3.3.18 \(\int \frac {\sin ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [218]

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

[Out]

1/2*(a^2-6*b^2)*x/a^4+3*b*sin(d*x+c)/a^3/d-3/2*cos(d*x+c)*sin(d*x+c)/a^2/d+cos(d*x+c)^2*sin(d*x+c)/a/d/(b+a*co
s(d*x+c))-2*b*(2*a^2-3*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]
time = 0.39, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 = {3957, 2968, 3127, 3129, 3102, 2814, 2738, 214} \begin {gather*} \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 {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 {\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \end {gather*}

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 214

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

Rule 2738

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 2814

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 2968

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 3102

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 3127

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*Si
n[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 3129

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

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 ) \text {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*}

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Mathematica [A]
time = 0.55, size = 178, normalized size = 1.17 \begin {gather*} \frac {\frac {16 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {4 a^2 b c-24 b^3 c+4 a^2 b d x-24 b^3 d x+4 a \left (a^2-6 b^2\right ) (c+d x) \cos (c+d x)-a \left (a^2-24 b^2\right ) \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} \end {gather*}

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 [A]
time = 0.15, size = 199, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}-3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2}+2 b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 b a -\frac {1}{2} a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}-6 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}}{d}\)	\(199\)
default	\(\frac {\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}-3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2}+2 b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 b a -\frac {1}{2} a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}-6 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}}{d}\)	\(199\)
risch	\(\frac {x}{2 a^{2}}-\frac {3 x \,b^{2}}{a^{4}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{a^{3} d}+\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{a^{3} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {2 i b^{2} \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{4} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}\)	\(439\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [A]
time = 4.62, size = 551, normalized size = 3.62 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

