\(\int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx\) [1236]

Optimal result

Integrand size = 26, antiderivative size = 75 \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=d^2 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {\left (b^2-4 a c\right ) d^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}} \]

[Out]

1/2*(-4*a*c+b^2)*d^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(1/2)+d^2*(2*c*x+b)*(c*x^2+b*x+a)^(1
/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {706, 635, 212} \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {d^2 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+d^2 (b+2 c x) \sqrt {a+b x+c x^2} \]

[In]

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

[Out]

d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2] + ((b^2 - 4*a*c)*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2
])])/(2*Sqrt[c])

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 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=d^2 \left ((b+2 c x) \sqrt {a+x (b+c x)}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}\right ) \]

[In]

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

[Out]

d^2*((b + 2*c*x)*Sqrt[a + x*(b + c*x)] + ((b^2 - 4*a*c)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])
])/Sqrt[c])

Maple [A] (verified)

Time = 3.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88

method	result	size
risch	\(d^{2} \left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}+\frac {\left (\frac {b^{2}}{2}-2 a c \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{2}}{\sqrt {c}}\)	\(66\)
default	\(d^{2} \left (\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+4 c^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+4 b c \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\right )\)	\(199\)

[In]

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

[Out]

d^2*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)+(1/2*b^2-2*a*c)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)*d^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.60 \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} d^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, c^{2} d^{2} x + b c d^{2}\right )} \sqrt {c x^{2} + b x + a}}{4 \, c}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (2 \, c^{2} d^{2} x + b c d^{2}\right )} \sqrt {c x^{2} + b x + a}}{2 \, c}\right ] \]

[In]

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

[Out]

[-1/4*((b^2 - 4*a*c)*sqrt(c)*d^2*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^2*d^2*x + b*c*d^2)*sqrt(c*x^2 + b*x + a))/c, -1/2*((b^2 - 4*a*c)*sqrt(-c)*d^2*arctan(1/2*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^2*d^2*x + b*c*d^2)*sqrt(c*x^2 + b*x
+ a))/c]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (68) = 136\).

Time = 0.60 (sec) , antiderivative size = 233, normalized size of antiderivative = 3.11 \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (b d^{2} + 2 c d^{2} x\right ) \sqrt {a + b x + c x^{2}} + \left (- 2 a c d^{2} + \frac {b^{2} d^{2}}{2}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \cdot \left (\frac {4 c^{2} d^{2} \left (a + b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 8 a c^{2} d^{2} + 4 b^{2} c d^{2}\right )}{3 b^{2}} + \frac {\sqrt {a + b x} \left (4 a^{2} c^{2} d^{2} - 4 a b^{2} c d^{2} + b^{4} d^{2}\right )}{b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {4 c^{2} d^{2} x^{3}}{3 \sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((b*d**2 + 2*c*d**2*x)*sqrt(a + b*x + c*x**2) + (-2*a*c*d**2 + b**2*d**2/2)*Piecewise((log(b + 2*sqr
t(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*(4*c**2*d**2*(a + b*x)**(5/2)/(5*b**2) + (a + b*x)**(3/2)*(-8*a*c**2*d
**2 + 4*b**2*c*d**2)/(3*b**2) + sqrt(a + b*x)*(4*a**2*c**2*d**2 - 4*a*b**2*c*d**2 + b**4*d**2)/b**2)/b, Ne(b,
0)), (4*c**2*d**2*x**3/(3*sqrt(a)), True))

Maxima [F(-2)]

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

[In]

integrate((2*c*d*x+b*d)^2/(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 = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx={\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a} - \frac {{\left (b^{2} d^{2} - 4 \, a c d^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} \]

[In]

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

[Out]

(2*c*d^2*x + b*d^2)*sqrt(c*x^2 + b*x + a) - 1/2*(b^2*d^2 - 4*a*c*d^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*sqrt(c) + b))/sqrt(c)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

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

[Out]

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