3.1192 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=195 \[ -\frac{(5 A-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{(5 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 A \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{4 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.41414, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4221, 3042, 2978, 2748, 2636, 2639, 2641} \[ -\frac{(5 A-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{(5 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 A \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{4 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 1.37652, size = 275, normalized size = 1.41 \[ \frac{e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt{\sec (c+d x)} \left (i \left (4 A e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-19 A e^{i (c+d x)}-29 A e^{2 i (c+d x)}-31 A e^{3 i (c+d x)}-12 A e^{4 i (c+d x)}-5 A-C e^{i (c+d x)}+C e^{2 i (c+d x)}-C e^{3 i (c+d x)}+C\right )-(5 A-C) \left (1+e^{i (c+d x)}\right )^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{12 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + E^(I*(c + d*x)))*(-((5*A - C)*(1 + E^(I*(c + d*x)))^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]) + I*
(-5*A + C - 19*A*E^(I*(c + d*x)) - C*E^(I*(c + d*x)) - 29*A*E^((2*I)*(c + d*x)) + C*E^((2*I)*(c + d*x)) - 31*A
*E^((3*I)*(c + d*x)) - C*E^((3*I)*(c + d*x)) - 12*A*E^((4*I)*(c + d*x)) + 4*A*E^(I*(c + d*x))*(1 + E^(I*(c + d
*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))*Sqrt[Sec[c + d*
x]])/(12*a^2*d*E^((2*I)*(c + d*x))*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 1.206, size = 452, normalized size = 2.3 \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)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2,x)

[Out]

-1/6*(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)*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF
(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-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)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*
c)-48*A*(-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)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)*(43*A+C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(37*A+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)/s
in(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)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^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{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(3/2)/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^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)*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**2,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} + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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