\(\int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx\) [75]

Optimal result

Integrand size = 40, antiderivative size = 86 \[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=-\frac {2 \sqrt {\frac {f^2 (c+d x)}{d+c f^2}} \operatorname {EllipticPi}\left (\frac {2 b}{b+a f^2},\arcsin \left (\frac {\sqrt {1-f^2 x}}{\sqrt {2}}\right ),\frac {2 d}{d+c f^2}\right )}{\left (b+a f^2\right ) \sqrt {c+d x}} \]

[Out]

-2*EllipticPi(1/2*(-f^2*x+1)^(1/2)*2^(1/2),2*b/(a*f^2+b),2^(1/2)*(d/(c*f^2+d))^(1/2))*(f^2*(d*x+c)/(c*f^2+d))^
(1/2)/(a*f^2+b)/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {174, 552, 551} \[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=-\frac {2 \sqrt {\frac {f^2 (c+d x)}{c f^2+d}} \operatorname {EllipticPi}\left (\frac {2 b}{a f^2+b},\arcsin \left (\frac {\sqrt {1-f^2 x}}{\sqrt {2}}\right ),\frac {2 d}{c f^2+d}\right )}{\left (a f^2+b\right ) \sqrt {c+d x}} \]

[In]

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

[Out]

(-2*Sqrt[(f^2*(c + d*x))/(d + c*f^2)]*EllipticPi[(2*b)/(b + a*f^2), ArcSin[Sqrt[1 - f^2*x]/Sqrt[2]], (2*d)/(d
+ c*f^2)])/((b + a*f^2)*Sqrt[c + d*x])

Rule 174

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 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], 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 552

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

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{\sqrt {2-x^2} \left (b+a f^2-b x^2\right ) \sqrt {c+\frac {d}{f^2}-\frac {d x^2}{f^2}}} \, dx,x,\sqrt {1-f^2 x}\right )\right ) \\ & = -\frac {\left (2 \sqrt {\frac {f^2 (c+d x)}{d+c f^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^2} \left (b+a f^2-b x^2\right ) \sqrt {1-\frac {d x^2}{\left (c+\frac {d}{f^2}\right ) f^2}}} \, dx,x,\sqrt {1-f^2 x}\right )}{\sqrt {c+d x}} \\ & = -\frac {2 \sqrt {\frac {f^2 (c+d x)}{d+c f^2}} \Pi \left (\frac {2 b}{b+a f^2};\sin ^{-1}\left (\frac {\sqrt {1-f^2 x}}{\sqrt {2}}\right )|\frac {2 d}{d+c f^2}\right )}{\left (b+a f^2\right ) \sqrt {c+d x}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.94 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\frac {2 i (c+d x) \sqrt {\frac {d \left (-1+f^2 x\right )}{f^2 (c+d x)}} \sqrt {\frac {d \left (1+f^2 x\right )}{f^2 (c+d x)}} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-c-\frac {d}{f^2}}}{\sqrt {c+d x}}\right ),\frac {-d+c f^2}{d+c f^2}\right )-\operatorname {EllipticPi}\left (\frac {(b c-a d) f^2}{b \left (d+c f^2\right )},i \text {arcsinh}\left (\frac {\sqrt {-c-\frac {d}{f^2}}}{\sqrt {c+d x}}\right ),\frac {-d+c f^2}{d+c f^2}\right )\right )}{(-b c+a d) \sqrt {-c-\frac {d}{f^2}} \sqrt {1-f^4 x^2}} \]

[In]

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

[Out]

((2*I)*(c + d*x)*Sqrt[(d*(-1 + f^2*x))/(f^2*(c + d*x))]*Sqrt[(d*(1 + f^2*x))/(f^2*(c + d*x))]*(EllipticF[I*Arc
Sinh[Sqrt[-c - d/f^2]/Sqrt[c + d*x]], (-d + c*f^2)/(d + c*f^2)] - EllipticPi[((b*c - a*d)*f^2)/(b*(d + c*f^2))
, I*ArcSinh[Sqrt[-c - d/f^2]/Sqrt[c + d*x]], (-d + c*f^2)/(d + c*f^2)]))/((-(b*c) + a*d)*Sqrt[-c - d/f^2]*Sqrt
[1 - f^4*x^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(83)=166\).

