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

Optimal. Leaf size=155 \[ -\frac{(5 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(5 A-C) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (\cos (c+d x)+1)}-\frac{4 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{4 A \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.323075, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3042, 2978, 2748, 2636, 2639, 2641} \[ -\frac{(5 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(5 A-C) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (\cos (c+d x)+1)}-\frac{4 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{4 A \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 6.63269, size = 834, normalized size = 5.38 \[ -\frac{2 i A \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (\frac{2 e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt{e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt{e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)+1}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac{2 \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt{e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt{e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)+1}}{d \left (-1+e^{2 i d x}\right ) \sin (c)-i d \left (1+e^{2 i d x}\right ) \cos (c)}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (c+d x) a+a)^2}+\frac{\sqrt{\cos (c+d x)} \left (\frac{2 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{2 (A+C) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{8 A \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{8 A \cot \left (\frac{c}{2}\right ) \sec (c)}{d}+\frac{8 A \sec (c) \sec (c+d x) \sin (d x)}{d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (c+d x) a+a)^2}+\frac{10 A \csc \left (\frac{c}{2}\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (\frac{c}{2}\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (c+d x) a+a)^2 \sqrt{\cot ^2(c)+1}}-\frac{2 C \csc \left (\frac{c}{2}\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (\frac{c}{2}\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (c+d x) a+a)^2 \sqrt{\cot ^2(c)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)
*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x
)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1
+ E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(
2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] +
I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Cos[c
+ d*x])^2 + (10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]
]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*S
in[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]
) - (2*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c
/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - A
rcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (Cos[c
/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((8*A*Cot[c/2]*Sec[c])/d + (8*A*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d
+ (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (8*A*Sec[c]*Sec[c + d*x]*Sin[d*
x])/d + (2*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2

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Maple [B]  time = 0.152, size = 452, normalized size = 2.9 \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)/cos(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)/cos(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 )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^3 + a^2*cos(d*x +
c)^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)/cos(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{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \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+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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