3.12.100 \(\int \frac {-1+k x^2}{(a+b x) \sqrt {(1-x) x (1-k x)} (b+a k x)} \, dx\)

Optimal. Leaf size=88 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {a+b} \sqrt {k x^3+(-k-1) x^2+x} \sqrt {a k+b}}{\sqrt {a} \sqrt {b} (x-1) (k x-1)}\right )}{\sqrt {a} \sqrt {b} \sqrt {a+b} \sqrt {a k+b}} \]

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Rubi [C]  time = 6.83, antiderivative size = 253, normalized size of antiderivative = 2.88, number of steps used = 14, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {6718, 6688, 6742, 714, 115, 934, 12, 168, 537} \begin {gather*} \frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k x} \Pi \left (-\frac {a}{b};\sin ^{-1}\left (\sqrt {-k} \sqrt {-x}\right )|\frac {1}{k}\right )}{a b \sqrt {-k} \sqrt {x-x^2} \sqrt {(1-x) x (1-k x)}}+\frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k x} \Pi \left (-\frac {b}{a k};\sin ^{-1}\left (\sqrt {-k} \sqrt {-x}\right )|\frac {1}{k}\right )}{a b \sqrt {-k} \sqrt {x-x^2} \sqrt {(1-x) x (1-k x)}}+\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k x} F\left (\left .\sin ^{-1}\left (\sqrt {x}\right )\right |k\right )}{a b \sqrt {(1-x) x (1-k x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 115

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (GtQ[-(b/d), 0] || LtQ[-(b/f), 0])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

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 714

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Int[(d + e*x)^m/(Sqrt[b*x]*Sqrt[1
+ (c*x)/b]), x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4] && LtQ[
c, 0] && RationalQ[b]

Rule 934

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x])/Sqrt[a + b*x + c*x^2], Int[1/((d +
e*x)*Sqrt[f + g*x]*Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x]), x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6718

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n*z^q)^FracP
art[p])/(v^(m*FracPart[p])*w^(n*FracPart[p])*z^(q*FracPart[p])), Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; Free
Q[{a, m, n, p, q}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !FreeQ[w, x] &&  !FreeQ[z, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [C]  time = 1.88, size = 169, normalized size = 1.92 \begin {gather*} \frac {2 i (x-1)^{3/2} \sqrt {\frac {x}{x-1}} \sqrt {\frac {1-k x}{k-k x}} \left (a b (k-1) F\left (i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|\frac {k-1}{k}\right )+a (a k+b) \Pi \left (\frac {a+b}{b};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|\frac {k-1}{k}\right )+b (a+b) \Pi \left (\frac {b}{a k}+1;i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|\frac {k-1}{k}\right )\right )}{a b (a+b) \sqrt {(x-1) x (k x-1)} (a k+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(-1 + x)^(3/2)*Sqrt[x/(-1 + x)]*Sqrt[(1 - k*x)/(k - k*x)]*(a*b*(-1 + k)*EllipticF[I*ArcSinh[1/Sqrt[-1 +
x]], (-1 + k)/k] + a*(b + a*k)*EllipticPi[(a + b)/b, I*ArcSinh[1/Sqrt[-1 + x]], (-1 + k)/k] + b*(a + b)*Ellip
ticPi[1 + b/(a*k), I*ArcSinh[1/Sqrt[-1 + x]], (-1 + k)/k]))/(a*b*(a + b)*(b + a*k)*Sqrt[(-1 + x)*x*(-1 + k*x)]
)

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IntegrateAlgebraic [A]  time = 0.30, size = 88, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {a+b} \sqrt {b+a k} \sqrt {x+(-1-k) x^2+k x^3}}{\sqrt {a} \sqrt {b} (-1+x) (-1+k x)}\right )}{\sqrt {a} \sqrt {b} \sqrt {a+b} \sqrt {b+a k}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[a + b]*Sqrt[b + a*k]*Sqrt[x + (-1 - k)*x^2 + k*x^3])/(Sqrt[a]*Sqrt[b]*(-1 + x)*(-1 + k*x))])/
(Sqrt[a]*Sqrt[b]*Sqrt[a + b]*Sqrt[b + a*k])

