\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx\) [479]

Optimal result

Integrand size = 33, antiderivative size = 125 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=-\frac {(A-3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

(A-B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))-(A-3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d+(A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/
cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3039, 4105, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}+\frac {(A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[In]

Int[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x]

[Out]

-(((A - 3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d)) + ((A - B)*Sqrt[Cos[c + d
*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a
*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx \\ & = \frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \frac {-\frac {1}{2} a (A-3 B)+\frac {1}{2} a (A-B) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^2} \\ & = \frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(A-3 B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a}+\frac {(A-B) \int \sqrt {\sec (c+d x)} \, dx}{2 a} \\ & = \frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {\left ((A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a}+\frac {\left ((A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a} \\ & = -\frac {(A-3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.94 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.38 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-6 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+\frac {6 \left ((A-2 B) \cos \left (\frac {1}{2} (c-d x)\right )-B \cos \left (\frac {1}{2} (3 c+d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sec (c+d x)}}+12 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-12 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}\right )}{6 a d (1+\cos (c+d x))} \]

[In]

Integrate[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x]

[Out]

(Cos[(c + d*x)/2]^2*((2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))
]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4,
-E^((2*I)*(c + d*x))]))/E^(I*d*x) - (6*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^(
(2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (6*((A - 2*B)*Cos[(c - d*x)/2] - B*Cos[(3*c + d*x)/2])*Cs
c[c/2]*Sec[c/2]*Sec[(c + d*x)/2])/Sqrt[Sec[c + d*x]] + 12*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[
Sec[c + d*x]] - 12*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]))/(6*a*d*(1 + Cos[c + d*x
]))

Maple [A] (verified)

Time = 4.71 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.95

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (-2 B +2 A \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A +B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(244\)

[In]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*a)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-
B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+(-2*B+2*A)*sin(1/2*d*x+1/2*
c)^4+(-A+B)*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/si
n(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.90 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {2 \, {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/2*(2*(A - B)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I
*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(-I*A + 3*I*B)*cos(d*x + c) + sqrt(2
)*(-I*A + 3*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + (sqrt(2)
*(I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(I*A - 3*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {\int \frac {A}{\cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))/sec(d*x+c)**(1/2),x)

[Out]

(Integral(A/(cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x) + Integral(B*cos(c + d*x)/(cos(c + d*x)
*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x))/a

Maxima [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)

Mupad [F(-1)]

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

[In]

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))),x)

[Out]

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))), x)