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

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

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Rubi [A]  time = 0.19, antiderivative size = 71, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.05, size = 56, normalized size = 0.79 \begin {gather*} -\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 112, normalized size = 1.58 \begin {gather*} \left (-a d^2-2 c\right ) \tanh ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {d x+1}}\right )-\frac {d \sqrt {1-d x} \left (\frac {a d (1-d x)}{d x+1}+a d-\frac {2 b (1-d x)}{d x+1}+2 b\right )}{\sqrt {d x+1} \left (\frac {1-d x}{d x+1}-1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 65, normalized size = 0.92 \begin {gather*} \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 {gather*}

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

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 >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2,2]%%%}] at parameters values [70,22]Warning, choosing root of [1,0,-4,0,%%%{4,[2,2]%%%}] at parameters va
lues [42,56]1/d*(-1/2*(a*d^3+2*c*d)*ln(abs(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))+2-1/2*(-2*sqrt(-d*x+1)+2*
sqrt(2))/sqrt(d*x+1)))+1/2*(a*d^3+2*c*d)*ln(abs(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-2-1/2*(-2*sqrt(-d*x+
1)+2*sqrt(2))/sqrt(d*x+1)))-(2*a*d^3*(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2)
)/sqrt(d*x+1))^3-4*b*d^2*(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1
))^3+8*a*d^3*(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1))+16*b*d^2*
(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1)))/((2*sqrt(d*x+1)/(-2*s
qrt(-d*x+1)+2*sqrt(2))-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1))^2-4)^2)

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maple [C]  time = 0.02, size = 108, normalized size = 1.52 \begin {gather*} -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (a \,d^{2} x^{2} \arctanh \left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right )+2 c \,x^{2} \arctanh \left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right )+2 \sqrt {-d^{2} x^{2}+1}\, b x +\sqrt {-d^{2} x^{2}+1}\, a \right ) \mathrm {csgn}\relax (d )^{2}}{2 \sqrt {-d^{2} x^{2}+1}\, x^{2}} \end {gather*}

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 = 0.97, size = 98, normalized size = 1.38 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 5.85, size = 312, normalized size = 4.39 \begin {gather*} c\,\left (\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )-\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\right )-\frac {\frac {a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {a\,d^2}{2}+\frac {15\,a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^4}{2\,{\left (\sqrt {d\,x+1}-1\right )}^4}}{\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {32\,{\left (\sqrt {1-d\,x}-1\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}}+\frac {a\,d^2\,\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )}{2}-\frac {a\,d^2\,\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2}-\frac {b\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{x}+\frac {a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{32\,{\left (\sqrt {d\,x+1}-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(log(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 - 1) - log(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1))
) - ((a*d^2*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (a*d^2)/2 + (15*a*d^2*((1 - d*x)^(1/2) - 1)^4)/
(2*((d*x + 1)^(1/2) - 1)^4))/((16*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (32*((1 - d*x)^(1/2) - 1)
^4)/((d*x + 1)^(1/2) - 1)^4 + (16*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6) + (a*d^2*log(((1 - d*x)^(1
/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 - 1))/2 - (a*d^2*log(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/2 - (b*(
1 - d*x)^(1/2)*(d*x + 1)^(1/2))/x + (a*d^2*((1 - d*x)^(1/2) - 1)^2)/(32*((d*x + 1)^(1/2) - 1)^2)

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sympy [C]  time = 80.29, size = 218, normalized size = 3.07 \begin {gather*} \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 {gather*}

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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