3.1189 \(\int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=199 \[ \frac{(5 A+7 C) \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{(3 A+5 C) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{(A+C) \sin (c+d x)}{d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac{(3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 (5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.290666, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4221, 3042, 2748, 2635, 2641, 2639} \[ \frac{(5 A+7 C) \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{(3 A+5 C) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{(A+C) \sin (c+d x)}{d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac{(3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 (5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.88615, size = 458, normalized size = 2.3 \[ -\frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{2 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left ((60 A+83 C) \cos \left (\frac{1}{2} (c-d x)\right )+(30 A+43 C) \cos \left (\frac{1}{2} (3 c+d x)\right )+C \sin (c) \left (7 \sin \left (\frac{3}{2} (c+d x)\right )-3 \sin \left (\frac{5}{2} (c+d x)\right )\right )\right )}{\sqrt{\sec (c+d x)}}+60 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+120 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+84 \sqrt{2} C \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+200 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{60 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]^2*((60*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) + (84*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) + 120*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]] + 200*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + (2*Csc[c/2]*Sec[c/2]
*Sec[(c + d*x)/2]*((60*A + 83*C)*Cos[(c - d*x)/2] + (30*A + 43*C)*Cos[(3*c + d*x)/2] + C*Sin[c]*(7*Sin[(3*(c +
d*x))/2] - 3*Sin[(5*(c + d*x))/2])))/Sqrt[Sec[c + d*x]]))/(60*a*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 1.223, size = 276, normalized size = 1.4 \begin{align*}{\frac{1}{15\,da}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( 15\,A{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +45\,A{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +25\,C{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +63\,C{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) -48\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+56\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 30\,A+30\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -15\,A-23\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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))/sec(d*x+c)^(3/2),x)

[Out]

1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^(1/2)*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-48*C*sin(1/2*
d*x+1/2*c)^8+56*C*sin(1/2*d*x+1/2*c)^6+(30*A+30*C)*sin(1/2*d*x+1/2*c)^4+(-15*A-23*C)*sin(1/2*d*x+1/2*c)^2)/a/c
os(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)/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{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \sec \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)/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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

[Out]

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