3.515 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=213 \[ \frac{(11 A-7 B+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

[Out]

((11*A - 7*B + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*S
qrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((7*A - 3
*B + 3*C)*Sin[c + d*x])/(6*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((19*A - 15*B + 3*C)*Sin[c + d*x
])/(6*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.598038, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{(11 A-7 B+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

((11*A - 7*B + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*S
qrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((7*A - 3
*B + 3*C)*Sin[c + d*x])/(6*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((19*A - 15*B + 3*C)*Sin[c + d*x
])/(6*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{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{2} a (7 A-3 B+3 C)-2 a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{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 \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{4} a^2 (19 A-15 B+3 C)+\frac{1}{2} a^2 (7 A-3 B+3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{3 a^3 (11 A-7 B+3 C)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(11 A-7 B+3 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 \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(11 A-7 B+3 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{(11 A-7 B+3 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 \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B+3 C) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.87591, size = 1207, normalized size = 5.67 \[ \text{result too large to display} \]

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)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

(8*C*Cos[c/2 + (d*x)/2]^3*Sin[c/2 + (d*x)/2])/(3*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(
3/2)) - ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(1 - 2*Sin[c/2 + (d*x)/2]))/(6*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 + S
in[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(1 + 2*Sin[c/2 + (d
*x)/2]))/(6*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (16*
C*Cos[c/2 + (d*x)/2]^3*Sin[c/2 + (d*x)/2])/(3*d*(a*(1 + Cos[c + d*x]))^(3/2)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])
- ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(5*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] +
(1 + Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 +
(d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/(d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((A - B + C)*Cos[c/2 + (d*x)/2]^3*
(5*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 + (d*x)/2])/((1 + Sin[c/
2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2]
)))/(d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((A + 3*B - 7*C)*Cot[c/2 + (d*x)/2]^3*Csc[c/2 + (d*x)/2]^2*(-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*Hypergeometric2F1[2, 7/2, 9/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Si
n[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*Sin[c/2 + (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(63*d*(a*(1 + Cos[c + d*
x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))

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Maple [B]  time = 0.115, size = 471, normalized size = 2.2 \begin{align*} \text{result too large to display} \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))^(3/2),x)

[Out]

-1/12/d*(a*(1+cos(d*x+c)))^(1/2)*(33*A*arcsin((-1+cos(d*x+c))/sin(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)-21*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*sin(d*x+c)*cos(d*x+c)^2+9*C*arcsin((-1+cos(d*x+c))/sin(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)+33*A*arcsin((-1+cos(d*x+c))/sin(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)-21*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*co
s(d*x+c)+9*C*arcsin((-1+cos(d*x+c))/sin(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
)-38*A*cos(d*x+c)^3+30*B*cos(d*x+c)^3-6*C*cos(d*x+c)^3+14*A*cos(d*x+c)^2-6*B*cos(d*x+c)^2+6*C*cos(d*x+c)^2+32*
A*cos(d*x+c)-24*B*cos(d*x+c)-8*A)/a^2/sin(d*x+c)/(1+cos(d*x+c))/cos(d*x+c)^(3/2)

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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)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.43174, size = 618, normalized size = 2.9 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\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 (19 \, A - 15 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (A - B\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 )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(2)*((11*A - 7*B + 3*C)*cos(d*x + c)^4 + 2*(11*A - 7*B + 3*C)*cos(d*x + c)^3 + (11*A - 7*B + 3*C)*
cos(d*x + c)^2)*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*((19*A - 15*B + 3*C)*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) - 4*A)*sq
rt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*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))**(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{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))^(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)^(5/2)), x)