\(\int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx\) [181]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=-\frac {3 b C \sqrt {a+b x+c x^2}}{4 c^2}+\frac {C x \sqrt {a+b x+c x^2}}{2 c}+\frac {\left (8 A c^2+3 b^2 C-4 a c C\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1675, 654, 635, 212} \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {\left (-4 a c C+8 A c^2+3 b^2 C\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2}}-\frac {3 b C \sqrt {a+b x+c x^2}}{4 c^2}+\frac {C x \sqrt {a+b x+c x^2}}{2 c} \]

[In]

Int[(A + C*x^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(-3*b*C*Sqrt[a + b*x + c*x^2])/(4*c^2) + (C*x*Sqrt[a + b*x + c*x^2])/(2*c) + ((8*A*c^2 + 3*b^2*C - 4*a*c*C)*Ar
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*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 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]

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 1675

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {C (-3 b+2 c x) \sqrt {a+x (b+c x)}}{4 c^2}+\frac {\left (8 A c^2+3 b^2 C-4 a c C\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{4 c^{5/2}} \]

[In]

Integrate[(A + C*x^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(C*(-3*b + 2*c*x)*Sqrt[a + x*(b + c*x)])/(4*c^2) + ((8*A*c^2 + 3*b^2*C - 4*a*c*C)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a
] + Sqrt[a + x*(b + c*x)])])/(4*c^(5/2))

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {C \left (-2 c x +3 b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{2}}+\frac {\left (8 A \,c^{2}-4 C a c +3 C \,b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}\)	\(76\)
default	\(\frac {A \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+C \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(138\)

[In]

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

[Out]

-1/4*C*(-2*c*x+3*b)/c^2*(c*x^2+b*x+a)^(1/2)+1/8*(8*A*c^2-4*C*a*c+3*C*b^2)/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^
2+b*x+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.95 \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {{\left (3 \, C b^{2} - 4 \, C a c + 8 \, A c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (2 \, C c^{2} x - 3 \, C b c\right )} \sqrt {c x^{2} + b x + a}}{16 \, c^{3}}, -\frac {{\left (3 \, C b^{2} - 4 \, C a c + 8 \, A c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (2 \, C c^{2} x - 3 \, C b c\right )} \sqrt {c x^{2} + b x + a}}{8 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/16*((3*C*b^2 - 4*C*a*c + 8*A*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*(2*C*c^2*x - 3*C*b*c)*sqrt(c*x^2 + b*x + a))/c^3, -1/8*((3*C*b^2 - 4*C*a*c + 8*A*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*(2*C*c^2*x - 3*C*
b*c)*sqrt(c*x^2 + b*x + a))/c^3]

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.82 \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (- \frac {3 C b}{4 c^{2}} + \frac {C x}{2 c}\right ) \sqrt {a + b x + c x^{2}} + \left (A - \frac {C a}{2 c} + \frac {3 C b^{2}}{8 c^{2}}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 A \sqrt {a + b x} + \frac {2 C \left (a^{2} \sqrt {a + b x} - \frac {2 a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x\right )^{\frac {5}{2}}}{5}\right )}{b^{2}}}{b} & \text {for}\: b \neq 0 \\\frac {A x + \frac {C x^{3}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

integrate((C*x**2+A)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Piecewise(((-3*C*b/(4*c**2) + C*x/(2*c))*sqrt(a + b*x + c*x**2) + (A - C*a/(2*c) + 3*C*b**2/(8*c**2))*Piecewis
e((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)), Ne(c, 0)), ((2*A*sqrt(a + b*x) + 2*C*(a**2*sqrt(a + b*x) - 2*a*(a +
b*x)**(3/2)/3 + (a + b*x)**(5/2)/5)/b**2)/b, Ne(b, 0)), ((A*x + C*x**3/3)/sqrt(a), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(1/2),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 = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (\frac {2 \, C x}{c} - \frac {3 \, C b}{c^{2}}\right )} - \frac {{\left (3 \, C b^{2} - 4 \, C a c + 8 \, A c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*C*x/c - 3*C*b/c^2) - 1/8*(3*C*b^2 - 4*C*a*c + 8*A*c^2)*log(abs(2*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C x^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {C\,x^2+A}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

int((A + C*x^2)/(a + b*x + c*x^2)^(1/2),x)

[Out]

int((A + C*x^2)/(a + b*x + c*x^2)^(1/2), x)