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

Optimal. Leaf size=158 \[ \frac{(9 A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{3/2} d}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(-3*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(3/2)*d) + ((9*A + C)*ArcTanh[(Sqrt[a]*Sin[
c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Tan[c + d*x])/(2*d*(a + a*Cos[
c + d*x])^(3/2)) + ((3*A + C)*Tan[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.491487, antiderivative size = 158, 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 = {3042, 2984, 2985, 2649, 206, 2773} \[ \frac{(9 A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{3/2} d}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(3/2)*d) + ((9*A + C)*ArcTanh[(Sqrt[a]*Sin[
c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Tan[c + d*x])/(2*d*(a + a*Cos[
c + d*x])^(3/2)) + ((3*A + C)*Tan[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (a (3 A+C)-\frac{1}{2} a (3 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-3 a^2 A+\frac{1}{2} a^2 (3 A+C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^3}\\ &=-\frac{(A+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}-\frac{(3 A) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^2}+\frac{(9 A+C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{(3 A) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a d}-\frac{(9 A+C) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{2 a d}\\ &=-\frac{3 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac{(9 A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A+C) \tan (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.21139, size = 167, normalized size = 1.06 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (2 (9 A+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 \sqrt{2} A \cos ^2\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \sin \left (\frac{1}{2} (c+d x)\right ) (2 A \sec (c+d x)+3 A+C)}{\sin ^2\left (\frac{1}{2} (c+d x)\right )-1}\right )}{d (a (\cos (c+d x)+1))^{3/2} (2 A+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(2*(9*A + C)*ArcTanh[Sin[(c + d*x)/2]] + (12*Sqrt[2]
*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 - 2*(3*A + C + 2*A*Sec[c + d*x])*Sin[(c + d*x)/2])/(-1
+ Sin[(c + d*x)/2]^2)))/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(2*A + C + C*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.115, size = 746, normalized size = 4.7 \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)^2/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(18*A*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d
*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+2*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+
1/2*c))*cos(1/2*d*x+1/2*c)^4*a-12*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-12*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a
^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a-9*A*
ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-C*ln(2*
(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a+6*A*a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+6*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)
*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a+6*A*ln(-4*(a*2
^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*
cos(1/2*d*x+1/2*c)^2*a+2*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2-A*2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/cos(1/2*d*x+1/2*c)
/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(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)^2/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.88059, size = 830, normalized size = 5.25 \begin{align*} \frac{\sqrt{2}{\left ({\left (9 \, A + C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (9 \, A + C\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, A + C\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 6 \,{\left (A \cos \left (d x + c\right )^{3} + 2 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left ({\left (3 \, A + C\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\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)^2/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/8*(sqrt(2)*((9*A + C)*cos(d*x + c)^3 + 2*(9*A + C)*cos(d*x + c)^2 + (9*A + C)*cos(d*x + c))*sqrt(a)*log(-(a*
cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x +
c)^2 + 2*cos(d*x + c) + 1)) + 6*(A*cos(d*x + c)^3 + 2*A*cos(d*x + c)^2 + A*cos(d*x + c))*sqrt(a)*log((a*cos(d*
x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos
(d*x + c)^3 + cos(d*x + c)^2)) + 4*((3*A + C)*cos(d*x + c) + 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*
d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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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)**2/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 3.35156, size = 468, normalized size = 2.96 \begin{align*} -\frac{\frac{\sqrt{2}{\left (9 \, A \sqrt{a} + C \sqrt{a}\right )} \log \left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a^{2}} + \frac{12 \, A \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{a^{\frac{3}{2}}} - \frac{12 \, A \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{a^{\frac{3}{2}}} - \frac{16 \, \sqrt{2}{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt{a} - A a^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a} - \frac{2 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\sqrt{2} A a + \sqrt{2} C a\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}}}{8 \, d} \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)^2/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/8*(sqrt(2)*(9*A*sqrt(a) + C*sqrt(a))*log((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
)^2)/a^2 + 12*A*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) +
3)))/a^(3/2) - 12*A*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt
(2) - 3)))/a^(3/2) - 16*sqrt(2)*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqr
t(a) - A*a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*a) - 2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(sqrt
(2)*A*a + sqrt(2)*C*a)*tan(1/2*d*x + 1/2*c)/a^3)/d