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fricas [B]  time = 1.59, size = 639, normalized size = 7.26 \begin {gather*} \left [-\frac {\sqrt {-a^{2} b^{2} - a b^{3} - {\left (a^{3} b + a^{2} b^{2}\right )} k} \log \left (\frac {a^{2} b^{2} k^{2} x^{4} + a^{2} b^{2} - 2 \, {\left ({\left (3 \, a^{3} b + 4 \, a^{2} b^{2}\right )} k^{2} + {\left (4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} k\right )} x^{3} + {\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4} + {\left (a^{4} + 8 \, a^{3} b + 8 \, a^{2} b^{2}\right )} k^{2} + 4 \, {\left (2 \, a^{3} b + 5 \, a^{2} b^{2} + 2 \, a b^{3}\right )} k\right )} x^{2} - 4 \, {\left (a b k x^{2} + a b - {\left (2 \, a b + b^{2} + {\left (a^{2} + 2 \, a b\right )} k\right )} x\right )} \sqrt {-a^{2} b^{2} - a b^{3} - {\left (a^{3} b + a^{2} b^{2}\right )} k} \sqrt {k x^{3} - {\left (k + 1\right )} x^{2} + x} - 2 \, {\left (4 \, a^{2} b^{2} + 3 \, a b^{3} + {\left (3 \, a^{3} b + 4 \, a^{2} b^{2}\right )} k\right )} x}{a^{2} b^{2} k^{2} x^{4} + a^{2} b^{2} + 2 \, {\left (a^{3} b k^{2} + a b^{3} k\right )} x^{3} + {\left (a^{4} k^{2} + 4 \, a^{2} b^{2} k + b^{4}\right )} x^{2} + 2 \, {\left (a^{3} b k + a b^{3}\right )} x}\right )}{2 \, {\left (a^{2} b^{2} + a b^{3} + {\left (a^{3} b + a^{2} b^{2}\right )} k\right )}}, \frac {\arctan \left (\frac {{\left (a b k x^{2} + a b - {\left (2 \, a b + b^{2} + {\left (a^{2} + 2 \, a b\right )} k\right )} x\right )} \sqrt {a^{2} b^{2} + a b^{3} + {\left (a^{3} b + a^{2} b^{2}\right )} k} \sqrt {k x^{3} - {\left (k + 1\right )} x^{2} + x}}{2 \, {\left ({\left ({\left (a^{3} b + a^{2} b^{2}\right )} k^{2} + {\left (a^{2} b^{2} + a b^{3}\right )} k\right )} x^{3} - {\left (a^{2} b^{2} + a b^{3} + {\left (a^{3} b + a^{2} b^{2}\right )} k^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} k\right )} x^{2} + {\left (a^{2} b^{2} + a b^{3} + {\left (a^{3} b + a^{2} b^{2}\right )} k\right )} x\right )}}\right )}{\sqrt {a^{2} b^{2} + a b^{3} + {\left (a^{3} b + a^{2} b^{2}\right )} k}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a^2*b^2 - a*b^3 - (a^3*b + a^2*b^2)*k)*log((a^2*b^2*k^2*x^4 + a^2*b^2 - 2*((3*a^3*b + 4*a^2*b^2)*k
^2 + (4*a^2*b^2 + 3*a*b^3)*k)*x^3 + (8*a^2*b^2 + 8*a*b^3 + b^4 + (a^4 + 8*a^3*b + 8*a^2*b^2)*k^2 + 4*(2*a^3*b
+ 5*a^2*b^2 + 2*a*b^3)*k)*x^2 - 4*(a*b*k*x^2 + a*b - (2*a*b + b^2 + (a^2 + 2*a*b)*k)*x)*sqrt(-a^2*b^2 - a*b^3
- (a^3*b + a^2*b^2)*k)*sqrt(k*x^3 - (k + 1)*x^2 + x) - 2*(4*a^2*b^2 + 3*a*b^3 + (3*a^3*b + 4*a^2*b^2)*k)*x)/(a
^2*b^2*k^2*x^4 + a^2*b^2 + 2*(a^3*b*k^2 + a*b^3*k)*x^3 + (a^4*k^2 + 4*a^2*b^2*k + b^4)*x^2 + 2*(a^3*b*k + a*b^
3)*x))/(a^2*b^2 + a*b^3 + (a^3*b + a^2*b^2)*k), arctan(1/2*(a*b*k*x^2 + a*b - (2*a*b + b^2 + (a^2 + 2*a*b)*k)*
x)*sqrt(a^2*b^2 + a*b^3 + (a^3*b + a^2*b^2)*k)*sqrt(k*x^3 - (k + 1)*x^2 + x)/(((a^3*b + a^2*b^2)*k^2 + (a^2*b^
2 + a*b^3)*k)*x^3 - (a^2*b^2 + a*b^3 + (a^3*b + a^2*b^2)*k^2 + (a^3*b + 2*a^2*b^2 + a*b^3)*k)*x^2 + (a^2*b^2 +
a*b^3 + (a^3*b + a^2*b^2)*k)*x))/sqrt(a^2*b^2 + a*b^3 + (a^3*b + a^2*b^2)*k)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.17, size = 315, normalized size = 3.58

