3.2.63 \(\int \frac {\sqrt {-1+c^2 x^2}}{(d-c^2 d x^2)^{5/2}} \, dx\) [163]

Optimal. Leaf size=79 \[ \frac {x \sqrt {-1+c^2 x^2}}{2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\sqrt {-1+c^2 x^2} \tanh ^{-1}(c x)}{2 c d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 91, normalized size of antiderivative = 1.15, 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, 205, 214} \begin {gather*} \frac {x \sqrt {c^2 x^2-1}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {c^2 x^2-1} \tanh ^{-1}(c x)}{2 c d^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[-1 + c^2*x^2])/(2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (Sqrt[-1 + c^2*x^2]*ArcTanh[c*x])/(2*c*d^2*
Sqrt[d - c^2*d*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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 214

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

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+c^2 x^2}}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {\sqrt {-1+c^2 x^2} \int \frac {1}{\left (d-c^2 d x^2\right )^2} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {x \sqrt {-1+c^2 x^2}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {-1+c^2 x^2} \int \frac {1}{d-c^2 d x^2} \, dx}{2 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \sqrt {-1+c^2 x^2}}{2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {-1+c^2 x^2} \tanh ^{-1}(c x)}{2 c d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 57, normalized size = 0.72 \begin {gather*} \frac {-c x+\left (-1+c^2 x^2\right ) \tanh ^{-1}(c x)}{2 c d^2 \sqrt {-1+c^2 x^2} \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 94, normalized size = 1.19

method	result	size
default	\(\frac {\sqrt {-\left (c^{2} x^{2}-1\right ) d}\, \left (\ln \left (c x -1\right ) c^{2} x^{2}-\ln \left (c x +1\right ) c^{2} x^{2}+2 c x -\ln \left (c x -1\right )+\ln \left (c x +1\right )\right )}{4 \sqrt {c^{2} x^{2}-1}\, d^{3} \left (c x -1\right ) c \left (c x +1\right )}\)	\(94\)
risch	\(-\frac {x}{2 d^{2} \sqrt {c^{2} x^{2}-1}\, \sqrt {-\left (c^{2} x^{2}-1\right ) d}}-\frac {\sqrt {c^{2} x^{2}-1}\, \ln \left (c x -1\right )}{4 d^{2} \sqrt {-\left (c^{2} x^{2}-1\right ) d}\, c}+\frac {\sqrt {c^{2} x^{2}-1}\, \ln \left (-c x -1\right )}{4 d^{2} \sqrt {-\left (c^{2} x^{2}-1\right ) d}\, c}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4/(c^2*x^2-1)^(1/2)*(-(c^2*x^2-1)*d)^(1/2)*(ln(c*x-1)*c^2*x^2-ln(c*x+1)*c^2*x^2+2*c*x-ln(c*x-1)+ln(c*x+1))/d
^3/(c*x-1)/c/(c*x+1)

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Maxima [A]
time = 0.30, size = 70, normalized size = 0.89 \begin {gather*} -\frac {x}{2 \, {\left (c^{2} \sqrt {-d} d^{2} x^{2} - \sqrt {-d} d^{2}\right )}} - \frac {\sqrt {-d} \log \left (c x + 1\right )}{4 \, c d^{3}} + \frac {\sqrt {-d} \log \left (c x - 1\right )}{4 \, c d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(c^2*sqrt(-d)*d^2*x^2 - sqrt(-d)*d^2) - 1/4*sqrt(-d)*log(c*x + 1)/(c*d^3) + 1/4*sqrt(-d)*log(c*x - 1)/(
c*d^3)

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Fricas [A]
time = 1.08, size = 315, normalized size = 3.99 \begin {gather*} \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^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\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^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(5/2),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^3*x^4 - 2*c^3*d^3*x^2 + c*d^3), 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^3*x^4 - 2*c^3*d^3*x^2 + c*d^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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 {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((c*x - 1)*(c*x + 1))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c^2*x^2 - 1)/(-c^2*d*x^2 + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c^2\,x^2-1}}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*x^2 - 1)^(1/2)/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((c^2*x^2 - 1)^(1/2)/(d - c^2*d*x^2)^(5/2), x)

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