3.1291 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=205 \[ \frac{(5 A-3 B+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{(3 A-3 B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(5 A-3 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{(3 A-3 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

((3*A - 3*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + ((5*A - 3*B + 3*C)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - ((3*A - 3*B + C)*Sqrt[Sec[c + d*x]]*
Sin[c + d*x])/(a*d) + ((5*A - 3*B + 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((A - B + C)*Sec[c + d*x]^
(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

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Rubi [A]  time = 0.348067, antiderivative size = 205, 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, 2748, 2636, 2641, 2639} \[ \frac{(5 A-3 B+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{(3 A-3 B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(5 A-3 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{(3 A-3 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A - 3*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + ((5*A - 3*B + 3*C)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - ((3*A - 3*B + C)*Sqrt[Sec[c + d*x]]*
Sin[c + d*x])/(a*d) + ((5*A - 3*B + 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((A - B + C)*Sec[c + d*x]^
(3/2)*Sin[c + d*x])/(d*(a + a*Cos[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 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 2636

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

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

Mathematica [A]  time = 2.50083, size = 162, normalized size = 0.79 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \left (2 (5 A-3 B+3 C) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 (3 A-3 B+C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\tan \left (\frac{1}{2} (c+d x)\right ) (3 (3 A-3 B+C) \cos (2 (c+d x))+4 (2 A-3 B) \cos (c+d x)+5 A-9 B+3 C)\right )}{3 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]^(3/2)*(6*(3*A - 3*B + C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 2*(5*
A - 3*B + 3*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - (5*A - 9*B + 3*C + 4*(2*A - 3*B)*Cos[c + d*x] +
3*(3*A - 3*B + C)*Cos[2*(c + d*x)])*Tan[(c + d*x)/2]))/(3*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 3.162, size = 494, normalized size = 2.4 \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)^(5/2)/(a+a*cos(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*(2*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
))+(-2*A+2*B)*(-(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*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+(A-B+C)*(cos(1/2*
d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*s
in(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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{a \cos \left (d x + c\right ) + a}\,{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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a), x)

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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 )} \sec \left (d x + c\right )^{\frac{5}{2}}}{a \cos \left (d x + c\right ) + a}, 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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a), 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)**(5/2)/(a+a*cos(d*x+c)),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 )} \sec \left (d x + c\right )^{\frac{5}{2}}}{a \cos \left (d x + c\right ) + a}\,{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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a), x)