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

Optimal. Leaf size=267 \[ \frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-4 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac{\left (5 a^2-6 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{x \left (a^2-12 b^2\right )}{2 a^5}+\frac{\sin (c+d x) \cos ^3(c+d x)}{2 a d (a \cos (c+d x)+b)^2} \]

[Out]

((a^2 - 12*b^2)*x)/(2*a^5) - (b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a +
b]])/(a^5*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b*(11*a^2 - 12*b^2)*Sin[c + d*x])/(2*a^4*(a^2 - b^2)*d) - ((5*a^2
- 6*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b^2)*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(2*a*d*(b + a*Cos[c
+ d*x])^2) + ((3*a^2 - 4*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))

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Rubi [A]  time = 0.940394, antiderivative size = 267, 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, 2889, 3048, 3049, 3023, 2735, 2659, 208} \[ \frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-4 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac{\left (5 a^2-6 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{x \left (a^2-12 b^2\right )}{2 a^5}+\frac{\sin (c+d x) \cos ^3(c+d x)}{2 a d (a \cos (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 - 12*b^2)*x)/(2*a^5) - (b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a +
b]])/(a^5*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b*(11*a^2 - 12*b^2)*Sin[c + d*x])/(2*a^4*(a^2 - b^2)*d) - ((5*a^2
- 6*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b^2)*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(2*a*d*(b + a*Cos[c
+ d*x])^2) + ((3*a^2 - 4*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(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)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^2(c+d x)}{(-b-a \cos (c+d x))^3} \, dx\\ &=-\int \frac{\cos ^3(c+d x) \left (1-\cos ^2(c+d x)\right )}{(-b-a \cos (c+d x))^3} \, dx\\ &=\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (a^2-b^2\right )-4 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{(-b-a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (2 \left (3 a^4-7 a^2 b^2+4 b^4\right )+a b \left (a^2-b^2\right ) \cos (c+d x)-2 \left (5 a^2-6 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\int \frac{2 b \left (5 a^4-11 a^2 b^2+6 b^4\right )-2 a \left (a^4-3 a^2 b^2+2 b^4\right ) \cos (c+d x)-2 b \left (11 a^2-12 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{4 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac{\int \frac{-2 a b \left (5 a^4-11 a^2 b^2+6 b^4\right )+2 \left (a^2-12 b^2\right ) \left (a^2-b^2\right )^2 \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{4 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2-12 b^2\right ) x}{2 a^5}+\frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\left (b \left (6 a^4-19 a^2 b^2+12 b^4\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{2 a^5 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2-12 b^2\right ) x}{2 a^5}+\frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\left (b \left (6 a^4-19 a^2 b^2+12 b^4\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^5 \left (a^2-b^2\right ) d}\\ &=\frac{\left (a^2-12 b^2\right ) x}{2 a^5}-\frac{b \left (6 a^4-19 a^2 b^2+12 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{b \left (11 a^2-12 b^2\right ) \sin (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a d (b+a \cos (c+d x))^2}+\frac{\left (3 a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.87882, size = 282, normalized size = 1.06 \[ \frac{\frac{b^2 \left (2 \left (-13 a^2 b^2+a^4+12 b^4\right ) (c+d x)+\left (22 a^3 b-24 a b^3\right ) \sin (c+d x)+\left (17 a^4-18 a^2 b^2\right ) \sin (2 (c+d x))\right )+4 a b \left (-13 a^2 b^2+a^4+12 b^4\right ) (c+d x) \cos (c+d x)-2 a^4 \left (a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x)+2 a^2 \left (a^2-b^2\right ) \cos ^2(c+d x) \left (\left (a^2-12 b^2\right ) (c+d x)+4 a b \sin (c+d x)\right )}{(a \cos (c+d x)+b)^2}+\frac{4 b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{4 a^5 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (4
*a*b*(a^4 - 13*a^2*b^2 + 12*b^4)*(c + d*x)*Cos[c + d*x] - 2*a^4*(a^2 - b^2)*Cos[c + d*x]^3*Sin[c + d*x] + 2*a^
2*(a^2 - b^2)*Cos[c + d*x]^2*((a^2 - 12*b^2)*(c + d*x) + 4*a*b*Sin[c + d*x]) + b^2*(2*(a^4 - 13*a^2*b^2 + 12*b
^4)*(c + d*x) + (22*a^3*b - 24*a*b^3)*Sin[c + d*x] + (17*a^4 - 18*a^2*b^2)*Sin[2*(c + d*x)]))/(b + a*Cos[c + d
*x])^2)/(4*a^5*(a - b)*(a + b)*d)

