3.357 \(\int \frac{a+b x^2}{x^5 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\)

Optimal. Leaf size=99 \[ \frac{\sqrt{c x-1} \sqrt{c x+1} \left (3 a c^2+4 b\right )}{8 x^2}+\frac{1}{8} c^2 \left (3 a c^2+4 b\right ) \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )+\frac{a \sqrt{c x-1} \sqrt{c x+1}}{4 x^4} \]

[Out]

(a*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*x^4) + ((4*b + 3*a*c^2)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(8*x^2) + (c^2*(4*b
+ 3*a*c^2)*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*x]])/8

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Rubi [A]  time = 0.0780453, antiderivative size = 99, normalized size of antiderivative = 1., 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 = {454, 103, 12, 92, 205} \[ \frac{\sqrt{c x-1} \sqrt{c x+1} \left (3 a c^2+4 b\right )}{8 x^2}+\frac{1}{8} c^2 \left (3 a c^2+4 b\right ) \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )+\frac{a \sqrt{c x-1} \sqrt{c x+1}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(x^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]

[Out]

(a*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*x^4) + ((4*b + 3*a*c^2)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(8*x^2) + (c^2*(4*b
+ 3*a*c^2)*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*x]])/8

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]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{a+b x^2}{x^5 \sqrt{-1+c x} \sqrt{1+c x}} \, dx &=\frac{a \sqrt{-1+c x} \sqrt{1+c x}}{4 x^4}+\frac{1}{4} \left (4 b+3 a c^2\right ) \int \frac{1}{x^3 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{a \sqrt{-1+c x} \sqrt{1+c x}}{4 x^4}+\frac{\left (4 b+3 a c^2\right ) \sqrt{-1+c x} \sqrt{1+c x}}{8 x^2}+\frac{1}{8} \left (4 b+3 a c^2\right ) \int \frac{c^2}{x \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{a \sqrt{-1+c x} \sqrt{1+c x}}{4 x^4}+\frac{\left (4 b+3 a c^2\right ) \sqrt{-1+c x} \sqrt{1+c x}}{8 x^2}+\frac{1}{8} \left (c^2 \left (4 b+3 a c^2\right )\right ) \int \frac{1}{x \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{a \sqrt{-1+c x} \sqrt{1+c x}}{4 x^4}+\frac{\left (4 b+3 a c^2\right ) \sqrt{-1+c x} \sqrt{1+c x}}{8 x^2}+\frac{1}{8} \left (c^3 \left (4 b+3 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}}{4 x^4}+\frac{\left (4 b+3 a c^2\right ) \sqrt{-1+c x} \sqrt{1+c x}}{8 x^2}+\frac{1}{8} c^2 \left (4 b+3 a c^2\right ) \tan ^{-1}\left (\sqrt{-1+c x} \sqrt{1+c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.09147, size = 102, normalized size = 1.03 \[ \frac{\left (c^2 x^2-1\right ) \left (a \left (3 c^2 x^2+2\right )+4 b x^2\right )-c^2 x^4 \sqrt{1-c^2 x^2} \left (3 a c^2+4 b\right ) \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{8 x^4 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)/(x^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]

[Out]

((-1 + c^2*x^2)*(4*b*x^2 + a*(2 + 3*c^2*x^2)) - c^2*(4*b + 3*a*c^2)*x^4*Sqrt[1 - c^2*x^2]*ArcTanh[Sqrt[1 - c^2
*x^2]])/(8*x^4*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

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Maple [A]  time = 0.02, size = 125, normalized size = 1.3 \begin{align*} -{\frac{1}{8\,{x}^{4}}\sqrt{cx-1}\sqrt{cx+1} \left ( 3\,\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){x}^{4}a{c}^{4}+4\,\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){x}^{4}b{c}^{2}-3\,\sqrt{{c}^{2}{x}^{2}-1}{x}^{2}a{c}^{2}-4\,\sqrt{{c}^{2}{x}^{2}-1}{x}^{2}b-2\,\sqrt{{c}^{2}{x}^{2}-1}a \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/x^5/(c*x-1)^(1/2)/(c*x+1)^(1/2),x)

[Out]

-1/8*(c*x-1)^(1/2)*(c*x+1)^(1/2)*(3*arctan(1/(c^2*x^2-1)^(1/2))*x^4*a*c^4+4*arctan(1/(c^2*x^2-1)^(1/2))*x^4*b*
c^2-3*(c^2*x^2-1)^(1/2)*x^2*a*c^2-4*(c^2*x^2-1)^(1/2)*x^2*b-2*(c^2*x^2-1)^(1/2)*a)/(c^2*x^2-1)^(1/2)/x^4