Time = 3.43 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.47

method	result	size
default	\(-\frac {2 \left (c \,f^{2}-d \right ) \Pi \left (\sqrt {\frac {\left (d x +c \right ) f^{2}}{c \,f^{2}-d}}, -\frac {\left (c \,f^{2}-d \right ) b}{f^{2} \left (a d -b c \right )}, \sqrt {\frac {c \,f^{2}-d}{c \,f^{2}+d}}\right ) \sqrt {-\frac {\left (f^{2} x +1\right ) d}{c \,f^{2}-d}}\, \sqrt {-\frac {\left (f^{2} x -1\right ) d}{c \,f^{2}+d}}\, \sqrt {\frac {\left (d x +c \right ) f^{2}}{c \,f^{2}-d}}\, \sqrt {f^{2} x +1}\, \sqrt {-f^{2} x +1}\, \sqrt {d x +c}}{f^{2} \left (a d -b c \right ) \left (d \,f^{4} x^{3}+c \,f^{4} x^{2}-d x -c \right )}\)	\(212\)
elliptic	\(\frac {2 \sqrt {-\left (f^{4} x^{2}-1\right ) \left (d x +c \right )}\, \left (\frac {c}{d}-\frac {1}{f^{2}}\right ) \sqrt {\frac {x +\frac {c}{d}}{\frac {c}{d}-\frac {1}{f^{2}}}}\, \sqrt {\frac {x -\frac {1}{f^{2}}}{-\frac {c}{d}-\frac {1}{f^{2}}}}\, \sqrt {\frac {x +\frac {1}{f^{2}}}{-\frac {c}{d}+\frac {1}{f^{2}}}}\, \Pi \left (\sqrt {\frac {x +\frac {c}{d}}{\frac {c}{d}-\frac {1}{f^{2}}}}, \frac {-\frac {c}{d}+\frac {1}{f^{2}}}{-\frac {c}{d}+\frac {a}{b}}, \sqrt {\frac {-\frac {c}{d}+\frac {1}{f^{2}}}{-\frac {c}{d}-\frac {1}{f^{2}}}}\right )}{\sqrt {d x +c}\, \sqrt {-f^{2} x +1}\, \sqrt {f^{2} x +1}\, b \sqrt {-d \,f^{4} x^{3}-c \,f^{4} x^{2}+d x +c}\, \left (-\frac {c}{d}+\frac {a}{b}\right )}\)	\(243\)

[In]

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

[Out]

-2*(c*f^2-d)*EllipticPi(((d*x+c)*f^2/(c*f^2-d))^(1/2),-(c*f^2-d)*b/f^2/(a*d-b*c),((c*f^2-d)/(c*f^2+d))^(1/2))*
(-(f^2*x+1)*d/(c*f^2-d))^(1/2)*(-(f^2*x-1)*d/(c*f^2+d))^(1/2)*((d*x+c)*f^2/(c*f^2-d))^(1/2)*(f^2*x+1)^(1/2)*(-
f^2*x+1)^(1/2)*(d*x+c)^(1/2)/f^2/(a*d-b*c)/(d*f^4*x^3+c*f^4*x^2-d*x-c)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\int \frac {1}{\left (a + b x\right ) \sqrt {c + d x} \sqrt {- f^{2} x + 1} \sqrt {f^{2} x + 1}}\, dx \]

[In]

integrate(1/(b*x+a)/(d*x+c)**(1/2)/(-f**2*x+1)**(1/2)/(f**2*x+1)**(1/2),x)

[Out]

Integral(1/((a + b*x)*sqrt(c + d*x)*sqrt(-f**2*x + 1)*sqrt(f**2*x + 1)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\int { \frac {1}{\sqrt {f^{2} x + 1} \sqrt {-f^{2} x + 1} {\left (b x + a\right )} \sqrt {d x + c}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(f^2*x + 1)*sqrt(-f^2*x + 1)*(b*x + a)*sqrt(d*x + c)), x)

Giac [F]

\[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\int { \frac {1}{\sqrt {f^{2} x + 1} \sqrt {-f^{2} x + 1} {\left (b x + a\right )} \sqrt {d x + c}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {1-f^2 x} \sqrt {1+f^2 x}} \, dx=\int \frac {1}{\left (a+b\,x\right )\,\sqrt {1-f^2\,x}\,\sqrt {x\,f^2+1}\,\sqrt {c+d\,x}} \,d x \]

[In]

int(1/((a + b*x)*(1 - f^2*x)^(1/2)*(f^2*x + 1)^(1/2)*(c + d*x)^(1/2)),x)

[Out]

int(1/((a + b*x)*(1 - f^2*x)^(1/2)*(f^2*x + 1)^(1/2)*(c + d*x)^(1/2)), x)