method	result	size
default	\(-\frac {2 \sqrt {-\left (x -\frac {1}{k}\right ) k}\, \sqrt {\frac {-1+x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticF \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{a b k \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}}+\frac {2 \sqrt {-\left (x -\frac {1}{k}\right ) k}\, \sqrt {\frac {-1+x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {b}{a k}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{a^{2} k^{2} \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {b}{a k}\right )}+\frac {2 \sqrt {-\left (x -\frac {1}{k}\right ) k}\, \sqrt {\frac {-1+x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {a}{b}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{b^{2} k \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {a}{b}\right )}\)	\(315\)
elliptic	\(-\frac {2 \sqrt {-k x +1}\, \sqrt {-\frac {1}{\frac {1}{k}-1}+\frac {x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticF \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{a b k \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}}+\frac {2 \sqrt {-k x +1}\, \sqrt {-\frac {1}{\frac {1}{k}-1}+\frac {x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {a}{b}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{\left (-a^{2} k +b^{2}\right ) k \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {a}{b}\right )}-\frac {2 \sqrt {-k x +1}\, \sqrt {-\frac {1}{\frac {1}{k}-1}+\frac {x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {a}{b}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right ) a^{2}}{\left (-a^{2} k +b^{2}\right ) \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {a}{b}\right ) b^{2}}+\frac {2 \sqrt {-k x +1}\, \sqrt {-\frac {1}{\frac {1}{k}-1}+\frac {x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {b}{a k}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right )}{\left (a^{2} k -b^{2}\right ) k \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {b}{a k}\right )}-\frac {2 \sqrt {-k x +1}\, \sqrt {-\frac {1}{\frac {1}{k}-1}+\frac {x}{\frac {1}{k}-1}}\, \sqrt {k x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k}\right ) k}, \frac {1}{k \left (\frac {1}{k}+\frac {b}{a k}\right )}, \sqrt {\frac {1}{k \left (\frac {1}{k}-1\right )}}\right ) b^{2}}{\left (a^{2} k -b^{2}\right ) k^{2} \sqrt {k \,x^{3}-k \,x^{2}-x^{2}+x}\, \left (\frac {1}{k}+\frac {b}{a k}\right ) a^{2}}\)	\(608\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a/b/k*(-(x-1/k)*k)^(1/2)*((-1+x)/(1/k-1))^(1/2)*(k*x)^(1/2)/(k*x^3-k*x^2-x^2+x)^(1/2)*EllipticF((-(x-1/k)*k
)^(1/2),(1/k/(1/k-1))^(1/2))+2/a^2/k^2*(-(x-1/k)*k)^(1/2)*((-1+x)/(1/k-1))^(1/2)*(k*x)^(1/2)/(k*x^3-k*x^2-x^2+
x)^(1/2)/(1/k+b/a/k)*EllipticPi((-(x-1/k)*k)^(1/2),1/k/(1/k+b/a/k),(1/k/(1/k-1))^(1/2))+2/b^2/k*(-(x-1/k)*k)^(
1/2)*((-1+x)/(1/k-1))^(1/2)*(k*x)^(1/2)/(k*x^3-k*x^2-x^2+x)^(1/2)/(1/k+a/b)*EllipticPi((-(x-1/k)*k)^(1/2),1/k/
(1/k+a/b),(1/k/(1/k-1))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 103, normalized size = 1.17 \begin {gather*} \frac {\ln \left (\frac {2\,\sqrt {x\,\left (k\,x-1\right )\,\left (x-1\right )}\,\sqrt {a\,b\,\left (a+b\right )\,\left (b+a\,k\right )}+b^2\,x\,1{}\mathrm {i}-a\,b\,1{}\mathrm {i}+a^2\,k\,x\,1{}\mathrm {i}+a\,b\,x\,2{}\mathrm {i}+a\,b\,k\,x\,2{}\mathrm {i}-a\,b\,k\,x^2\,1{}\mathrm {i}}{\left (b+a\,k\,x\right )\,\left (a+b\,x\right )}\right )\,1{}\mathrm {i}}{\sqrt {a\,b\,\left (a+b\right )\,\left (b+a\,k\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((b^2*x*1i - a*b*1i + 2*(x*(k*x - 1)*(x - 1))^(1/2)*(a*b*(a + b)*(b + a*k))^(1/2) + a^2*k*x*1i + a*b*x*2i
+ a*b*k*x*2i - a*b*k*x^2*1i)/((b + a*k*x)*(a + b*x)))*1i)/(a*b*(a + b)*(b + a*k))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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