3.155 \(\int \frac{a+b x+c x^2}{\sqrt{-1+d x} \sqrt{1+d x}} \, dx\)

Optimal. Leaf size=52 \[ \frac{\left (2 a d^2+c\right ) \cosh ^{-1}(d x)}{2 d^3}+\frac{\sqrt{d x-1} \sqrt{d x+1} (2 b+c x)}{2 d^2} \]

[Out]

((2*b + c*x)*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/(2*d^2) + ((c + 2*a*d^2)*ArcCosh[d*x])/(2*d^3)

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Rubi [B]  time = 0.0711419, antiderivative size = 135, normalized size of antiderivative = 2.6, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {901, 1815, 641, 217, 206} \[ \frac{\sqrt{d^2 x^2-1} \left (2 a d^2+c\right ) \tanh ^{-1}\left (\frac{d x}{\sqrt{d^2 x^2-1}}\right )}{2 d^3 \sqrt{d x-1} \sqrt{d x+1}}-\frac{b \left (1-d^2 x^2\right )}{d^2 \sqrt{d x-1} \sqrt{d x+1}}-\frac{c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt{d x-1} \sqrt{d x+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]

[Out]

-((b*(1 - d^2*x^2))/(d^2*Sqrt[-1 + d*x]*Sqrt[1 + d*x])) - (c*x*(1 - d^2*x^2))/(2*d^2*Sqrt[-1 + d*x]*Sqrt[1 + d
*x]) + ((c + 2*a*d^2)*Sqrt[-1 + d^2*x^2]*ArcTanh[(d*x)/Sqrt[-1 + d^2*x^2]])/(2*d^3*Sqrt[-1 + d*x]*Sqrt[1 + d*x
])

Rule 901

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Dist[((d + e*x)^FracPart[m]*(f + g*x)^FracPart[m])/(d*f + e*g*x^2)^FracPart[m], Int[(d*f + e*g*x^2)^m*(a + b*x
+ c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] && EqQ[e*f + d*g, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 217

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{\sqrt{-1+d x} \sqrt{1+d x}} \, dx &=\frac{\sqrt{-1+d^2 x^2} \int \frac{a+b x+c x^2}{\sqrt{-1+d^2 x^2}} \, dx}{\sqrt{-1+d x} \sqrt{1+d x}}\\ &=-\frac{c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}+\frac{\sqrt{-1+d^2 x^2} \int \frac{c+2 a d^2+2 b d^2 x}{\sqrt{-1+d^2 x^2}} \, dx}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}\\ &=-\frac{b \left (1-d^2 x^2\right )}{d^2 \sqrt{-1+d x} \sqrt{1+d x}}-\frac{c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}+\frac{\left (\left (c+2 a d^2\right ) \sqrt{-1+d^2 x^2}\right ) \int \frac{1}{\sqrt{-1+d^2 x^2}} \, dx}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}\\ &=-\frac{b \left (1-d^2 x^2\right )}{d^2 \sqrt{-1+d x} \sqrt{1+d x}}-\frac{c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}+\frac{\left (\left (c+2 a d^2\right ) \sqrt{-1+d^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1-d^2 x^2} \, dx,x,\frac{x}{\sqrt{-1+d^2 x^2}}\right )}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}\\ &=-\frac{b \left (1-d^2 x^2\right )}{d^2 \sqrt{-1+d x} \sqrt{1+d x}}-\frac{c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt{-1+d x} \sqrt{1+d x}}+\frac{\left (c+2 a d^2\right ) \sqrt{-1+d^2 x^2} \tanh ^{-1}\left (\frac{d x}{\sqrt{-1+d^2 x^2}}\right )}{2 d^3 \sqrt{-1+d x} \sqrt{1+d x}}\\ \end{align*}

Mathematica [B]  time = 0.210479, size = 126, normalized size = 2.42 \[ \frac{4 \sqrt{1-d x} \tanh ^{-1}\left (\sqrt{\frac{d x-1}{d x+1}}\right ) (d (a d-b)+c)+d \sqrt{-(d x-1)^2} \sqrt{d x+1} (2 b+c x)+2 \sqrt{d x-1} (2 b d-c) \sin ^{-1}\left (\frac{\sqrt{1-d x}}{\sqrt{2}}\right )}{2 d^3 \sqrt{1-d x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x + c*x^2)/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]

[Out]

(d*(2*b + c*x)*Sqrt[-(-1 + d*x)^2]*Sqrt[1 + d*x] + 2*(-c + 2*b*d)*Sqrt[-1 + d*x]*ArcSin[Sqrt[1 - d*x]/Sqrt[2]]
+ 4*(c + d*(-b + a*d))*Sqrt[1 - d*x]*ArcTanh[Sqrt[(-1 + d*x)/(1 + d*x)]])/(2*d^3*Sqrt[1 - d*x])

