3.479 \(\int \frac{\sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ -\frac{a (2 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{d \sqrt{\cos (c+d x)}}+\frac{\sqrt{a} (2 B+C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.357451, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3043, 2981, 2774, 216} \[ -\frac{a (2 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{d \sqrt{\cos (c+d x)}}+\frac{\sqrt{a} (2 B+C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

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

Rule 3043

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

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{\sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (A+B)-\frac{1}{2} a (2 A-C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{a}\\ &=-\frac{a (2 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{2} (2 B+C) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{a (2 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{(2 B+C) \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 )}{d}\\ &=\frac{\sqrt{a} (2 B+C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}-\frac{a (2 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.30405, size = 104, normalized size = 0.86 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) (2 A+C \cos (c+d x))+\sqrt{2} (2 B+C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}\right )}{2 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

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

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Maple [A]  time = 0.136, size = 210, normalized size = 1.7 \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+2\,A\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+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 ) \cos \left ( dx+c \right ) +C\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 ) \cos \left ( dx+c \right ) \right ){\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x)

[Out]

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

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Maxima [B]  time = 2.37409, size = 1397, normalized size = 11.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*(4*B*sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(
2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)) + (2*(cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) - 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(
a) + sqrt(a)*(arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*co
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) + 1))) + 1) - arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(co
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(
d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) + 1))) - 1) - arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*
cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + arctan2((cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)) - 1)))*C + 8*A*(sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(2)*sqrt(
a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x +
c) + 1) + 1)^(3/2)))/d

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Fricas [A]  time = 2.69673, size = 346, normalized size = 2.86 \begin{align*} \frac{{\left (C \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \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} +{\left (2 \, B + C\right )} \cos \left (d x + c\right )\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 )}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )} \left (A + B \cos{\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*(a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/cos(c + d*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{a \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)/cos(d*x + c)^(3/2), x)