3.504 \(\int \frac{\cos ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=208 \[ -\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2+6 b^2\right )}{2 a^4} \]

[Out]

((a^2 + 6*b^2)*x)/(2*a^4) - (2*b^3*(4*a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(
a - b)^(3/2)*(a + b)^(3/2)*d) - (b*(2*a^2 - 3*b^2)*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 3*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*d) + (b^2*Cos[c + d*x]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x
]))

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Rubi [A]  time = 0.58549, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3847, 4104, 3919, 3831, 2659, 208} \[ -\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2+6 b^2\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + 6*b^2)*x)/(2*a^4) - (2*b^3*(4*a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(
a - b)^(3/2)*(a + b)^(3/2)*d) - (b*(2*a^2 - 3*b^2)*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 3*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*d) + (b^2*Cos[c + d*x]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x
]))

Rule 3847

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)
*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (-a^2+3 b^2+a b \sec (c+d x)-2 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (-2 b \left (2 a^2-3 b^2\right )+a \left (a^2+b^2\right ) \sec (c+d x)+b \left (a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{-a^4-5 a^2 b^2+6 b^4-a b \left (a^2-3 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2+6 b^2\right ) x}{2 a^4}-\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b^3 \left (4 a^2-3 b^2\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2+6 b^2\right ) x}{2 a^4}-\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b^2 \left (4 a^2-3 b^2\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2+6 b^2\right ) x}{2 a^4}-\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 b^2 \left (4 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d}\\ &=\frac{\left (a^2+6 b^2\right ) x}{2 a^4}-\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.724508, size = 144, normalized size = 0.69 \[ \frac{2 \left (a^2+6 b^2\right ) (c+d x)-\frac{8 b^3 \left (3 b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+a^2 \sin (2 (c+d x))+\frac{4 a b^4 \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}-8 a b \sin (c+d x)}{4 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 + 6*b^2)*(c + d*x) - (8*b^3*(-4*a^2 + 3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^
2 - b^2)^(3/2) - 8*a*b*Sin[c + d*x] + (4*a*b^4*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + a^2*Sin[
2*(c + d*x)])/(4*a^4*d)

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Maple [A]  time = 0.083, size = 362, normalized size = 1.7 \begin{align*} -{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}b}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) b}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{2}}{d{a}^{4}}}-2\,{\frac{{b}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-8\,{\frac{{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+6\,{\frac{{b}^{5}}{d{a}^{4} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+b*sec(d*x+c))^2,x)

[Out]

-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^3*b+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*
c)*b+1/d/a^2*arctan(tan(1/2*d*x+1/2*c))+6/d/a^4*arctan(tan(1/2*d*x+1/2*c))*b^2-2/d*b^4/a^3/(a^2-b^2)*tan(1/2*d
*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-8/d*b^3/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arct
anh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+6/d*b^5/a^4/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*ta
n(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12801, size = 1445, normalized size = 6.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*d*x*cos(d*x + c) + (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*d*x
+ (4*a^2*b^4 - 3*b^6 + (4*a^3*b^3 - 3*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2
*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2
+ 2*a*b*cos(d*x + c) + b^2)) - (4*a^5*b^2 - 10*a^3*b^4 + 6*a*b^6 - (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)^2
+ 3*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a
^8*b - 2*a^6*b^3 + a^4*b^5)*d), 1/2*((a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*d*x*cos(d*x + c) + (a^6*b + 4*a^
4*b^3 - 11*a^2*b^5 + 6*b^7)*d*x - 2*(4*a^2*b^4 - 3*b^6 + (4*a^3*b^3 - 3*a*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*
arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (4*a^5*b^2 - 10*a^3*b^4 + 6*a*b^6
- (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)^2 + 3*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a
^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a^8*b - 2*a^6*b^3 + a^4*b^5)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**2/(a + b*sec(c + d*x))**2, x)

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Giac [A]  time = 1.28963, size = 356, normalized size = 1.71 \begin{align*} -\frac{\frac{4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}} + \frac{4 \,{\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{{\left (a^{2} + 6 \, b^{2}\right )}{\left (d x + c\right )}}{a^{4}} + \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*b^4*tan(1/2*d*x + 1/2*c)/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b
)) + 4*(4*a^2*b^3 - 3*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c)
- b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - a^4*b^2)*sqrt(-a^2 + b^2)) - (a^2 + 6*b^2)*(d*x + c)/a^4
+ 2*(a*tan(1/2*d*x + 1/2*c)^3 + 4*b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) + 4*b*tan(1/2*d*x + 1/2*c
))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3))/d