Integrand size = 14, antiderivative size = 112 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}} \] Output:
-3/64*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2+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)
Time = 0.68 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-3 b^2+20 a c+8 b c x+8 c^2 x^2\right )}{64 c^2}+\frac {3 \left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}} \] Input:
Integrate[(a + b*x + c*x^2)^(3/2),x]
Output:
((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*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + b*x + c*x^2])])/(64*c^(5/2))
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1087, 1087, 1092, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b x+c x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^2+b x+a}dx}{16 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c}\) |
Input:
Int[(a + b*x + c*x^2)^(3/2),x]
Output:
((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c* x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sq rt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(3/2))))/(16*c)
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] && (Gt Q[a, 0] || LtQ[b, 0])
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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Time = 0.84 (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 | \(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+40 a \,c^{2} x +2 b^{2} c x +20 a b c -3 b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{64 c^{2}}+\frac {3 \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}\) | \(110\) |
Input:
int((c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.09 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.47 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{256 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, c^{3}}\right ] \] Input:
integrate((c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
Output:
[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)*sqrt( 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)*a rctan(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 + 2 0*a*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^3]
Time = 1.10 (sec) , antiderivative size = 539, normalized size of antiderivative = 4.81 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:
integrate((c*x**2+b*x+a)**(3/2),x)
Output:
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, Tr ue)) + 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/(12*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, True))
Exception generated. \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{64} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c x + 3 \, b\right )} x + \frac {b^{2} c^{2} + 20 \, a c^{3}}{c^{3}}\right )} x - \frac {3 \, b^{3} c - 20 \, a b c^{2}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} \] Input:
integrate((c*x^2+b*x+a)^(3/2),x, algorithm="giac")
Output:
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)
Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}+\frac {\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \] Input:
int((a + b*x + c*x^2)^(3/2),x)
Output:
(((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)
Time = 0.19 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.25 \[ \int \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {40 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2}+80 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x -6 \sqrt {c \,x^{2}+b x +a}\, b^{3} c +4 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x +48 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{2}+32 \sqrt {c \,x^{2}+b x +a}\, c^{4} x^{3}+48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}-24 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c +3 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}}{128 c^{3}} \] Input:
int((c*x^2+b*x+a)^(3/2),x)
Output:
(40*sqrt(a + b*x + c*x**2)*a*b*c**2 + 80*sqrt(a + b*x + c*x**2)*a*c**3*x - 6*sqrt(a + b*x + c*x**2)*b**3*c + 4*sqrt(a + b*x + c*x**2)*b**2*c**2*x + 48*sqrt(a + b*x + c*x**2)*b*c**3*x**2 + 32*sqrt(a + b*x + c*x**2)*c**4*x** 3 + 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a *c - b**2))*a**2*c**2 - 24*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c + 3*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4)/(128*c**3)