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Maple [B]  time = 0.088, size = 729, normalized size = 2.7 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^
3*b+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*b-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)-12/d/a^5*arctan(tan(1/2*d*x+1/2*c))*b^2+1/d/a^3*arctan(tan(1/2*d*x+1/2*c))-6/d*b^2/a^2/(tan(1/2*d*x+1/2*c)
^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3-1/d*b^3/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+
1/2*c)^2*b-a-b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3+6/d*b^4/a^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2
/(a+b)*tan(1/2*d*x+1/2*c)^3+6/d*b^2/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)*tan(1/2*d*
x+1/2*c)-1/d*b^3/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)*tan(1/2*d*x+1/2*c)-6/d*b^4/a^
4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)*tan(1/2*d*x+1/2*c)-6/d*b/a/(a^2-b^2)/((a+b)*(a-b
))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+19/d*b^3/a^3/(a^2-b^2)/((a+b)*(a-b))^(1/2)*arct
anh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-12/d*b^5/a^5/(a^2-b^2)/((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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.48311, size = 2184, normalized size = 8.18 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(a^8 - 14*a^6*b^2 + 25*a^4*b^4 - 12*a^2*b^6)*d*x*cos(d*x + c)^2 + 4*(a^7*b - 14*a^5*b^3 + 25*a^3*b^5 -
12*a*b^7)*d*x*cos(d*x + c) + 2*(a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*d*x - (6*a^4*b^3 - 19*a^2*b^5 + 1
2*b^7 + (6*a^6*b - 19*a^4*b^3 + 12*a^2*b^5)*cos(d*x + c)^2 + 2*(6*a^5*b^2 - 19*a^3*b^4 + 12*a*b^6)*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)) + 2*(11*a^5*b^3 - 23*a^3*b^5
+ 12*a*b^7 - (a^8 - 2*a^6*b^2 + a^4*b^4)*cos(d*x + c)^3 + 4*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + (1
7*a^6*b^2 - 35*a^4*b^4 + 18*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 2*a^9*b^2 + a^7*b^4)*d*cos(d*x + c)^
2 + 2*(a^10*b - 2*a^8*b^3 + a^6*b^5)*d*cos(d*x + c) + (a^9*b^2 - 2*a^7*b^4 + a^5*b^6)*d), 1/2*((a^8 - 14*a^6*b
^2 + 25*a^4*b^4 - 12*a^2*b^6)*d*x*cos(d*x + c)^2 + 2*(a^7*b - 14*a^5*b^3 + 25*a^3*b^5 - 12*a*b^7)*d*x*cos(d*x
+ c) + (a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*d*x - (6*a^4*b^3 - 19*a^2*b^5 + 12*b^7 + (6*a^6*b - 19*a^4
*b^3 + 12*a^2*b^5)*cos(d*x + c)^2 + 2*(6*a^5*b^2 - 19*a^3*b^4 + 12*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*arcta
n(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (11*a^5*b^3 - 23*a^3*b^5 + 12*a*b^7 - (
a^8 - 2*a^6*b^2 + a^4*b^4)*cos(d*x + c)^3 + 4*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + (17*a^6*b^2 - 35*
a^4*b^4 + 18*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 2*a^9*b^2 + a^7*b^4)*d*cos(d*x + c)^2 + 2*(a^10*b -
2*a^8*b^3 + a^6*b^5)*d*cos(d*x + c) + (a^9*b^2 - 2*a^7*b^4 + a^5*b^6)*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 )^{3}}\, 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))**3,x)

[Out]

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

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Giac [B]  time = 1.51067, size = 815, normalized size = 3.05 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="giac")

[Out]

-1/2*(2*(6*a^4*b - 19*a^2*b^3 + 12*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^7 - a^5*b^2)*sqrt(-a^2 + b^2)) - 2*(a^5*tan(1/2
*d*x + 1/2*c)^7 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^7 - 18*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 + 7*a^2*b^3*tan(1/2*d*x +
1/2*c)^7 + 18*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 12*b^5*tan(1/2*d*x + 1/2*c)^7 - 3*a^5*tan(1/2*d*x + 1/2*c)^5 - 4
*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 14*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 37*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 18*a*b
^4*tan(1/2*d*x + 1/2*c)^5 - 36*b^5*tan(1/2*d*x + 1/2*c)^5 + 3*a^5*tan(1/2*d*x + 1/2*c)^3 - 4*a^4*b*tan(1/2*d*x
+ 1/2*c)^3 + 14*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 37*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 18*a*b^4*tan(1/2*d*x + 1
/2*c)^3 - 36*b^5*tan(1/2*d*x + 1/2*c)^3 - a^5*tan(1/2*d*x + 1/2*c) + 4*a^4*b*tan(1/2*d*x + 1/2*c) + 18*a^3*b^2
*tan(1/2*d*x + 1/2*c) + 7*a^2*b^3*tan(1/2*d*x + 1/2*c) - 18*a*b^4*tan(1/2*d*x + 1/2*c) - 12*b^5*tan(1/2*d*x +
1/2*c))/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/2*c)^2 - a
- b)^2) - (a^2 - 12*b^2)*(d*x + c)/a^5)/d