3.1342 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=181 \[ \frac {\sqrt {2} (A-B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {(2 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {C \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4221, 3045, 2982, 2782, 205, 2774, 216} \[ \frac {\sqrt {2} (A-B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {(2 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {C \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*B - C)*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 + C)*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) + (C*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[S
ec[c + d*x]])

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 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]

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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 132, normalized size = 0.73 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 (A-B+C) \tan ^{-1}\left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )+\sqrt {2} (2 B-C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 C \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)}\right )}{d \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 10.96, size = 171, normalized size = 0.94 \[ \frac {\sqrt {a \cos \left (d x + c\right ) + a} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left ({\left (2 \, B - C\right )} \cos \left (d x + c\right ) + 2 \, B - C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {\sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right ) + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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)*C*sqrt(cos(d*x + c))*sin(d*x + c) - ((2*B - C)*cos(d*x + c) + 2*B - C)*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 + C)*a*cos(d*x + c) +
(A - B + C)*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {a \cos \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.57, size = 247, normalized size = 1.36 \[ \frac {\left (-C \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 B \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}+2 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-2 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+2 C \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right ) \sqrt {2}}{2 d \sin \left (d x +c \right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(-C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-2*B*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)/cos(d*x+c))+C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*2^(1/2)+2*A*arc
sin((-1+cos(d*x+c))/sin(d*x+c))-2*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))+2*C*arcsin((-1+cos(d*x+c))/sin(d*x+c)))
*(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)
*2^(1/2)/a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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