\(\int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx\) [413]

Optimal result

Integrand size = 22, antiderivative size = 108 \[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(b,c)\right ),\frac {2 \sqrt {b^2+c^2}}{a+\sqrt {b^2+c^2}}\right ) \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}}{e \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \]

[Out]

2*(cos(1/2*d+1/2*e*x-1/2*arctan(b,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(b,c))*EllipticF(sin(1/2*d+1/2*e*x-
1/2*arctan(b,c)),2^(1/2)*((b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2))*((a+b*cos(e*x+d)+c*sin(e*x+d))/(a+(b^2+c
^2)^(1/2)))^(1/2)/e/(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3206, 2740} \[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {2 \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}} \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(b,c)\right ),\frac {2 \sqrt {b^2+c^2}}{a+\sqrt {b^2+c^2}}\right )}{e \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \]

[In]

Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]],x]

[Out]

(2*EllipticF[(d + e*x - ArcTan[b, c])/2, (2*Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[(a + b*Cos[d + e*x] +
c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])])/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])

Rule 2740

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

Rule 3206

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - ArcTan[b, c]]], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}} \int \frac {1}{\sqrt {\frac {a}{a+\sqrt {b^2+c^2}}+\frac {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}{a+\sqrt {b^2+c^2}}}} \, dx}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \\ & = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(b,c)\right ),\frac {2 \sqrt {b^2+c^2}}{a+\sqrt {b^2+c^2}}\right ) \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}}{e \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 0.68 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.64 \[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {3}{2},\frac {a+\sqrt {1+\frac {b^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {b}{c}\right )\right )}{a-\sqrt {1+\frac {b^2}{c^2}} c},\frac {a+\sqrt {1+\frac {b^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {b}{c}\right )\right )}{a+\sqrt {1+\frac {b^2}{c^2}} c}\right ) \sec \left (d+e x+\arctan \left (\frac {b}{c}\right )\right ) \sqrt {-\frac {\sqrt {1+\frac {b^2}{c^2}} c \left (-1+\sin \left (d+e x+\arctan \left (\frac {b}{c}\right )\right )\right )}{a+\sqrt {1+\frac {b^2}{c^2}} c}} \sqrt {\frac {\sqrt {1+\frac {b^2}{c^2}} c \left (1+\sin \left (d+e x+\arctan \left (\frac {b}{c}\right )\right )\right )}{-a+\sqrt {1+\frac {b^2}{c^2}} c}} \sqrt {a+\sqrt {1+\frac {b^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {b}{c}\right )\right )}}{\sqrt {1+\frac {b^2}{c^2}} c e} \]

[In]

Integrate[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]],x]

[Out]

(2*AppellF1[1/2, 1/2, 1/2, 3/2, (a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(a - Sqrt[1 + b^2/c^2]*c)
, (a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(a + Sqrt[1 + b^2/c^2]*c)]*Sec[d + e*x + ArcTan[b/c]]*S
qrt[-((Sqrt[1 + b^2/c^2]*c*(-1 + Sin[d + e*x + ArcTan[b/c]]))/(a + Sqrt[1 + b^2/c^2]*c))]*Sqrt[(Sqrt[1 + b^2/c
^2]*c*(1 + Sin[d + e*x + ArcTan[b/c]]))/(-a + Sqrt[1 + b^2/c^2]*c)]*Sqrt[a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x +
ArcTan[b/c]]])/(Sqrt[1 + b^2/c^2]*c*e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(137)=274\).

Time = 1.35 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.73

method	result	size
default	\(\frac {2 \left (a +\sqrt {b^{2}+c^{2}}\right ) \sqrt {\frac {\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+a}{a +\sqrt {b^{2}+c^{2}}}}\, \sqrt {\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )+1\right ) \sqrt {b^{2}+c^{2}}}{-a +\sqrt {b^{2}+c^{2}}}}\, \sqrt {-\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right ) \sqrt {b^{2}+c^{2}}}{a +\sqrt {b^{2}+c^{2}}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+a}{a +\sqrt {b^{2}+c^{2}}}}, \sqrt {-\frac {a +\sqrt {b^{2}+c^{2}}}{-a +\sqrt {b^{2}+c^{2}}}}\right )}{\sqrt {b^{2}+c^{2}}\, \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+a \sqrt {b^{2}+c^{2}}}{\sqrt {b^{2}+c^{2}}}}\, e}\)	\(295\)

[In]

int(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(a+(b^2+c^2)^(1/2))*(((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(a+(b^2+c^2)^(1/2)))^(1/2)*((sin(e*x+d-arct
an(-b,c))+1)*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)*(-(sin(e*x+d-arctan(-b,c))-1)*(b^2+c^2)^(1/2)/(a+(b^2
+c^2)^(1/2)))^(1/2)*EllipticF((((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(a+(b^2+c^2)^(1/2)))^(1/2),(-(a+(b^
2+c^2)^(1/2))/(-a+(b^2+c^2)^(1/2)))^(1/2))/(b^2+c^2)^(1/2)/cos(e*x+d-arctan(-b,c))/((b^2*sin(e*x+d-arctan(-b,c
))+c^2*sin(e*x+d-arctan(-b,c))+a*(b^2+c^2)^(1/2))/(b^2+c^2)^(1/2))^(1/2)/e