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Maxima [A]  time = 1.41214, size = 120, normalized size = 1.21 \begin{align*} -\frac{3}{8} \, a c^{4} \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - \frac{1}{2} \, b c^{2} \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{3 \, \sqrt{c^{2} x^{2} - 1} a c^{2}}{8 \, x^{2}} + \frac{\sqrt{c^{2} x^{2} - 1} b}{2 \, x^{2}} + \frac{\sqrt{c^{2} x^{2} - 1} a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="maxima")

[Out]

-3/8*a*c^4*arcsin(1/(sqrt(c^2)*abs(x))) - 1/2*b*c^2*arcsin(1/(sqrt(c^2)*abs(x))) + 3/8*sqrt(c^2*x^2 - 1)*a*c^2
/x^2 + 1/2*sqrt(c^2*x^2 - 1)*b/x^2 + 1/4*sqrt(c^2*x^2 - 1)*a/x^4

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Fricas [A]  time = 1.52103, size = 186, normalized size = 1.88 \begin{align*} \frac{2 \,{\left (3 \, a c^{4} + 4 \, b c^{2}\right )} x^{4} \arctan \left (-c x + \sqrt{c x + 1} \sqrt{c x - 1}\right ) +{\left ({\left (3 \, a c^{2} + 4 \, b\right )} x^{2} + 2 \, a\right )} \sqrt{c x + 1} \sqrt{c x - 1}}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="fricas")

[Out]

1/8*(2*(3*a*c^4 + 4*b*c^2)*x^4*arctan(-c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + ((3*a*c^2 + 4*b)*x^2 + 2*a)*sqrt(c
*x + 1)*sqrt(c*x - 1))/x^4

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Sympy [C]  time = 52.0269, size = 148, normalized size = 1.49 \begin{align*} - \frac{a c^{4}{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{11}{4}, \frac{13}{4}, 1 & 3, 3, \frac{7}{2} \\\frac{5}{2}, \frac{11}{4}, 3, \frac{13}{4}, \frac{7}{2} & 0 \end{matrix} \middle |{\frac{1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} + \frac{i a c^{4}{G_{6, 6}^{2, 6}\left (\begin{matrix} 2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}, 3, 1 & \\\frac{9}{4}, \frac{11}{4} & 2, \frac{5}{2}, \frac{5}{2}, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} - \frac{b 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 b 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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/x**5/(c*x-1)**(1/2)/(c*x+1)**(1/2),x)

[Out]

-a*c**4*meijerg(((11/4, 13/4, 1), (3, 3, 7/2)), ((5/2, 11/4, 3, 13/4, 7/2), (0,)), 1/(c**2*x**2))/(4*pi**(3/2)
) + I*a*c**4*meijerg(((2, 9/4, 5/2, 11/4, 3, 1), ()), ((9/4, 11/4), (2, 5/2, 5/2, 0)), exp_polar(2*I*pi)/(c**2
*x**2))/(4*pi**(3/2)) - b*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*b*c**2*meijerg(((1, 5/4, 3/2, 7/4, 2, 1), ()), ((5/4, 7/4), (1, 3/2, 3/2, 0)), exp_po
lar(2*I*pi)/(c**2*x**2))/(4*pi**(3/2))

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Giac [B]  time = 1.16859, size = 362, normalized size = 3.66 \begin{align*} -\frac{{\left (3 \, a c^{5} + 4 \, b c^{3}\right )} \arctan \left (\frac{1}{2} \,{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{2}\right ) + \frac{2 \,{\left (3 \, a c^{5}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{14} + 4 \, b c^{3}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{14} + 44 \, a c^{5}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{10} + 16 \, b c^{3}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{10} - 176 \, a c^{5}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{6} - 64 \, b c^{3}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{6} - 192 \, a c^{5}{\left (\sqrt{c x + 1} - \sqrt{c x - 1}\right )}^{2} - 256 \, b 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 )}^{4}}}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="giac")

[Out]

-1/4*((3*a*c^5 + 4*b*c^3)*arctan(1/2*(sqrt(c*x + 1) - sqrt(c*x - 1))^2) + 2*(3*a*c^5*(sqrt(c*x + 1) - sqrt(c*x
- 1))^14 + 4*b*c^3*(sqrt(c*x + 1) - sqrt(c*x - 1))^14 + 44*a*c^5*(sqrt(c*x + 1) - sqrt(c*x - 1))^10 + 16*b*c^
3*(sqrt(c*x + 1) - sqrt(c*x - 1))^10 - 176*a*c^5*(sqrt(c*x + 1) - sqrt(c*x - 1))^6 - 64*b*c^3*(sqrt(c*x + 1) -
sqrt(c*x - 1))^6 - 192*a*c^5*(sqrt(c*x + 1) - sqrt(c*x - 1))^2 - 256*b*c^3*(sqrt(c*x + 1) - sqrt(c*x - 1))^2)
/((sqrt(c*x + 1) - sqrt(c*x - 1))^4 + 4)^4)/c