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Maple [C]  time = 0.016, size = 120, normalized size = 2.3 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{2\,{d}^{3}}\sqrt{dx-1}\sqrt{dx+1} \left ({\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-1}xc+2\,{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-1}b+2\,\ln \left ( \left ({\it csgn} \left ( d \right ) \sqrt{{d}^{2}{x}^{2}-1}+dx \right ){\it csgn} \left ( d \right ) \right ) a{d}^{2}+\ln \left ( \left ({\it csgn} \left ( d \right ) \sqrt{{d}^{2}{x}^{2}-1}+dx \right ){\it csgn} \left ( d \right ) \right ) c \right ){\frac{1}{\sqrt{{d}^{2}{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*x-1)^(1/2)*(d*x+1)^(1/2)*(csgn(d)*d*(d^2*x^2-1)^(1/2)*x*c+2*csgn(d)*d*(d^2*x^2-1)^(1/2)*b+2*ln((csgn(d)
*(d^2*x^2-1)^(1/2)+d*x)*csgn(d))*a*d^2+ln((csgn(d)*(d^2*x^2-1)^(1/2)+d*x)*csgn(d))*c)*csgn(d)/d^3/(d^2*x^2-1)^
(1/2)

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Maxima [B]  time = 1.35955, size = 142, normalized size = 2.73 \begin{align*} \frac{a \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - 1} \sqrt{d^{2}}\right )}{\sqrt{d^{2}}} + \frac{\sqrt{d^{2} x^{2} - 1} c x}{2 \, d^{2}} + \frac{\sqrt{d^{2} x^{2} - 1} b}{d^{2}} + \frac{c \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - 1} \sqrt{d^{2}}\right )}{2 \, \sqrt{d^{2}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.10158, size = 150, normalized size = 2.88 \begin{align*} \frac{{\left (c d x + 2 \, b d\right )} \sqrt{d x + 1} \sqrt{d x - 1} -{\left (2 \, a d^{2} + c\right )} \log \left (-d x + \sqrt{d x + 1} \sqrt{d x - 1}\right )}{2 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 20.7389, size = 277, normalized size = 5.33 \begin{align*} \frac{a{G_{6, 6}^{6, 2}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} & \frac{1}{2}, \frac{1}{2}, 1, 1 \\0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 0 & \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d} - \frac{i a{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, 1 & \\- \frac{1}{4}, \frac{1}{4} & - \frac{1}{2}, 0, 0, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d} + \frac{b{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{4} & 0, 0, \frac{1}{2}, 1 \\- \frac{1}{2}, - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, 0 & \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{2}} + \frac{i b{G_{6, 6}^{2, 6}\left (\begin{matrix} -1, - \frac{3}{4}, - \frac{1}{2}, - \frac{1}{4}, 0, 1 & \\- \frac{3}{4}, - \frac{1}{4} & -1, - \frac{1}{2}, - \frac{1}{2}, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{2}} + \frac{c{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{4} & - \frac{1}{2}, - \frac{1}{2}, 0, 1 \\-1, - \frac{3}{4}, - \frac{1}{2}, - \frac{1}{4}, 0, 0 & \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{3}} - \frac{i c{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{3}{2}, - \frac{5}{4}, -1, - \frac{3}{4}, - \frac{1}{2}, 1 & \\- \frac{5}{4}, - \frac{3}{4} & - \frac{3}{2}, -1, -1, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d) - I*a
*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), exp_polar(2*I*pi)/(d**2*x**2))/(4
*pi**(3/2)*d) + b*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), 1/(d**2*x**2))/(4
*pi**(3/2)*d**2) + I*b*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), exp_po
lar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**2) + c*meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2,
-1/4, 0, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**3) - I*c*meijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/
4, -3/4), (-3/2, -1, -1, 0)), exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**3)

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Giac [A]  time = 2.5708, size = 104, normalized size = 2. \begin{align*} \frac{{\left ({\left (d x + 1\right )} c d^{4} + 2 \, b d^{5} - c d^{4}\right )} \sqrt{d x + 1} \sqrt{d x - 1} - 2 \,{\left (2 \, a d^{6} + c d^{4}\right )} \log \left ({\left | -\sqrt{d x + 1} + \sqrt{d x - 1} \right |}\right )}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(((d*x + 1)*c*d^4 + 2*b*d^5 - c*d^4)*sqrt(d*x + 1)*sqrt(d*x - 1) - 2*(2*a*d^6 + c*d^4)*log(abs(-sqrt(d*x
+ 1) + sqrt(d*x - 1))))/d