c x^2)^{3/2} \, dx\) [13]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}} \]

[Out]

-1/16*b*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^2+1/5*(c*x^2+b*x)^(5/2)/c-3/128*b^5*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))
/c^(7/2)+3/128*b^3*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {654, 626, 634, 212} \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}+\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c} \]

[In]

Int[x*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - (b*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (b*x + c*x^2)
^(5/2)/(5*c) - (3*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c} \\ & = -\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 b^3\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^2} \\ & = \frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 b^5\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^3} \\ & = \frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 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^3} \\ & = \frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.07 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^4-10 b^3 c x+8 b^2 c^2 x^2+176 b c^3 x^3+128 c^4 x^4\right )+\frac {30 b^5 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{640 c^{7/2}} \]

[In]

Integrate[x*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4 - 10*b^3*c*x + 8*b^2*c^2*x^2 + 176*b*c^3*x^3 + 128*c^4*x^4) + (30*b^5*ArcT
anh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(640*c^(7/2))

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {3 \left (\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{5}-\left (\sqrt {c}\, b^{4}-\frac {2 c^{\frac {3}{2}} b^{3} x}{3}+\frac {8 c^{\frac {5}{2}} b^{2} x^{2}}{15}+\frac {176 c^{\frac {7}{2}} b \,x^{3}}{15}+\frac {128 c^{\frac {9}{2}} x^{4}}{15}\right ) \sqrt {x \left (c x +b \right )}\right )}{128 c^{\frac {7}{2}}}\)	\(84\)
risch	\(\frac {\left (128 c^{4} x^{4}+176 b \,c^{3} x^{3}+8 b^{2} c^{2} x^{2}-10 b^{3} c x +15 b^{4}\right ) x \left (c x +b \right )}{640 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {3 b^{5} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}\)	\(95\)
default	\(\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\)	\(110\)

[In]

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

[Out]

-3/128/c^(7/2)*(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+1
76/15*c^(7/2)*b*x^3+128/15*c^(9/2)*x^4)*(x*(c*x+b))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.74 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\left [\frac {15 \, b^{5} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{1280 \, c^{4}}, \frac {15 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{640 \, c^{4}}\right ] \]

[In]

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

[Out]

[1/1280*(15*b^5*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*
c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqr
t(-c)/(c*x)) + (128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4]

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.31 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} - \frac {3 b^{5} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {3 b^{4}}{128 c^{3}} - \frac {b^{3} x}{64 c^{2}} + \frac {b^{2} x^{2}}{80 c} + \frac {11 b x^{3}}{40} + \frac {c x^{4}}{5}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {7}{2}}}{7 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-3*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**3) + sqrt(b*x + c*x**2)*(3*b**4/(128*c**3) - b
**3*x/(64*c**2) + b**2*x**2/(80*c) + 11*b*x**3/40 + c*x**4/5), Ne(c, 0)), (2*(b*x)**(7/2)/(7*b**2), Ne(b, 0)),
(0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.13 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\frac {3 \, \sqrt {c x^{2} + b x} b^{3} x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x}{8 \, c} - \frac {3 \, b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{2} + b x} b^{4}}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}}}{5 \, c} \]

[In]

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

[Out]

3/64*sqrt(c*x^2 + b*x)*b^3*x/c^2 - 1/8*(c*x^2 + b*x)^(3/2)*b*x/c - 3/256*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*
x)*sqrt(c))/c^(7/2) + 3/128*sqrt(c*x^2 + b*x)*b^4/c^3 - 1/16*(c*x^2 + b*x)^(3/2)*b^2/c^2 + 1/5*(c*x^2 + b*x)^(
5/2)/c

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\frac {3 \, b^{5} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} + \frac {1}{640} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x + 11 \, b\right )} x + \frac {b^{2}}{c}\right )} x - \frac {5 \, b^{3}}{c^{2}}\right )} x + \frac {15 \, b^{4}}{c^{3}}\right )} \]

[In]

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

[Out]

3/256*b^5*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2) + 1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(
8*c*x + 11*b)*x + b^2/c)*x - 5*b^3/c^2)*x + 15*b^4/c^3)

Mupad [B] (verification not implemented)

Time = 9.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.07 \[ \int x \left (b x+c x^2\right )^{3/2} \, dx=\frac {{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}-\frac {b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c} \]

[In]

int(x*(b*x + c*x^2)^(3/2),x)

[Out]

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