3.86 \(\int \frac{\sin ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=69 \[ \frac{2 \sin (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{2 \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac{5 x}{2 a^2} \]

[Out]

(-5*x)/(2*a^2) + (2*Sin[c + d*x])/(a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (2*Sin[c + d*x])/(a^2*d*(1
+ Cos[c + d*x]))

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Rubi [A]  time = 0.316197, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2874, 2950, 2709, 2637, 2635, 8, 2648} \[ \frac{2 \sin (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{2 \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac{5 x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/(2*a^2) + (2*Sin[c + d*x])/(a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (2*Sin[c + d*x])/(a^2*d*(1
+ Cos[c + d*x]))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2950

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2648

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac{\int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))}{-a-a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{\int (-a+a \cos (c+d x))^2 \cot ^2(c+d x) \, dx}{a^4}\\ &=\frac{\int \left (-2+2 \cos (c+d x)-\cos ^2(c+d x)+\frac{2}{1+\cos (c+d x)}\right ) \, dx}{a^2}\\ &=-\frac{2 x}{a^2}-\frac{\int \cos ^2(c+d x) \, dx}{a^2}+\frac{2 \int \cos (c+d x) \, dx}{a^2}+\frac{2 \int \frac{1}{1+\cos (c+d x)} \, dx}{a^2}\\ &=-\frac{2 x}{a^2}+\frac{2 \sin (c+d x)}{a^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{2 \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{\int 1 \, dx}{2 a^2}\\ &=-\frac{5 x}{2 a^2}+\frac{2 \sin (c+d x)}{a^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{2 \sin (c+d x)}{a^2 d (1+\cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.321016, size = 121, normalized size = 1.75 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (-25 \sin \left (c+\frac{d x}{2}\right )-21 \sin \left (c+\frac{3 d x}{2}\right )-21 \sin \left (2 c+\frac{3 d x}{2}\right )+3 \sin \left (2 c+\frac{5 d x}{2}\right )+3 \sin \left (3 c+\frac{5 d x}{2}\right )+60 d x \cos \left (c+\frac{d x}{2}\right )-119 \sin \left (\frac{d x}{2}\right )+60 d x \cos \left (\frac{d x}{2}\right )\right )}{48 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c/2]*Sec[(c + d*x)/2]*(60*d*x*Cos[(d*x)/2] + 60*d*x*Cos[c + (d*x)/2] - 119*Sin[(d*x)/2] - 25*Sin[c + (d*
x)/2] - 21*Sin[c + (3*d*x)/2] - 21*Sin[2*c + (3*d*x)/2] + 3*Sin[2*c + (5*d*x)/2] + 3*Sin[3*c + (5*d*x)/2]))/(4
8*a^2*d)

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Maple [A]  time = 0.086, size = 103, normalized size = 1.5 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2}}}+5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+3\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2*tan(1/2*d*x+1/2*c)+5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+3/d/a^2/(1+tan(1/2*d*x+1/2*
c)^2)^2*tan(1/2*d*x+1/2*c)-5/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.53877, size = 189, normalized size = 2.74 \begin{align*} \frac{\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{5 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{2 \, \sin \left (d x + c\right )}{a^{2}{\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 2
*sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.70279, size = 158, normalized size = 2.29 \begin{align*} -\frac{5 \, d x \cos \left (d x + c\right ) + 5 \, d x +{\left (\cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(5*d*x*cos(d*x + c) + 5*d*x + (cos(d*x + c)^2 - 3*cos(d*x + c) - 8)*sin(d*x + c))/(a^2*d*cos(d*x + c) + a
^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sin ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.32468, size = 101, normalized size = 1.46 \begin{align*} -\frac{\frac{5 \,{\left (d x + c\right )}}{a^{2}} - \frac{4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{2 \,{\left (5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, \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^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(5*(d*x + c)/a^2 - 4*tan(1/2*d*x + 1/2*c)/a^2 - 2*(5*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2*c))/((t
an(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d