3.19 \(\int \frac{a+b x+c x^2}{x^3 \sqrt{1-d x} \sqrt{1+d x}} \, dx\)

Optimal. Leaf size=71 \[ -\frac{1}{2} \left (a d^2+2 c\right ) \tanh ^{-1}\left (\sqrt{1-d^2 x^2}\right )-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x} \]

[Out]

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

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Rubi [A]  time = 0.184015, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1609, 1807, 807, 266, 63, 208} \[ -\frac{1}{2} \left (a d^2+2 c\right ) \tanh ^{-1}\left (\sqrt{1-d^2 x^2}\right )-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1609

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[P
x*(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d,
0] && EqQ[m, n] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{x^3 \sqrt{1-d x} \sqrt{1+d x}} \, dx &=\int \frac{a+b x+c x^2}{x^3 \sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{1}{2} \int \frac{-2 b-\left (2 c+a d^2\right ) x}{x^2 \sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x}-\frac{1}{2} \left (-2 c-a d^2\right ) \int \frac{1}{x \sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x}-\frac{1}{4} \left (-2 c-a d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-d^2 x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x}-\frac{1}{2} \left (a+\frac{2 c}{d^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{d^2}-\frac{x^2}{d^2}} \, dx,x,\sqrt{1-d^2 x^2}\right )\\ &=-\frac{a \sqrt{1-d^2 x^2}}{2 x^2}-\frac{b \sqrt{1-d^2 x^2}}{x}-\frac{1}{2} \left (2 c+a d^2\right ) \tanh ^{-1}\left (\sqrt{1-d^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0466222, size = 56, normalized size = 0.79 \[ -\frac{\sqrt{1-d^2 x^2} (a+2 b x)}{2 x^2}-\frac{1}{2} \left (a d^2+2 c\right ) \tanh ^{-1}\left (\sqrt{1-d^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0., size = 108, normalized size = 1.5 \begin{align*} -{\frac{ \left ({\it csgn} \left ( d \right ) \right ) ^{2}}{2\,{x}^{2}}\sqrt{-dx+1}\sqrt{dx+1} \left ({\it Artanh} \left ({\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){x}^{2}a{d}^{2}+2\,{\it Artanh} \left ({\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){x}^{2}c+2\,\sqrt{-{d}^{2}{x}^{2}+1}xb+\sqrt{-{d}^{2}{x}^{2}+1}a \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)/x^3/(-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)^2*(arctanh(1/(-d^2*x^2+1)^(1/2))*x^2*a*d^2+2*arctanh(1/(-d^2*x^2+1)^
(1/2))*x^2*c+2*(-d^2*x^2+1)^(1/2)*x*b+(-d^2*x^2+1)^(1/2)*a)/(-d^2*x^2+1)^(1/2)/x^2

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Maxima [A]  time = 3.97374, size = 132, normalized size = 1.86 \begin{align*} -\frac{1}{2} \, a d^{2} \log \left (\frac{2 \, \sqrt{-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - c \log \left (\frac{2 \, \sqrt{-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - \frac{\sqrt{-d^{2} x^{2} + 1} b}{x} - \frac{\sqrt{-d^{2} x^{2} + 1} a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/x^3/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.01834, size = 154, normalized size = 2.17 \begin{align*} \frac{{\left (a d^{2} + 2 \, c\right )} x^{2} \log \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{x}\right ) -{\left (2 \, b x + a\right )} \sqrt{d x + 1} \sqrt{-d x + 1}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/x^3/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 34.2892, size = 218, normalized size = 3.07 \begin{align*} \frac{i a d^{2}{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{7}{4}, \frac{9}{4}, 1 & 2, 2, \frac{5}{2} \\\frac{3}{2}, \frac{7}{4}, 2, \frac{9}{4}, \frac{5}{2} & 0 \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} - \frac{a d^{2}{G_{6, 6}^{2, 6}\left (\begin{matrix} 1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2, 1 & \\\frac{5}{4}, \frac{7}{4} & 1, \frac{3}{2}, \frac{3}{2}, 0 \end{matrix} \middle |{\frac{e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} + \frac{i b d{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{5}{4}, \frac{7}{4}, 1 & \frac{3}{2}, \frac{3}{2}, 2 \\1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2 & 0 \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} + \frac{b d{G_{6, 6}^{2, 6}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2}, 1 & \\\frac{3}{4}, \frac{5}{4} & \frac{1}{2}, 1, 1, 0 \end{matrix} \middle |{\frac{e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} + \frac{i c{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4}, 1 & 1, 1, \frac{3}{2} \\\frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2} & 0 \end{matrix} \middle |{\frac{1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} - \frac{c{G_{6, 6}^{2, 6}\left (\begin{matrix} 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1 & \\\frac{1}{4}, \frac{3}{4} & 0, \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/x**3/(-d*x+1)**(1/2)/(d*x+1)**(1/2),x)

[Out]

I*a*d**2*meijerg(((7/4, 9/4, 1), (2, 2, 5/2)), ((3/2, 7/4, 2, 9/4, 5/2), (0,)), 1/(d**2*x**2))/(4*pi**(3/2)) -
a*d**2*meijerg(((1, 5/4, 3/2, 7/4, 2, 1), ()), ((5/4, 7/4), (1, 3/2, 3/2, 0)), exp_polar(-2*I*pi)/(d**2*x**2)
)/(4*pi**(3/2)) + I*b*d*meijerg(((5/4, 7/4, 1), (3/2, 3/2, 2)), ((1, 5/4, 3/2, 7/4, 2), (0,)), 1/(d**2*x**2))/
(4*pi**(3/2)) + b*d*meijerg(((1/2, 3/4, 1, 5/4, 3/2, 1), ()), ((3/4, 5/4), (1/2, 1, 1, 0)), exp_polar(-2*I*pi)
/(d**2*x**2))/(4*pi**(3/2)) + I*c*meijerg(((3/4, 5/4, 1), (1, 1, 3/2)), ((1/2, 3/4, 1, 5/4, 3/2), (0,)), 1/(d*
*2*x**2))/(4*pi**(3/2)) - c*meijerg(((0, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)), exp_polar(
-2*I*pi)/(d**2*x**2))/(4*pi**(3/2))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/x^3/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError