3.103 \(\int \frac{\sin ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=97 \[ \frac{3 \sin (c+d x)}{a^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{19 \sin (c+d x)}{3 a^3 d (\cos (c+d x)+1)}-\frac{2 \sin (c+d x)}{3 a^3 d (\cos (c+d x)+1)^2}-\frac{11 x}{2 a^3} \]

[Out]

(-11*x)/(2*a^3) + (3*Sin[c + d*x])/(a^3*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (2*Sin[c + d*x])/(3*a^3*d
*(1 + Cos[c + d*x])^2) + (19*Sin[c + d*x])/(3*a^3*d*(1 + Cos[c + d*x]))

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Rubi [A]  time = 0.309781, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2874, 2966, 2637, 2635, 8, 2650, 2648} \[ \frac{3 \sin (c+d x)}{a^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{19 \sin (c+d x)}{3 a^3 d (\cos (c+d x)+1)}-\frac{2 \sin (c+d x)}{3 a^3 d (\cos (c+d x)+1)^2}-\frac{11 x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*x)/(2*a^3) + (3*Sin[c + d*x])/(a^3*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (2*Sin[c + d*x])/(3*a^3*d
*(1 + Cos[c + d*x])^2) + (19*Sin[c + d*x])/(3*a^3*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 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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 2650

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

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))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^2(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac{\int \frac{\cos ^3(c+d x) (-a+a \cos (c+d x))}{(-a-a \cos (c+d x))^2} \, dx}{a^2}\\ &=-\frac{\int \left (\frac{5}{a}-\frac{3 \cos (c+d x)}{a}+\frac{\cos ^2(c+d x)}{a}+\frac{2}{a (1+\cos (c+d x))^2}-\frac{7}{a (1+\cos (c+d x))}\right ) \, dx}{a^2}\\ &=-\frac{5 x}{a^3}-\frac{\int \cos ^2(c+d x) \, dx}{a^3}-\frac{2 \int \frac{1}{(1+\cos (c+d x))^2} \, dx}{a^3}+\frac{3 \int \cos (c+d x) \, dx}{a^3}+\frac{7 \int \frac{1}{1+\cos (c+d x)} \, dx}{a^3}\\ &=-\frac{5 x}{a^3}+\frac{3 \sin (c+d x)}{a^3 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{2 \sin (c+d x)}{3 a^3 d (1+\cos (c+d x))^2}+\frac{7 \sin (c+d x)}{a^3 d (1+\cos (c+d x))}-\frac{\int 1 \, dx}{2 a^3}-\frac{2 \int \frac{1}{1+\cos (c+d x)} \, dx}{3 a^3}\\ &=-\frac{11 x}{2 a^3}+\frac{3 \sin (c+d x)}{a^3 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{2 \sin (c+d x)}{3 a^3 d (1+\cos (c+d x))^2}+\frac{19 \sin (c+d x)}{3 a^3 d (1+\cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.417204, size = 177, normalized size = 1.82 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (1326 \sin \left (c+\frac{d x}{2}\right )-2012 \sin \left (c+\frac{3 d x}{2}\right )-498 \sin \left (2 c+\frac{3 d x}{2}\right )-135 \sin \left (2 c+\frac{5 d x}{2}\right )-135 \sin \left (3 c+\frac{5 d x}{2}\right )+15 \sin \left (3 c+\frac{7 d x}{2}\right )+15 \sin \left (4 c+\frac{7 d x}{2}\right )+1980 d x \cos \left (c+\frac{d x}{2}\right )+660 d x \cos \left (c+\frac{3 d x}{2}\right )+660 d x \cos \left (2 c+\frac{3 d x}{2}\right )-3216 \sin \left (\frac{d x}{2}\right )+1980 d x \cos \left (\frac{d x}{2}\right )\right )}{960 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c/2]*Sec[(c + d*x)/2]^3*(1980*d*x*Cos[(d*x)/2] + 1980*d*x*Cos[c + (d*x)/2] + 660*d*x*Cos[c + (3*d*x)/2]
+ 660*d*x*Cos[2*c + (3*d*x)/2] - 3216*Sin[(d*x)/2] + 1326*Sin[c + (d*x)/2] - 2012*Sin[c + (3*d*x)/2] - 498*Sin
[2*c + (3*d*x)/2] - 135*Sin[2*c + (5*d*x)/2] - 135*Sin[3*c + (5*d*x)/2] + 15*Sin[3*c + (7*d*x)/2] + 15*Sin[4*c
+ (7*d*x)/2]))/(960*a^3*d)

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Maple [A]  time = 0.092, size = 122, normalized size = 1.3 \begin{align*} -{\frac{1}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+6\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3}}}+7\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+5\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-11\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

-1/3/d/a^3*tan(1/2*d*x+1/2*c)^3+6/d/a^3*tan(1/2*d*x+1/2*c)+7/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*
c)^3+5/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-11/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.79722, size = 221, normalized size = 2.28 \begin{align*} \frac{\frac{3 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{18 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac{33 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, 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))^3,x, algorithm="maxima")

[Out]

1/3*(3*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x + c)^2
/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (18*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d
*x + c)^3/(cos(d*x + c) + 1)^3)/a^3 - 33*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.81664, size = 259, normalized size = 2.67 \begin{align*} -\frac{33 \, d x \cos \left (d x + c\right )^{2} + 66 \, d x \cos \left (d x + c\right ) + 33 \, d x +{\left (3 \, \cos \left (d x + c\right )^{3} - 12 \, \cos \left (d x + c\right )^{2} - 71 \, \cos \left (d x + c\right ) - 52\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/6*(33*d*x*cos(d*x + c)^2 + 66*d*x*cos(d*x + c) + 33*d*x + (3*cos(d*x + c)^3 - 12*cos(d*x + c)^2 - 71*cos(d*
x + c) - 52)*sin(d*x + c))/(a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + a^3*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 ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

Integral(sin(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.30563, size = 130, normalized size = 1.34 \begin{align*} -\frac{\frac{33 \,{\left (d x + c\right )}}{a^{3}} - \frac{6 \,{\left (7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, \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}} + \frac{2 \,{\left (a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 18 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{9}}}{6 \, 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))^3,x, algorithm="giac")

[Out]

-1/6*(33*(d*x + c)/a^3 - 6*(7*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2
*a^3) + 2*(a^6*tan(1/2*d*x + 1/2*c)^3 - 18*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d