3.1202 \(\int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=249 \[ \frac{(A+11 C) \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac{(9 A+119 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A+C) \sin (c+d x)}{5 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^3}-\frac{2 C \sin (c+d x)}{3 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[Out]

-((9*A + 119*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 11*C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) - ((A + C)*Sin[c + d*x])/(5*d*(a + a*Co
s[c + d*x])^3*Sec[c + d*x]^(7/2)) - (2*C*Sin[c + d*x])/(3*a*d*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)) - ((9
*A + 119*C)*Sin[c + d*x])/(30*d*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^(3/2)) + ((A + 11*C)*Sin[c + d*x])/(2*a^
3*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.597213, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4221, 3042, 2977, 2748, 2639, 2635, 2641} \[ \frac{(A+11 C) \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac{(9 A+119 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A+C) \sin (c+d x)}{5 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^3}-\frac{2 C \sin (c+d x)}{3 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((9*A + 119*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 11*C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) - ((A + C)*Sin[c + d*x])/(5*d*(a + a*Co
s[c + d*x])^3*Sec[c + d*x]^(7/2)) - (2*C*Sin[c + d*x])/(3*a*d*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)) - ((9
*A + 119*C)*Sin[c + d*x])/(30*d*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^(3/2)) + ((A + 11*C)*Sin[c + d*x])/(2*a^
3*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 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (\frac{1}{2} a (3 A-7 C)+\frac{1}{2} a (3 A+13 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 C \sin (c+d x)}{3 a 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 (-25 a^2 C+\frac{3}{2} a^2 (3 A+23 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 C \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right ) \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}{4} a^3 (9 A+119 C)+\frac{45}{4} a^3 (A+11 C) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 C \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 (A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{4 a^3}-\frac{\left ((9 A+119 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{(9 A+119 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{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 C \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}+\frac{\left ((A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac{(9 A+119 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{(A+11 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{2 a^3 d}-\frac{(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 C \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.15037, size = 573, normalized size = 2.3 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left ((156 A+1961 C) \cos \left (\frac{1}{2} (c-d x)\right )+(114 A+1609 C) \cos \left (\frac{1}{2} (3 c+d x)\right )+90 A \cos \left (\frac{1}{2} (c+3 d x)\right )+45 A \cos \left (\frac{1}{2} (5 c+3 d x)\right )+27 A \cos \left (\frac{1}{2} (3 c+5 d x)\right )+1165 C \cos \left (\frac{1}{2} (c+3 d x)\right )+620 C \cos \left (\frac{1}{2} (5 c+3 d x)\right )+292 C \cos \left (\frac{1}{2} (3 c+5 d x)\right )+65 C \cos \left (\frac{1}{2} (7 c+5 d x)\right )+5 C \cos \left (\frac{1}{2} (5 c+7 d x)\right )-5 C \cos \left (\frac{1}{2} (9 c+7 d x)\right )\right )}{8 \sqrt{\sec (c+d x)}}+18 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+60 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+238 \sqrt{2} C \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+660 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*((18*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (238*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 +
E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometri
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (((156*A + 1961*C)*Cos[(c - d*x)/2] + (114*A + 1609*C)
*Cos[(3*c + d*x)/2] + 90*A*Cos[(c + 3*d*x)/2] + 1165*C*Cos[(c + 3*d*x)/2] + 45*A*Cos[(5*c + 3*d*x)/2] + 620*C*
Cos[(5*c + 3*d*x)/2] + 27*A*Cos[(3*c + 5*d*x)/2] + 292*C*Cos[(3*c + 5*d*x)/2] + 65*C*Cos[(7*c + 5*d*x)/2] + 5*
C*Cos[(5*c + 7*d*x)/2] - 5*C*Cos[(9*c + 7*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5)/(8*Sqrt[Sec[c + d*x]]
) + 60*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 660*C*Sqrt[Cos[c + d*x]]*EllipticF[
(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 1.295, size = 465, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(5/2),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)*(160*C*cos(1/2*d*x+1/2*c)^10+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))+468*C*cos(1/2*d*x+1/2*c)^8+330*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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+714*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))-198*A
*cos(1/2*d*x+1/2*c)^6-1058*C*cos(1/2*d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4+474*C*cos(1/2*d*x+1/2*c)^4-27*A*c
os(1/2*d*x+1/2*c)^2-47*C*cos(1/2*d*x+1/2*c)^2+3*A+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+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(5/2),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{C \cos \left (d x + c\right )^{2} + A}{{\left (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}\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)/((a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3)*sec(d*
x + c)^(5/2)), 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+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3/sec(d*x+c)**(5/2),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} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \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+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2)), x)