3.507 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=143 \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]

[Out]

(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]])])/(Sq
rt[a]*d) + (2*A*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - (2*(A - 3*B)*Sin[c + d*x])/(
3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.377353, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3043, 2984, 12, 2782, 205} \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

(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]])])/(Sq
rt[a]*d) + (2*A*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - (2*(A - 3*B)*Sin[c + d*x])/(
3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*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] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx &=\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{-\frac{1}{2} a (A-3 B)+\frac{1}{2} a (2 A+3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{3 a}\\ &=\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{4 \int \frac{3 a^2 (A-B+C)}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^2}\\ &=\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+(A-B+C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(2 a (A-B+C)) \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{\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{a} d}+\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.78888, size = 701, normalized size = 4.9 \[ \frac{2 (A-B+C) \cot \left (\frac{c}{2}+\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (-12 \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac{7}{2};1,\frac{9}{2};-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}\right )-12 \left (3 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+4\right ) \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}\right )+7 \sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}} \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3 \left (8 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-20 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+15\right ) \left (\left (3-7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}-3 \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \tanh ^{-1}\left (\sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right )\right )\right )}{63 d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{7/2} \sqrt{a (\cos (c+d x)+1)}}+\frac{8 B \sin \left (\frac{c}{2}+\frac{d x}{2}\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )} \sqrt{a (\cos (c+d x)+1)}}+\frac{4 B \sin \left (\frac{c}{2}+\frac{d x}{2}\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2} \sqrt{a (\cos (c+d x)+1)}}-\frac{8 C \sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2} \sqrt{a (\cos (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

(4*B*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2])/(3*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)
) - (8*C*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2]^3)/(3*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)
^(3/2)) + (8*B*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2])/(3*d*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[1 - 2*Sin[c/2 + (d*
x)/2]^2]) + (2*(A - B + C)*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^4*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{
2, 2, 7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeo
metric2F1[2, 7/2, 9/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c
/2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 -
2*Sin[c/2 + (d*x)/2]^2)^3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2
)*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Si
n[c/2 + (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(63*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)
/2]^2)^(7/2))

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Maple [B]  time = 0.161, size = 379, normalized size = 2.7 \begin{align*} -{\frac{1}{3\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 3\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-3\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+3\,C\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+3\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}\cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-3\,B\cos \left ( dx+c \right ) \sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+3\,C\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}\cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+2\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -6\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -2\,A\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \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)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x)

[Out]

-1/3/d*(a*(1+cos(d*x+c)))^(1/2)*(3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)-3*B*cos(d*x+c)^2*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
+3*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*A*arcsin((-1+
cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-3*B*cos(d*x+c)*2^(1/2)*arcsin((-1
+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(
d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+2*A*sin(d*x+c)*cos(d*x+c)-6*B*sin(d*x+c)*cos(d*x+c)-2*A*sin(d*x+c))/a
/(1+cos(d*x+c))/cos(d*x+c)^(3/2)

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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+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08113, size = 458, normalized size = 3.2 \begin{align*} -\frac{2 \,{\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac{3 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{a}}\right )}{\sqrt{a}}}{3 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\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)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/3*(2*((A - 3*B)*cos(d*x + c) - A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*sqrt(2)*((A
- B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*cos(d*x + c)^2)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos
(d*x + c))*sin(d*x + c)/((cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x
+ c)^2)

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Sympy [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+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt{a \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac{5}{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)/cos(d*x+c)^(5/2)/(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(a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)