\(\int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx\) [526]

Optimal result

Integrand size = 35, antiderivative size = 181 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {(2 A-B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {B \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

B*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+(2*A-B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2
))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)-(A-B)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a
+a*cos(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3040, 3062, 3061, 2861, 211, 2853, 222} \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {(2 A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {B \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

((2*A - B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqr
t[a]*d) - (Sqrt[2]*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])
]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) + (B*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c
+ d*x]])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 2861

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

Rule 3040

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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e
, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3062

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.56 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.58 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {i e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (B-B e^{i (c+d x)}+B e^{2 i (c+d x)}-B e^{3 i (c+d x)}-(2 A-B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} B e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+2 \sqrt {2} A e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-\sqrt {2} B e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+2 A e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-B e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\sec (c+d x)}}{4 d \sqrt {a (1+\cos (c+d x))}} \]

[In]

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

[Out]

((I/4)*(1 + E^(I*(c + d*x)))*(B - B*E^(I*(c + d*x)) + B*E^((2*I)*(c + d*x)) - B*E^((3*I)*(c + d*x)) - (2*A - B
)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*B*E^(I*(c + d*x))*Sqrt[1 +
E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 2*Sqrt[2]*A*E^(I
*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))
])] - Sqrt[2]*B*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 +
E^((2*I)*(c + d*x))])] + 2*A*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x
))]] - B*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[Sec[c + d*
x]])/(d*E^((2*I)*(c + d*x))*Sqrt[a*(1 + Cos[c + d*x])])

Maple [A] (verified)

Time = 20.34 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {\left (B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 A \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-B \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-2 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}\)	\(196\)
parts	\(\frac {A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}-\frac {B \left (-\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}\)	\(251\)

[In]

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

[Out]

1/2/d*(B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+2*A*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2))-B*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2*A*arcsin(cot(d*x+c)-csc(d*x+
c))-2*B*arcsin(cot(d*x+c)-csc(d*x+c)))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(1/2)/(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*2^(1/2)/a

Fricas [A] (verification not implemented)

none

Time = 1.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.93 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a \cos \left (d x + c\right ) + a} B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right ) + 2 \, A - B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right ) + a d} \]

[In]

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

[Out]

(sqrt(a*cos(d*x + c) + a)*B*sqrt(cos(d*x + c))*sin(d*x + c) - ((2*A - B)*cos(d*x + c) + 2*A - B)*sqrt(a)*arcta
n(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + sqrt(2)*((A - B)*a*cos(d*x + c) + (A -
B)*a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*
x + c) + a*d)

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*cos(c + d*x))/(sqrt(a*(cos(c + d*x) + 1))*sqrt(sec(c + d*x))), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imaginary; found %i

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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