3.1305 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=233 \[ -\frac{(9 A+B-C) \sin (c+d x)}{10 d \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A+B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(9 A+B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^3} \]

[Out]

((9*A + B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((3*A + B + C)*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B + C)*Sin[c + d*x])/(5*d*(a
+ a*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]) - ((6*A - B - 4*C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2*Sqrt[S
ec[c + d*x]]) - ((9*A + B - C)*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x])*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.658936, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4221, 3041, 2978, 2748, 2641, 2639} \[ -\frac{(9 A+B-C) \sin (c+d x)}{10 d \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A+B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(9 A+B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9*A + B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((3*A + B + C)*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B + C)*Sin[c + d*x])/(5*d*(a
+ a*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]) - ((6*A - B - 4*C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2*Sqrt[S
ec[c + d*x]]) - ((9*A + B - C)*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x])*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 2978

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

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]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (9 A+B-C)-\frac{1}{2} a (3 A-3 B-7 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sqrt{\sec (c+d x)}}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a^2 (21 A+4 B+C)-\frac{1}{2} a^2 (6 A-B-4 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sqrt{\sec (c+d x)}}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{(9 A+B-C) \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{4} a^3 (3 A+B+C)+\frac{3}{4} a^3 (9 A+B-C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sqrt{\sec (c+d x)}}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{(9 A+B-C) \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)}}+\frac{\left ((9 A+B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}+\frac{\left ((3 A+B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac{(9 A+B-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}+\frac{(3 A+B+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sqrt{\sec (c+d x)}}-\frac{(6 A-B-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{(9 A+B-C) \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.47224, size = 188, normalized size = 0.81 \[ \frac{2 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (10 (3 A+B+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 (9 A+B-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{1}{8} \left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) (4 (33 A+2 B-7 C) \cos (c+d x)+3 (9 A+B-C) \cos (2 (c+d x))+117 A-7 B-13 C)\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^6*Sqrt[Sec[c + d*x]]*(6*(9*A + B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 10*(3
*A + B + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((117*A - 7*B - 13*C + 4*(33*A + 2*B - 7*C)*Cos[c +
d*x] + 3*(9*A + B - C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^5*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/8))/(1
5*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [B]  time = 1.252, size = 624, normalized size = 2.7 \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)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*cos(1/2*d*x+1/2*c)^8-30*A*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*A
*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))+12*B*cos(1/2*d*x+1/2*c)^8-10*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+
1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-12*C*cos(1/2*d*x+1/2*c)^8-10*C*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*
c)^5-6*C*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2
*d*x+1/2*c),2^(1/2))-138*A*cos(1/2*d*x+1/2*c)^6-22*B*cos(1/2*d*x+1/2*c)^6+2*C*cos(1/2*d*x+1/2*c)^6+24*A*cos(1/
2*d*x+1/2*c)^4+6*B*cos(1/2*d*x+1/2*c)^4+24*C*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)^2+7*B*cos(1/2*d*x+1/2
*c)^2-17*C*cos(1/2*d*x+1/2*c)^2+3*A-3*B+3*C)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(-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)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, 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)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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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{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{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)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

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