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

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

[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

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Rubi [A]  time = 0.0747374, antiderivative size = 57, normalized size of antiderivative = 1., 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 = {454, 63, 217, 206} \[ \frac{a \sqrt{d x-c} \sqrt{c+d x}}{c^2 x}+\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{d} \]

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 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 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 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 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])

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) \operatorname{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) \operatorname{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 = 0.0389778, size = 90, normalized size = 1.58 \[ \frac{\sqrt{d^2 x^2-c^2} \left (\frac{a \sqrt{d^2 x^2-c^2}}{c^2 x}+\frac{b \tanh ^{-1}\left (\frac{d x}{\sqrt{d^2 x^2-c^2}}\right )}{d}\right )}{\sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.02, size = 97, normalized size = 1.7 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{{c}^{2}dx}\sqrt{dx-c}\sqrt{dx+c} \left ( \ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) xb{c}^{2}+{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a \right ){\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}} \end{align*}

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)

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.52417, size = 143, normalized size = 2.51 \begin{align*} -\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{align*}

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)

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Sympy [C]  time = 21.4911, size = 165, normalized size = 2.89 \begin{align*} - \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{align*}

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)

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Giac [A]  time = 1.18928, size = 89, normalized size = 1.56 \begin{align*} \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{align*}

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