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

Optimal. Leaf size=152 \[ -\frac{2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2-6 b^2\right )}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]

[Out]

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

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Rubi [A]  time = 0.570967, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2889, 3048, 3050, 3023, 2735, 2659, 208} \[ -\frac{2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2-6 b^2\right )}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 - 6*b^2)*x)/(2*a^4) - (2*b*(2*a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqr
t[a - b]*Sqrt[a + b]*d) + (3*b*Sin[c + d*x])/(a^3*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (Cos[c + d*x]
^2*Sin[c + d*x])/(a*d*(b + a*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 2889

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

Rule 3048

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

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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{\sin ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \frac{\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\int \frac{3 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)-6 b \left (a^2-b^2\right ) \cos ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac{\int \frac{-3 a b \left (a^2-b^2\right )+\left (a^2-6 b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{-b-a \cos (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{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^4}\\ &=\frac{\left (a^2-6 b^2\right ) x}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\left (2 b \left (2 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac{\left (a^2-6 b^2\right ) x}{2 a^4}-\frac{2 b \left (2 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 \sqrt{a-b} \sqrt{a+b} d}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.1514, size = 178, normalized size = 1.17 \[ \frac{\frac{-a \left (a^2-24 b^2\right ) \sin (c+d x)+4 a \left (a^2-6 b^2\right ) (c+d x) \cos (c+d x)+6 a^2 b \sin (2 (c+d x))+4 a^2 b c+4 a^2 b d x+a^3 (-\sin (3 (c+d x)))-24 b^3 c-24 b^3 d x}{a \cos (c+d x)+b}+\frac{16 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{8 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*b*(2*a^2 - 3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (4*a^2*b*c - 24
*b^3*c + 4*a^2*b*d*x - 24*b^3*d*x + 4*a*(a^2 - 6*b^2)*(c + d*x)*Cos[c + d*x] - a*(a^2 - 24*b^2)*Sin[c + d*x] +
6*a^2*b*Sin[2*(c + d*x)] - a^3*Sin[3*(c + d*x)])/(b + a*Cos[c + d*x]))/(8*a^4*d)

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Maple [B]  time = 0.072, size = 325, normalized size = 2.1 \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{b \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}}}+4\,{\frac{b\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}}}-{\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}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{2}}{d{a}^{4}}}+{\frac{1}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \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 ) }}-4\,{\frac{b}{d{a}^{2}\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}^{3}}{d{a}^{4}\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(sin(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+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/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)-6/d/a^4*arctan(tan(1/2*d*x+1/2*c))*b^2+1/d/a^2*arctan(tan(1/2*d*x+1/2*c))-2/d*b^2/a^3*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)-4/d*b/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2
*c)/((a+b)*(a-b))^(1/2))+6/d*b^3/a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(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(sin(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 = 2.03398, size = 1204, normalized size = 7.92 \begin{align*} \left [\frac{{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x -{\left (2 \, a^{2} b^{2} - 3 \, b^{4} +{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) +{\left (6 \, a^{3} b^{2} - 6 \, a b^{4} -{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}, \frac{{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x - 2 \,{\left (2 \, a^{2} b^{2} - 3 \, b^{4} +{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) +{\left (6 \, a^{3} b^{2} - 6 \, a b^{4} -{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^5 - 7*a^3*b^2 + 6*a*b^4)*d*x*cos(d*x + c) + (a^4*b - 7*a^2*b^3 + 6*b^5)*d*x - (2*a^2*b^2 - 3*b^4 + (2
*a^3*b - 3*a*b^3)*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*sqr
t(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))
+ (6*a^3*b^2 - 6*a*b^4 - (a^5 - a^3*b^2)*cos(d*x + c)^2 + 3*(a^4*b - a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a
^7 - a^5*b^2)*d*cos(d*x + c) + (a^6*b - a^4*b^3)*d), 1/2*((a^5 - 7*a^3*b^2 + 6*a*b^4)*d*x*cos(d*x + c) + (a^4*
b - 7*a^2*b^3 + 6*b^5)*d*x - 2*(2*a^2*b^2 - 3*b^4 + (2*a^3*b - 3*a*b^3)*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))) + (6*a^3*b^2 - 6*a*b^4 - (a^5 - a^3*b^2)*co
s(d*x + c)^2 + 3*(a^4*b - a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d*cos(d*x + c) + (a^6*b - a^4*
b^3)*d)]

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

[Out]

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

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Giac [A]  time = 1.30816, size = 324, normalized size = 2.13 \begin{align*} -\frac{\frac{4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} a^{3}} - \frac{{\left (a^{2} - 6 \, b^{2}\right )}{\left (d x + c\right )}}{a^{4}} + \frac{4 \,{\left (2 \, a^{2} b - 3 \, b^{3}\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 )}}{\sqrt{-a^{2} + b^{2}} 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(sin(d*x+c)^2/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*b^2*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)*a^3) - (a^2 -
6*b^2)*(d*x + c)/a^4 + 4*(2*a^2*b - 3*b^3)*(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)))/(sqrt(-a^2 + b^2)*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