3.1301 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=220 \[ \frac{(2 A-5 B+10 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(2 A-5 B+10 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(A-4 B+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[Out]

-(((A - 4*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + ((2*A - 5*B + 1
0*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A - B + C)*Sin[c + d*x])/(
3*d*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)) - ((A - 4*B + 7*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])*Se
c[c + d*x]^(3/2)) + ((2*A - 5*B + 10*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.496793, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4221, 3041, 2977, 2748, 2639, 2635, 2641} \[ \frac{(2 A-5 B+10 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(2 A-5 B+10 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(A-4 B+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - 4*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + ((2*A - 5*B + 1
0*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A - B + C)*Sin[c + d*x])/(
3*d*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)) - ((A - 4*B + 7*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])*Se
c[c + d*x]^(3/2)) + ((2*A - 5*B + 10*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]])

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]

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 2977

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx\\ &=-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a (A+5 B-5 C)+\frac{3}{2} a (A-B+3 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \left (-\frac{3}{2} a^2 (A-4 B+7 C)+\frac{3}{2} a^2 (2 A-5 B+10 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}-\frac{\left ((A-4 B+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}+\frac{\left ((2 A-5 B+10 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}\\ &=-\frac{(A-4 B+7 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(2 A-5 B+10 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{\left ((2 A-5 B+10 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac{(A-4 B+7 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{(2 A-5 B+10 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(A-4 B+7 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(2 A-5 B+10 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.74424, size = 183, normalized size = 0.83 \[ -\frac{2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (-2 (2 A-5 B+10 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 (A-4 B+7 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{1}{2} \left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((3 A-6 B+13 C) \cos (c+d x)+2 A-5 B+C \cos (2 (c+d x))+11 C)\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(6*(A - 4*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] - 2*
(2*A - 5*B + 10*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((2*A - 5*B + 11*C + (3*A - 6*B + 13*C)*Cos[
c + d*x] + C*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/2))/(3*a^2*d*(1 +
Cos[c + d*x])^2)

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Maple [A]  time = 1.332, size = 472, 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)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(3/2),x)

[Out]

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+10*C*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))+21*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))-5*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+10
*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+16*C*s
in(1/2*d*x+1/2*c)^8+(-12*A+24*B-76*C)*sin(1/2*d*x+1/2*c)^6+(16*A-34*B+84*C)*sin(1/2*d*x+1/2*c)^4+(-5*A+11*B-25
*C)*sin(1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/
2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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 )}^{2} \sec \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))^2/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}, x\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))^2/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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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)/(a+a*cos(d*x+c))**2/sec(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 )}^{2} \sec \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))^2/sec(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)^2*sec(d*x + c)^(3/2)), x)