\(\int (b x+c x^2)^{3/2} \, dx\) [298]

Optimal result

Integrand size = 13, antiderivative size = 89 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}+\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {626, 634, 212} \[ \int \left (b x+c x^2\right )^{3/2} \, dx=\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c} \]

[In]

Int[(b*x + c*x^2)^(3/2),x]

[Out]

(-3*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^2) + ((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) + (3*b^4*ArcTanh[(Sq
rt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-3 b^3+2 b^2 c x+24 b c^2 x^2+16 c^3 x^3\right )+\frac {6 b^4 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{64 c^{5/2}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^3 + 2*b^2*c*x + 24*b*c^2*x^2 + 16*c^3*x^3) + (6*b^4*ArcTanh[(Sqrt[c]*Sqrt[x]
)/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(64*c^(5/2))

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{4}}{64}-\frac {3 \left (\sqrt {c}\, b^{3}-\frac {2 c^{\frac {3}{2}} b^{2} x}{3}-8 c^{\frac {5}{2}} b \,x^{2}-\frac {16 c^{\frac {7}{2}} x^{3}}{3}\right ) \sqrt {x \left (c x +b \right )}}{64}}{c^{\frac {5}{2}}}\)	\(73\)
risch	\(-\frac {\left (-16 c^{3} x^{3}-24 b \,c^{2} x^{2}-2 b^{2} c x +3 b^{3}\right ) x \left (c x +b \right )}{64 c^{2} \sqrt {x \left (c x +b \right )}}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {5}{2}}}\)	\(84\)
default	\(\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\)	\(87\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.91 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=\left [\frac {3 \, b^{4} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{128 \, c^{3}}, -\frac {3 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{64 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/128*(3*b^4*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(16*c^4*x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*
x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3, -1/64*(3*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (16*c^4*
x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3]

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.89 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=b \left (\begin {cases} \frac {b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{2}} + \sqrt {b x + c x^{2}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x}{12 c} + \frac {x^{2}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x}{96 c^{2}} + \frac {b x^{2}}{24 c} + \frac {x^{3}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

b*Piecewise((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**2) + sqrt(b*x + c*x**2)*(-b**2/(8*c**2) + b*x/(1
2*c) + x**2/3), Ne(c, 0)), (2*(b*x)**(5/2)/(5*b**2), Ne(b, 0)), (0, True)) + c*Piecewise((-5*b**4*Piecewise((l
og(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 = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} x - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} x}{32 \, c} + \frac {3 \, b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{8 \, c} \]

[In]

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

[Out]

1/4*(c*x^2 + b*x)^(3/2)*x - 3/32*sqrt(c*x^2 + b*x)*b^2*x/c + 3/128*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c))/c^(5/2) - 3/64*sqrt(c*x^2 + b*x)*b^3/c^2 + 1/8*(c*x^2 + b*x)^(3/2)*b/c

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=-\frac {3 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} + \frac {1}{64} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, c x + 3 \, b\right )} x + \frac {b^{2}}{c}\right )} x - \frac {3 \, b^{3}}{c^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \left (b x+c x^2\right )^{3/2} \, dx=\frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {3\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c} \]

[In]

int((b*x + c*x^2)^(3/2),x)

[Out]

((b*x + c*x^2)^(3/2)*(b/2 + c*x))/(4*c) - (3*b^2*((b*x + c*x^2)^(1/2)*(x/2 + b/(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)