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

Optimal. Leaf size=60 \[ \frac {1}{2} \left (a c^2+2 b\right ) \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {a \sqrt {c x-1} \sqrt {c x+1}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {454, 92, 205} \[ \frac {1}{2} \left (a c^2+2 b\right ) \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {a \sqrt {c x-1} \sqrt {c x+1}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

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

Rule 454

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(a1*a2*e*
(m + 1)), x] + Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*e^n*(m + 1)), Int[(e*x)^(m + n)*(a1
+ b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, p}, x] && EqQ[non2, n/2] && Eq
Q[a2*b1 + a1*b2, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1
])) &&  !ILtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 77, normalized size = 1.28 \[ \frac {x^2 \sqrt {c^2 x^2-1} \left (a c^2+2 b\right ) \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )+a \left (c^2 x^2-1\right )}{2 x^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-1 + c^2*x^2) + (2*b + a*c^2)*x^2*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(2*x^2*Sqrt[-1 + c*x]*Sqr
t[1 + c*x])

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fricas [A]  time = 1.07, size = 57, normalized size = 0.95 \[ \frac {2 \, {\left (a c^{2} + 2 \, b\right )} x^{2} \arctan \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) + \sqrt {c x + 1} \sqrt {c x - 1} a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 114, normalized size = 1.90 \[ -\frac {{\left (a c^{3} + 2 \, b c\right )} \arctan \left (\frac {1}{2} \, {\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{2}\right ) + \frac {2 \, {\left (a c^{3} {\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{6} - 4 \, a c^{3} {\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{2}\right )}}{{\left ({\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{4} + 4\right )}^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*c^3 + 2*b*c)*arctan(1/2*(sqrt(c*x + 1) - sqrt(c*x - 1))^2) + 2*(a*c^3*(sqrt(c*x + 1) - sqrt(c*x - 1))^6 -
4*a*c^3*(sqrt(c*x + 1) - sqrt(c*x - 1))^2)/((sqrt(c*x + 1) - sqrt(c*x - 1))^4 + 4)^2)/c

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maple [A]  time = 0.08, size = 84, normalized size = 1.40 \[ -\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (a \,c^{2} x^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )+2 b \,x^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )-\sqrt {c^{2} x^{2}-1}\, a \right )}{2 \sqrt {c^{2} x^{2}-1}\, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.23, size = 45, normalized size = 0.75 \[ -\frac {1}{2} \, a c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - b \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\sqrt {c^{2} x^{2} - 1} a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.67, size = 297, normalized size = 4.95 \[ \frac {\frac {a\,c^2\,1{}\mathrm {i}}{32}+\frac {a\,c^2\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,{\left (\sqrt {c\,x+1}-1\right )}^2}-\frac {a\,c^2\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,{\left (\sqrt {c\,x+1}-1\right )}^4}}{\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {c\,x+1}-1\right )}^4}+\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {c\,x+1}-1\right )}^6}}-b\,\left (\ln \left (\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )\right )\,1{}\mathrm {i}-\frac {a\,c^2\,\ln \left (\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{2}+\frac {a\,c^2\,\ln \left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )\,1{}\mathrm {i}}{2}+\frac {a\,c^2\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,{\left (\sqrt {c\,x+1}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 63.62, size = 141, normalized size = 2.35 \[ - \frac {a c^{2} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4}, 1 & 2, 2, \frac {5}{2} \\\frac {3}{2}, \frac {7}{4}, 2, \frac {9}{4}, \frac {5}{2} & 0 \end {matrix} \middle | {\frac {1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i a c^{2} {G_{6, 6}^{2, 6}\left (\begin {matrix} 1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2, 1 & \\\frac {5}{4}, \frac {7}{4} & 1, \frac {3}{2}, \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {b {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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