\(\int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx\) [574]

Optimal result

Integrand size = 33, antiderivative size = 260 \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=-\frac {(A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a (a-b) b (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \]

[Out]

b*(A*b-B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2)/d/(b+a*sec(d*x+c))-(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/
cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/(a^2-b^2)/d-(A*b-
B*a)*(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)*se
c(d*x+c)^(1/2)/b/(a^2-b^2)/d+(3*A*a^2*b-A*b^3-B*a^3-B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/(a-b)/b/(a+b)^2/d

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3039, 4114, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\frac {b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \left (a^2-b^2\right )}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}+\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a b d (a-b) (a+b)^2} \]

[In]

Int[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + b*Cos[c + d*x])^2,x]

[Out]

-(((A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*(a^2 - b^2)*d)) - ((A*b - a
*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*(a^2 - b^2)*d) + ((3*a^2*A*b - A*b^3 -
a^3*B - a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*(a - b)*
b*(a + b)^2*d) + (b*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b + 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 2884

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

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 3934

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4114

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

Rule 4191

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 6.01 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {4 a b (-A b+a B) \sin (c+d x)}{a^2-b^2}+\frac {4 \cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (a A b^2-a^2 b B-a A b^2 \sec ^2(c+d x)+a^2 b B \sec ^2(c+d x)-a b (-A b+a B) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+(a-b) b (2 a A+A b+a B) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-3 a^2 A b \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+A b^3 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+a^3 B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+a b^2 B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{(a-b) b (a+b)}}{4 a^2 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + b*Cos[c + d*x])^2,x]

[Out]

((-4*a*b*(-(A*b) + a*B)*Sin[c + d*x])/(a^2 - b^2) + (4*Cos[c + d*x]*Cot[c + d*x]*(b + a*Sec[c + d*x])*(a*A*b^2
- a^2*b*B - a*A*b^2*Sec[c + d*x]^2 + a^2*b*B*Sec[c + d*x]^2 - a*b*(-(A*b) + a*B)*EllipticE[ArcSin[Sqrt[Sec[c
+ d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + (a - b)*b*(2*a*A + A*b + a*B)*EllipticF[ArcSin[Sqrt[S
ec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 3*a^2*A*b*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c +
d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + A*b^3*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -
1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + a^3*B*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Se
c[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + a*b^2*B*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*
x]]*Sqrt[-Tan[c + d*x]^2]))/((a - b)*b*(a + b)))/(4*a^2*d*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(720\) vs. \(2(324)=648\).

Time = 7.46 (sec) , antiderivative size = 721, normalized size of antiderivative = 2.77

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 (-\frac {4 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{\left (-2 a b +2 b^{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 )}}+\frac {2 \left (A b -B a \right ) \left (-\frac {b^{2} \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 )}}{a \left (a^{2}-b^{2}\right ) \left (2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b \right )}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 a \left (a +b \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 )}}-\frac {b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 \left (a^{2}-b^{2}\right ) a \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 )}}+\frac {b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 \left (a^{2}-b^{2}\right ) a \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 )}}-\frac {3 a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{\left (a^{2}-b^{2}\right ) \left (-2 a b +2 b^{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 )}}+\frac {b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{a \left (a^{2}-b^{2}\right ) \left (-2 a b +2 b^{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 )}}\right )}{b}\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}\)	\(721\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^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)*(-4*B/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1
/2*c),-2*b/(a-b),2^(1/2))+2*(A*b-B*a)/b*(-1/a*b^2/(a^2-b^2)*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)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(a+b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+
1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1
/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b
^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2
*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

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

[Out]

Integral((A + B*cos(c + d*x))*sqrt(sec(c + d*x))/(a + b*cos(c + d*x))**2, x)

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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