\(\int (a+b x+c x^2)^{3/2} \, dx\) [2346]

Optimal result

Integrand size = 14, antiderivative size = 112 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}} \]

[Out]

1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/128*(-4*a*c+b^2)^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^
(5/2)-3/64*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {626, 635, 212} \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c} \]

[In]

Int[(a + b*x + c*x^2)^(3/2),x]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^2) + ((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(8*c) +
(3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*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 635

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

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-3 b^2+20 a c+8 b c x+8 c^2 x^2\right )}{64 c^2}+\frac {3 \left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}} \]

[In]

Integrate[(a + b*x + c*x^2)^(3/2),x]

[Out]

((b + 2*c*x)*Sqrt[a + b*x + c*x^2]*(-3*b^2 + 20*a*c + 8*b*c*x + 8*c^2*x^2))/(64*c^2) + (3*(-b^2 + 4*a*c)^2*Arc
Tanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + b*x + c*x^2])])/(64*c^(5/2))

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\)	\(104\)
risch	\(\frac {\left (16 c^{3} x^{3}+24 x^{2} b \,c^{2}+40 a \,c^{2} x +2 b^{2} c x +20 a b c -3 b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{64 c^{2}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.47 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{256 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c
*x + b)*sqrt(c) - 4*a*c) + 4*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x)*sqr
t(c*x^2 + b*x + a))/c^3, -1/128*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2
*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2
+ 20*a*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^3]

Sympy [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 539, normalized size of antiderivative = 4.81 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=a \left (\begin {cases} \left (\frac {a}{2} - \frac {b^{2}}{8 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 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 ) + \left (\frac {b}{4 c} + \frac {x}{2}\right ) \sqrt {a + b x + c x^{2}} & \text {for}\: c \neq 0 \\\frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \left (- \frac {a b}{12 c} - \frac {b \left (\frac {a}{3} - \frac {b^{2}}{8 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 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 ) + \sqrt {a + b x + c x^{2}} \left (\frac {b x}{12 c} + \frac {x^{2}}{3} + \frac {\frac {a}{3} - \frac {b^{2}}{8 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (- \frac {a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x\right )^{\frac {5}{2}}}{5}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \left (- \frac {a \left (\frac {a}{4} - \frac {5 b^{2}}{48 c}\right )}{2 c} - \frac {b \left (- \frac {a b}{12 c} - \frac {3 b \left (\frac {a}{4} - \frac {5 b^{2}}{48 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 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 ) + \sqrt {a + b x + c x^{2}} \left (\frac {b x^{2}}{24 c} + \frac {x^{3}}{4} + \frac {x \left (\frac {a}{4} - \frac {5 b^{2}}{48 c}\right )}{2 c} + \frac {- \frac {a b}{12 c} - \frac {3 b \left (\frac {a}{4} - \frac {5 b^{2}}{48 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {a^{2} \left (a + b x\right )^{\frac {3}{2}}}{3} - \frac {2 a \left (a + b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a + b x\right )^{\frac {7}{2}}}{7}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a*Piecewise(((a/2 - b**2/(8*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b
**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + (b/(4*c) + x/2)*sqrt(a + b*
x + c*x**2), Ne(c, 0)), (2*(a + b*x)**(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True)) + b*Piecewise(((-a*b/(12*c) -
b*(a/3 - b**2/(8*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2
/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*(b*x/(1
2*c) + x**2/3 + (a/3 - b**2/(8*c))/c), Ne(c, 0)), (2*(-a*(a + b*x)**(3/2)/3 + (a + b*x)**(5/2)/5)/b**2, Ne(b,
0)), (sqrt(a)*x**2/2, True)) + c*Piecewise(((-a*(a/4 - 5*b**2/(48*c))/(2*c) - b*(-a*b/(12*c) - 3*b*(a/4 - 5*b*
*2/(48*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*
c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*(b*x**2/(24
*c) + x**3/4 + x*(a/4 - 5*b**2/(48*c))/(2*c) + (-a*b/(12*c) - 3*b*(a/4 - 5*b**2/(48*c))/(4*c))/c), Ne(c, 0)),
(2*(a**2*(a + b*x)**(3/2)/3 - 2*a*(a + b*x)**(5/2)/5 + (a + b*x)**(7/2)/7)/b**3, Ne(b, 0)), (sqrt(a)*x**3/3, T
rue))

Maxima [F(-2)]

Exception generated. \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{64} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c x + 3 \, b\right )} x + \frac {b^{2} c^{2} + 20 \, a c^{3}}{c^{3}}\right )} x - \frac {3 \, b^{3} c - 20 \, a b c^{2}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/64*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c*x + 3*b)*x + (b^2*c^2 + 20*a*c^3)/c^3)*x - (3*b^3*c - 20*a*b*c^2)/c^3) -
3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}+\frac {\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \]

[In]

int((a + b*x + c*x^2)^(3/2),x)

[Out]

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