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

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

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Rubi [B]  time = 0.07, antiderivative size = 135, normalized size of antiderivative = 2.60, 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} \begin {gather*} \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}} \end {gather*}

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 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])

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 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 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]

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*}

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Mathematica [B]  time = 0.22, size = 126, normalized size = 2.42 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.14, size = 112, normalized size = 2.15 \begin {gather*} \frac {\left (2 a d^2+c\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-1}}{\sqrt {d x+1}}\right )}{d^3}-\frac {\sqrt {d x-1} \left (\frac {2 b d (d x-1)}{d x+1}-2 b d-\frac {c (d x-1)}{d x+1}-c\right )}{d^3 \sqrt {d x+1} \left (\frac {d x-1}{d x+1}-1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 61, normalized size = 1.17 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.25, size = 80, normalized size = 1.54 \begin {gather*} \frac {\sqrt {d x + 1} \sqrt {d x - 1} {\left (\frac {{\left (d x + 1\right )} c}{d^{2}} + \frac {2 \, b d^{5} - c d^{4}}{d^{6}}\right )} - \frac {2 \, {\left (2 \, a d^{2} + c\right )} \log \left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}{d^{2}}}{2 \, d} \end {gather*}

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]

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

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maple [C]  time = 0.02, size = 120, normalized size = 2.31 \begin {gather*} \frac {\sqrt {d x -1}\, \sqrt {d x +1}\, \left (2 a \,d^{2} \ln \left (\left (d x +\sqrt {d^{2} x^{2}-1}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+\sqrt {d^{2} x^{2}-1}\, c d x \,\mathrm {csgn}\relax (d )+2 \sqrt {d^{2} x^{2}-1}\, b d \,\mathrm {csgn}\relax (d )+c \ln \left (\left (d x +\sqrt {d^{2} x^{2}-1}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )\right ) \mathrm {csgn}\relax (d )}{2 \sqrt {d^{2} x^{2}-1}\, d^{3}} \end {gather*}

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

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maxima [B]  time = 0.43, size = 90, normalized size = 1.73 \begin {gather*} \frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \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} d\right )}{2 \, d^{3}} \end {gather*}

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)*d)/d + 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)*d)/d^3

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mupad [B]  time = 12.40, size = 312, normalized size = 6.00 \begin {gather*} \frac {b\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}+\frac {2\,c\,\mathrm {atanh}\left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-\frac {\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {2\,c\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\sqrt {d\,x+1}-1}}{d^3-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*atanh(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1)))/d^3 - ((14*c*((d*x - 1)^(1/2) - 1i)^3)/((d*x + 1)^(1
/2) - 1)^3 + (14*c*((d*x - 1)^(1/2) - 1i)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*c*((d*x - 1)^(1/2) - 1i)^7)/((d*x +
1)^(1/2) - 1)^7 + (2*c*((d*x - 1)^(1/2) - 1i))/((d*x + 1)^(1/2) - 1))/(d^3 - (4*d^3*((d*x - 1)^(1/2) - 1i)^2)/
((d*x + 1)^(1/2) - 1)^2 + (6*d^3*((d*x - 1)^(1/2) - 1i)^4)/((d*x + 1)^(1/2) - 1)^4 - (4*d^3*((d*x - 1)^(1/2) -
1i)^6)/((d*x + 1)^(1/2) - 1)^6 + (d^3*((d*x - 1)^(1/2) - 1i)^8)/((d*x + 1)^(1/2) - 1)^8) - (4*a*atan((d*((d*x
- 1)^(1/2) - 1i))/(((d*x + 1)^(1/2) - 1)*(-d^2)^(1/2))))/(-d^2)^(1/2) + (b*(d*x - 1)^(1/2)*(d*x + 1)^(1/2))/d
^2

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sympy [C]  time = 48.29, size = 277, normalized size = 5.33 \begin {gather*} \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 {gather*}

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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