3.16.75 \(\int \frac {c+b x^2+c k^2 x^4}{\sqrt {(1-x^2) (1-k^2 x^2)} (-1+k^2 x^4)} \, dx\)

Optimal. Leaf size=108 \[ \frac {(-b-2 c k) \tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{4 (k-1) k}+\frac {(b-2 c k) \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{2 k (k+1)} \]

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Rubi [C]  time = 1.62, antiderivative size = 192, normalized size of antiderivative = 1.78, number of steps used = 8, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6719, 6725, 419, 537} \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} (b-2 c k) \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} (b+2 c k) \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + b*x^2 + c*k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

(c*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*EllipticF[ArcSin[x], k^2])/Sqrt[(1 - x^2)*(1 - k^2*x^2)] + ((b - 2*c*k)*Sqr
t[1 - x^2]*Sqrt[1 - k^2*x^2]*EllipticPi[-k, ArcSin[x], k^2])/(2*k*Sqrt[(1 - x^2)*(1 - k^2*x^2)]) - ((b + 2*c*k
)*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*EllipticPi[k, ArcSin[x], k^2])/(2*k*Sqrt[(1 - x^2)*(1 - k^2*x^2)])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {c+b x^2+c k^2 x^4}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (\frac {c}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}}+\frac {2 c+b x^2}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {2 c+b x^2}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (c \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (-\frac {b+2 c k}{2 k \sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}}+\frac {b-2 c k}{2 k \sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((b-2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left ((b+2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {(b-2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {(b+2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.51, size = 93, normalized size = 0.86 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left ((b-2 c k) \Pi \left (-k;\sin ^{-1}(x)|k^2\right )-(b+2 c k) \Pi \left (k;\sin ^{-1}(x)|k^2\right )+2 c k F\left (\sin ^{-1}(x)|k^2\right )\right )}{2 k \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + b*x^2 + c*k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

(Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*(2*c*k*EllipticF[ArcSin[x], k^2] + (b - 2*c*k)*EllipticPi[-k, ArcSin[x], k^2]
- (b + 2*c*k)*EllipticPi[k, ArcSin[x], k^2]))/(2*k*Sqrt[(-1 + x^2)*(-1 + k^2*x^2)])

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IntegrateAlgebraic [A]  time = 3.27, size = 108, normalized size = 1.00 \begin {gather*} \frac {(-b-2 c k) \tan ^{-1}\left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 (-1+k) k}+\frac {(b-2 c k) \tan ^{-1}\left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + b*x^2 + c*k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

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

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fricas [A]  time = 1.81, size = 107, normalized size = 0.99 \begin {gather*} \frac {{\left (2 \, c k^{2} - {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) + {\left (2 \, c k^{2} + {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{4 \, {\left (k^{3} - k\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*k^2*x^4+b*x^2+c)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c k^{2} x^{4} + b x^{2} + c}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*k^2*x^4+b*x^2+c)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="giac")

[Out]

integrate((c*k^2*x^4 + b*x^2 + c)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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maple [C]  time = 0.14, size = 174, normalized size = 1.61

method	result	size
default	\(\frac {c \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {\left (-2 c k +b \right ) \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\left (2 c k +b \right ) \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\)	\(174\)
elliptic	\(\frac {\left (\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right ) c}{-2+2 k}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right ) b}{4 k \left (-1+k \right )}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right ) c}{2+2 k}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right ) b}{4 k \left (1+k \right )}\right ) \sqrt {2}}{2}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*k^2*x^4+b*x^2+c)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x,method=_RETURNVERBOSE)

[Out]

c*(-x^2+1)^(1/2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticF(x,k)+1/2*(-2*c*k+b)/k*(-x^2+1)^(1/
2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticPi(x,-k,k)-1/2*(2*c*k+b)/k*(-x^2+1)^(1/2)*(-k^2*x^
2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticPi(x,k,k)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c k^{2} x^{4} + b x^{2} + c}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*k^2*x^4+b*x^2+c)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="maxima")

[Out]

integrate((c*k^2*x^4 + b*x^2 + c)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c\,k^2\,x^4+b\,x^2+c}{\left (k^2\,x^4-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + b*x^2 + c*k^2*x^4)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

int((c + b*x^2 + c*k^2*x^4)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b x^{2} + c k^{2} x^{4} + c}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*k**2*x**4+b*x**2+c)/((-x**2+1)*(-k**2*x**2+1))**(1/2)/(k**2*x**4-1),x)

[Out]

Integral((b*x**2 + c*k**2*x**4 + c)/(sqrt((x - 1)*(x + 1)*(k*x - 1)*(k*x + 1))*(k*x**2 - 1)*(k*x**2 + 1)), x)

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