3.4.64 \(\int \frac {a+b x^2}{x^2 \sqrt {-c+d x} \sqrt {c+d x}} \, dx\) [364]

Optimal. Leaf size=57 \[ \frac {a \sqrt {-c+d x} \sqrt {c+d x}}{c^2 x}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {465, 65, 223, 212} \begin {gather*} \frac {a \sqrt {d x-c} \sqrt {c+d x}}{c^2 x}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 465

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

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 57, normalized size = 1.00 \begin {gather*} \frac {a \sqrt {-c+d x} \sqrt {c+d x}}{c^2 x}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.30, size = 97, normalized size = 1.70

method	result	size
default	\(\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (\ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{2} x +\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right ) d a \right ) \mathrm {csgn}\left (d \right )}{c^{2} \sqrt {d^{2} x^{2}-c^{2}}\, d x}\)	\(97\)
risch	\(-\frac {a \left (-d x +c \right ) \sqrt {d x +c}}{c^{2} x \sqrt {d x -c}}+\frac {b \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 55, normalized size = 0.96 \begin {gather*} \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a}{c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 2.18, size = 68, normalized size = 1.19 \begin {gather*} -\frac {b c^{2} x \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right ) - a d^{2} x - \sqrt {d x + c} \sqrt {d x - c} a d}{c^{2} d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 27.99, size = 165, normalized size = 2.89 \begin {gather*} - \frac {a 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 {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} - \frac {i a 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 {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} + \frac {b {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 {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i b {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 {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.67, size = 66, normalized size = 1.16 \begin {gather*} \frac {\frac {16 \, a d^{2}}{{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}} - b \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 2.94, size = 77, normalized size = 1.35 \begin {gather*} \frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}{\sqrt {-d^2}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )}{\sqrt {-d^2}}+\frac {a\,\sqrt {c+d\,x}\,\sqrt {d\,x-c}}{c^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b*atan((d*((-c)^(1/2) - (d*x - c)^(1/2)))/((-d^2)^(1/2)*((c + d*x)^(1/2) - c^(1/2)))))/(-d^2)^(1/2) + (a*(c
+ d*x)^(1/2)*(d*x - c)^(1/2))/(c^2*x)

________________________________________________________________________________________