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

Optimal. Leaf size=161 \[ -\frac{(7 A-3 B-C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-((7*A - 3*B - C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqr
t[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((5*A - B +
C)*Sin[c + d*x])/(2*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.418086, antiderivative size = 161, 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 = {3041, 2984, 12, 2782, 205} \[ -\frac{(7 A-3 B-C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A - 3*B - C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqr
t[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((5*A - B +
C)*Sin[c + d*x])/(2*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 3041

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

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{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{2} a (5 A-B+C)-a (A-B-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int -\frac{a^2 (7 A-3 B-C)}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{a^3}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(7 A-3 B-C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(7 A-3 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 )}{2 d}\\ &=-\frac{(7 A-3 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 )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B+C) \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.64982, size = 455, normalized size = 2.83 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{(A+3 B-7 C) \csc ^3\left (\frac{1}{2} (c+d x)\right ) \left (5 (4 \cos (c+d x)+\cos (2 (c+d x))+1) \left (-\cos (c+d x)+\cos (c+d x) \sqrt{2-2 \sec (c+d x)} \tanh ^{-1}\left (\sqrt{\sin ^2\left (\frac{1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )+1\right )-2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x) \tan (c+d x) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\sec (c+d x) \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 \cos ^{\frac{3}{2}}(c+d x)}+\frac{5 (A-B+C) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )-1\right )}{\sqrt{\cos (c+d x)} \left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{5 (A-B+C) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+1\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )-1\right ) \sqrt{\cos (c+d x)}}-\frac{20 (A-B+C) \sqrt{\cos (c+d x)}}{\sin \left (\frac{1}{2} (c+d x)\right )-1}-\frac{20 (A-B+C) \sqrt{\cos (c+d x)}}{\sin \left (\frac{1}{2} (c+d x)\right )+1}+30 (A-B+C) \tan ^{-1}\left (\frac{1-2 \sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )-30 (A-B+C) \tan ^{-1}\left (\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )+1}{\sqrt{\cos (c+d x)}}\right )+\frac{80 C \sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )}{10 d (a (\cos (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*(30*(A - B + C)*ArcTan[(1 - 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - 30*(A - B + C)*ArcTa
n[(1 + 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - (20*(A - B + C)*Sqrt[Cos[c + d*x]])/(-1 + Sin[(c + d*x)/2]) +
(80*C*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]] - (20*(A - B + C)*Sqrt[Cos[c + d*x]])/(1 + Sin[(c + d*x)/2]) + (5*
(A - B + C)*(-1 + 2*Sin[(c + d*x)/2]))/(Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^2) - (5*(A -
B + C)*(1 + 2*Sin[(c + d*x)/2]))/(Sqrt[Cos[c + d*x]]*(-1 + Sin[(c + d*x)/2])) + ((A + 3*B - 7*C)*Csc[(c + d*x)
/2]^3*(5*(1 + 4*Cos[c + d*x] + Cos[2*(c + d*x)])*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)
/2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]) - 2*Hypergeometric2F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin[(c + d*x
)/2]^2)]*Sin[(c + d*x)/2]^4*Sin[c + d*x]*Tan[c + d*x]))/(2*Cos[c + d*x]^(3/2))))/(10*d*(a*(1 + Cos[c + d*x]))^
(3/2))

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Maple [B]  time = 0.149, size = 438, normalized size = 2.7 \begin{align*}{\frac{-1+\cos \left ( dx+c \right ) }{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}} \left ( 10\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+18\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}-2\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}-2\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}-7\,A\sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}-18\,A\cos \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+3\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}\sin \left ( dx+c \right ) +C\sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}-8\,A \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+2\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}+2\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-2\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\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+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/4/d*(-1+cos(d*x+c))*(10*A*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+18*A*cos(d*x+c)^3*(cos(d*x+c)/(1+co
s(d*x+c)))^(5/2)-2*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-2*B*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)
))^(3/2)-7*A*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-18*A*cos(d*x+c)*(cos(d*x+c)/(1
+cos(d*x+c)))^(5/2)+3*B*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)+C*2^(1/2)*arcsin((-
1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-8*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+2*B*cos(d*x+c)^2*(cos(
d*x+c)/(1+cos(d*x+c)))^(3/2)+2*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-2*C*cos(d*x+c)^3*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2))*(a*(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(5/2)/sin(d*x+c)^3/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
/a^2

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Maxima [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)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.68446, size = 552, normalized size = 3.43 \begin{align*} -\frac{\sqrt{2}{\left ({\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{2} +{\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (5 \, A - B + C\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\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)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*((7*A - 3*B - C)*cos(d*x + c)^3 + 2*(7*A - 3*B - C)*cos(d*x + c)^2 + (7*A - 3*B - C)*cos(d*x + c
))*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)
^2 + a*cos(d*x + c))) - 2*((5*A - B + C)*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d
*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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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)**(3/2)/(a+a*cos(d*x+c))**(3/2),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}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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