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 506, normalized size of antiderivative = 4.69 \[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {\sqrt {2} \sqrt {b + i \, c} {\left (i \, b + c\right )} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} b^{2} - 3 \, b^{4} - 4 \, a^{2} c^{2} + 6 i \, b c^{3} + 3 \, c^{4} - 2 i \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} c\right )}}{3 \, {\left (b^{4} + 2 \, b^{2} c^{2} + c^{4}\right )}}, -\frac {8 \, {\left (8 \, a^{3} b^{3} - 9 \, a b^{5} + 27 \, a b c^{4} - 9 i \, a c^{5} + 2 i \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} c^{3} - 6 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} c^{2} - 3 i \, {\left (8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} c\right )}}{27 \, {\left (b^{6} + 3 \, b^{4} c^{2} + 3 \, b^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b - 2 i \, a c + 3 \, {\left (b^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (i \, b^{2} + i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (b^{2} + c^{2}\right )}}\right ) + \sqrt {2} \sqrt {b - i \, c} {\left (-i \, b + c\right )} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} b^{2} - 3 \, b^{4} - 4 \, a^{2} c^{2} - 6 i \, b c^{3} + 3 \, c^{4} + 2 i \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} c\right )}}{3 \, {\left (b^{4} + 2 \, b^{2} c^{2} + c^{4}\right )}}, -\frac {8 \, {\left (8 \, a^{3} b^{3} - 9 \, a b^{5} + 27 \, a b c^{4} + 9 i \, a c^{5} - 2 i \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} c^{3} - 6 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} c^{2} + 3 i \, {\left (8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} c\right )}}{27 \, {\left (b^{6} + 3 \, b^{4} c^{2} + 3 \, b^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b + 2 i \, a c + 3 \, {\left (b^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (-i \, b^{2} - i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (b^{2} + c^{2}\right )}}\right )}{{\left (b^{2} + c^{2}\right )} e} \]

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2),x, algorithm="fricas")

[Out]

(sqrt(2)*sqrt(b + I*c)*(I*b + c)*weierstrassPInverse(4/3*(4*a^2*b^2 - 3*b^4 - 4*a^2*c^2 + 6*I*b*c^3 + 3*c^4 -
2*I*(4*a^2*b - 3*b^3)*c)/(b^4 + 2*b^2*c^2 + c^4), -8/27*(8*a^3*b^3 - 9*a*b^5 + 27*a*b*c^4 - 9*I*a*c^5 + 2*I*(4
*a^3 + 9*a*b^2)*c^3 - 6*(4*a^3*b - 3*a*b^3)*c^2 - 3*I*(8*a^3*b^2 - 9*a*b^4)*c)/(b^6 + 3*b^4*c^2 + 3*b^2*c^4 +
c^6), 1/3*(2*a*b - 2*I*a*c + 3*(b^2 + c^2)*cos(e*x + d) - 3*(I*b^2 + I*c^2)*sin(e*x + d))/(b^2 + c^2)) + sqrt(
2)*sqrt(b - I*c)*(-I*b + c)*weierstrassPInverse(4/3*(4*a^2*b^2 - 3*b^4 - 4*a^2*c^2 - 6*I*b*c^3 + 3*c^4 + 2*I*(
4*a^2*b - 3*b^3)*c)/(b^4 + 2*b^2*c^2 + c^4), -8/27*(8*a^3*b^3 - 9*a*b^5 + 27*a*b*c^4 + 9*I*a*c^5 - 2*I*(4*a^3
+ 9*a*b^2)*c^3 - 6*(4*a^3*b - 3*a*b^3)*c^2 + 3*I*(8*a^3*b^2 - 9*a*b^4)*c)/(b^6 + 3*b^4*c^2 + 3*b^2*c^4 + c^6),
1/3*(2*a*b + 2*I*a*c + 3*(b^2 + c^2)*cos(e*x + d) - 3*(-I*b^2 - I*c^2)*sin(e*x + d))/(b^2 + c^2)))/((b^2 + c^
2)*e)

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\int \frac {1}{\sqrt {a + b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )}}}\, dx \]

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))**(1/2),x)

[Out]

Integral(1/sqrt(a + b*cos(d + e*x) + c*sin(d + e*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}} \,d x } \]

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*cos(e*x + d) + c*sin(e*x + d) + a), x)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}} \,d x } \]

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*cos(e*x + d) + c*sin(e*x + d) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\int \frac {1}{\sqrt {a+b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )}} \,d x \]

[In]

int(1/(a + b*cos(d + e*x) + c*sin(d + e*x))^(1/2),x)

[Out]

int(1/(a + b*cos(d + e*x) + c*sin(d + e*x))^(1/2), x)