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

Optimal. Leaf size=192 \[ \frac{(2 a B+2 A b-b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{\sqrt{2} (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{b B \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

[Out]

((2*A*b + 2*a*B - b*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]])/(Sqrt[a]*d) + (Sqrt[2]*(a - b)*(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*B*Sin[c + d*x])/(d*Sqrt[a + a*Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.665248, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 54, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4221, 3045, 2982, 2782, 205, 2774, 216} \[ \frac{(2 a B+2 A b-b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{\sqrt{2} (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{b B \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b + 2*a*B - b*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]])/(Sqrt[a]*d) + (Sqrt[2]*(a - b)*(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*B*Sin[c + d*x])/(d*Sqrt[a + a*Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 3045

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

Rule 2982

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 2782

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 205

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

Rule 2774

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 216

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]

Rubi steps

\begin{align*} \int \frac{\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)}}{\sqrt{a+a \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{b 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{1}{2} a (2 a A+b B)+\frac{1}{2} a (2 A b+2 a B-b B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{a}\\ &=\frac{b B \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\left ((a-b) (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+2 a B-b 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 B \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left (2 a (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{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+2 a B-b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{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+2 a B-b B) \sin ^{-1}\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) (A-B) \tan ^{-1}\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 B \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.449816, size = 143, normalized size = 0.74 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (\sqrt{2} (2 a B+2 A b-b B) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (a-b) (A-B) \tan ^{-1}\left (\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )+2 b B \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)}\right )}{d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(Sqrt[2]*(2*A*b + 2*a*B - b*B)*ArcSin[Sqrt[2]*Sin[(c +
d*x)/2]] + 2*(a - b)*(A - B)*ArcTan[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]] + 2*b*B*Sqrt[Cos[c + d*x]]*Sin[(c +
d*x)/2]))/(d*Sqrt[a*(1 + Cos[c + d*x])])

________________________________________________________________________________________

Maple [A]  time = 0.799, size = 317, normalized size = 1.7 \begin{align*} -{\frac{\sqrt{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{2}} \left ( B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}b\sin \left ( dx+c \right ) +2\,A\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}b+2\,B\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}a-B\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}b-2\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) a+2\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) b+2\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) a-2\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) b \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \cos{\left (c + d x \right )}\right ) \left (a + b \cos{\left (c + d x \right )}\right ) \sqrt{\sec{\left (c + d x \right )}}}{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \cos \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt{\sec \left (d x + c\right )}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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