3.165 \(\int \frac{1}{\sqrt{-1+c^2 x^2} (d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{x \sqrt{c^2 x^2-1}}{2 \left (d-c^2 d x^2\right )^{3/2}}-\frac{\sqrt{c^2 x^2-1} \tanh ^{-1}(c x)}{2 c d \sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0168311, antiderivative size = 91, normalized size of antiderivative = 1.2, 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 = {23, 199, 208} \[ \frac{x \sqrt{d-c^2 d x^2}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt{c^2 x^2-1}}+\frac{\sqrt{d-c^2 d x^2} \tanh ^{-1}(c x)}{2 c d^2 \sqrt{c^2 x^2-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac{\sqrt{d-c^2 d x^2} \int \frac{1}{\left (d-c^2 d x^2\right )^2} \, dx}{\sqrt{-1+c^2 x^2}}\\ &=\frac{x \sqrt{d-c^2 d x^2}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt{-1+c^2 x^2}}+\frac{\sqrt{d-c^2 d x^2} \int \frac{1}{d-c^2 d x^2} \, dx}{2 d \sqrt{-1+c^2 x^2}}\\ &=\frac{x \sqrt{d-c^2 d x^2}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt{-1+c^2 x^2}}+\frac{\sqrt{d-c^2 d x^2} \tanh ^{-1}(c x)}{2 c d^2 \sqrt{-1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0287236, size = 57, normalized size = 0.75 \[ \frac{\left (1-c^2 x^2\right ) \tanh ^{-1}(c x)+c x}{2 c d \sqrt{c^2 x^2-1} \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 94, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( cx-1 \right ){x}^{2}{c}^{2}-\ln \left ( cx+1 \right ){x}^{2}{c}^{2}+2\,cx-\ln \left ( cx-1 \right ) +\ln \left ( cx+1 \right ) }{4\,{d}^{2}c \left ( cx-1 \right ) \left ( cx+1 \right ) }\sqrt{- \left ({c}^{2}{x}^{2}-1 \right ) d}{\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(c^2*x^2-1)^(1/2)*(-(c^2*x^2-1)*d)^(1/2)*(ln(c*x-1)*x^2*c^2-ln(c*x+1)*x^2*c^2+2*c*x-ln(c*x-1)+ln(c*x+1))/
d^2/c/(c*x-1)/(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} \sqrt{c^{2} x^{2} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-c^2*d*x^2 + d)^(3/2)*sqrt(c^2*x^2 - 1)), x)

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Fricas [B]  time = 2.03795, size = 664, normalized size = 8.74 \begin{align*} \left [-\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} c x +{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt{-d} \log \left (-\frac{c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \,{\left (c^{3} x^{3} + c x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} \sqrt{-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right )}{8 \,{\left (c^{5} d^{2} x^{4} - 2 \, c^{3} d^{2} x^{2} + c d^{2}\right )}}, -\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} c x -{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt{d} \arctan \left (\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} c \sqrt{d} x}{c^{4} d x^{4} - d}\right )}{4 \,{\left (c^{5} d^{2} x^{4} - 2 \, c^{3} d^{2} x^{2} + c d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*x + (c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c
^4*d*x^4 - 5*c^2*d*x^2 + 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c
^4*x^4 + 3*c^2*x^2 - 1)))/(c^5*d^2*x^4 - 2*c^3*d^2*x^2 + c*d^2), -1/4*(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1
)*c*x - (c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(d)*arctan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*sqrt(d)*x/(c^4*d*x
^4 - d)))/(c^5*d^2*x^4 - 2*c^3*d^2*x^2 + c*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\left (c x - 1\right ) \left (c x + 1\right )} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt((c*x - 1)*(c*x + 1))*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} \sqrt{c^{2} x^{2} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-c^2*d*x^2 + d)^(3/2)*sqrt(c^2*x^2 - 1)), x)