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 = 0.47, size = 240, normalized size = 1.58 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 3.57, size = 1655, normalized size = 10.89 \begin {gather*} \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2+6\,b^2\right )}{a^3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-a^2+3\,a\,b+6\,b^2\right )}{a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^2+3\,a\,b-6\,b^2\right )}{a^3}}{d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (3\,b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (a+3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}+\frac {\mathrm {atan}\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {8\,b}{a}+\frac {24\,b^2}{a^2}-\frac {24\,b^3}{a^3}+\frac {144\,b^4}{a^4}-\frac {144\,b^5}{a^5}-8}+\frac {8\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a-8\,b-\frac {24\,b^2}{a}+\frac {24\,b^3}{a^2}-\frac {144\,b^4}{a^3}+\frac {144\,b^5}{a^4}}-\frac {24\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,b-8\,a^2+24\,b^2-\frac {24\,b^3}{a}+\frac {144\,b^4}{a^2}-\frac {144\,b^5}{a^3}}+\frac {24\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{24\,a\,b^2+8\,a^2\,b-8\,a^3-24\,b^3+\frac {144\,b^4}{a}-\frac {144\,b^5}{a^2}}+\frac {144\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{24\,a\,b^3-8\,a^3\,b+8\,a^4-144\,b^4-24\,a^2\,b^2+\frac {144\,b^5}{a}}+\frac {144\,b^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{-8\,a^5+8\,a^4\,b+24\,a^3\,b^2-24\,a^2\,b^3+144\,a\,b^4-144\,b^5}\right )\,\left (a^2\,1{}\mathrm {i}-b^2\,6{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^4\,d}-\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^7-3\,a^6\,b+7\,a^5\,b^2+19\,a^4\,b^3-48\,a^3\,b^4-48\,a^2\,b^5+144\,a\,b^6-72\,b^7\right )}{a^6}+\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (\frac {8\,\left (2\,a^{12}-10\,a^{11}\,b+2\,a^{10}\,b^2+18\,a^9\,b^3-12\,a^8\,b^4\right )}{a^9}-\frac {8\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^6\,\left (a^6-a^4\,b^2\right )}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )\,1{}\mathrm {i}}{a^6-a^4\,b^2}+\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^7-3\,a^6\,b+7\,a^5\,b^2+19\,a^4\,b^3-48\,a^3\,b^4-48\,a^2\,b^5+144\,a\,b^6-72\,b^7\right )}{a^6}-\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (\frac {8\,\left (2\,a^{12}-10\,a^{11}\,b+2\,a^{10}\,b^2+18\,a^9\,b^3-12\,a^8\,b^4\right )}{a^9}+\frac {8\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^6\,\left (a^6-a^4\,b^2\right )}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )\,1{}\mathrm {i}}{a^6-a^4\,b^2}}{\frac {16\,\left (-2\,a^7\,b-4\,a^6\,b^2+33\,a^5\,b^3+18\,a^4\,b^4-153\,a^3\,b^5+54\,a^2\,b^6+162\,a\,b^7-108\,b^8\right )}{a^9}+\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^7-3\,a^6\,b+7\,a^5\,b^2+19\,a^4\,b^3-48\,a^3\,b^4-48\,a^2\,b^5+144\,a\,b^6-72\,b^7\right )}{a^6}+\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (\frac {8\,\left (2\,a^{12}-10\,a^{11}\,b+2\,a^{10}\,b^2+18\,a^9\,b^3-12\,a^8\,b^4\right )}{a^9}-\frac {8\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^6\,\left (a^6-a^4\,b^2\right )}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )}{a^6-a^4\,b^2}-\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^7-3\,a^6\,b+7\,a^5\,b^2+19\,a^4\,b^3-48\,a^3\,b^4-48\,a^2\,b^5+144\,a\,b^6-72\,b^7\right )}{a^6}-\frac {b\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (\frac {8\,\left (2\,a^{12}-10\,a^{11}\,b+2\,a^{10}\,b^2+18\,a^9\,b^3-12\,a^8\,b^4\right )}{a^9}+\frac {8\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^6\,\left (a^6-a^4\,b^2\right )}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )}{a^6-a^4\,b^2}}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,2{}\mathrm {i}}{d\,\left (a^6-a^4\,b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3*(a^2 + 6*b^2))/a^3 + (tan(c/2 + (d*x)/2)*(3*a*b - a^2 + 6*b^2))/a^3 - (tan(c/2 + (d*x
)/2)^5*(3*a*b + a^2 - 6*b^2))/a^3)/(d*(a + b + tan(c/2 + (d*x)/2)^2*(a + 3*b) - tan(c/2 + (d*x)/2)^4*(a - 3*b)
- tan(c/2 + (d*x)/2)^6*(a - b))) + (atan((8*tan(c/2 + (d*x)/2))/((8*b)/a + (24*b^2)/a^2 - (24*b^3)/a^3 + (144
*b^4)/a^4 - (144*b^5)/a^5 - 8) + (8*b*tan(c/2 + (d*x)/2))/(8*a - 8*b - (24*b^2)/a + (24*b^3)/a^2 - (144*b^4)/a
^3 + (144*b^5)/a^4) - (24*b^2*tan(c/2 + (d*x)/2))/(8*a*b - 8*a^2 + 24*b^2 - (24*b^3)/a + (144*b^4)/a^2 - (144*
b^5)/a^3) + (24*b^3*tan(c/2 + (d*x)/2))/(24*a*b^2 + 8*a^2*b - 8*a^3 - 24*b^3 + (144*b^4)/a - (144*b^5)/a^2) +
(144*b^4*tan(c/2 + (d*x)/2))/(24*a*b^3 - 8*a^3*b + 8*a^4 - 144*b^4 - 24*a^2*b^2 + (144*b^5)/a) + (144*b^5*tan(
c/2 + (d*x)/2))/(144*a*b^4 + 8*a^4*b - 8*a^5 - 144*b^5 - 24*a^2*b^3 + 24*a^3*b^2))*(a^2*1i - b^2*6i)*1i)/(a^4*
d) - (b*atan(((b*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((8*tan(c/2 + (d*x)/2)*(144*a*b^6 - 3*a^6*b + a^7 - 7
2*b^7 - 48*a^2*b^5 - 48*a^3*b^4 + 19*a^4*b^3 + 7*a^5*b^2))/a^6 + (b*((a + b)*(a - b))^(1/2)*((8*(2*a^12 - 10*a
^11*b - 12*a^8*b^4 + 18*a^9*b^3 + 2*a^10*b^2))/a^9 - (8*b*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(2*a^2 -
3*b^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2))*1i)/(a^6
- a^4*b^2) + (b*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((8*tan(c/2 + (d*x)/2)*(144*a*b^6 - 3*a^6*b + a^7 - 72
*b^7 - 48*a^2*b^5 - 48*a^3*b^4 + 19*a^4*b^3 + 7*a^5*b^2))/a^6 - (b*((a + b)*(a - b))^(1/2)*((8*(2*a^12 - 10*a^
11*b - 12*a^8*b^4 + 18*a^9*b^3 + 2*a^10*b^2))/a^9 + (8*b*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(2*a^2 - 3
*b^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2))*1i)/(a^6 -
a^4*b^2))/((16*(162*a*b^7 - 2*a^7*b - 108*b^8 + 54*a^2*b^6 - 153*a^3*b^5 + 18*a^4*b^4 + 33*a^5*b^3 - 4*a^6*b^
2))/a^9 + (b*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((8*tan(c/2 + (d*x)/2)*(144*a*b^6 - 3*a^6*b + a^7 - 72*b^
7 - 48*a^2*b^5 - 48*a^3*b^4 + 19*a^4*b^3 + 7*a^5*b^2))/a^6 + (b*((a + b)*(a - b))^(1/2)*((8*(2*a^12 - 10*a^11*
b - 12*a^8*b^4 + 18*a^9*b^3 + 2*a^10*b^2))/a^9 - (8*b*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^
2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2)))/(a^6 - a^4*b
^2) - (b*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((8*tan(c/2 + (d*x)/2)*(144*a*b^6 - 3*a^6*b + a^7 - 72*b^7 -
48*a^2*b^5 - 48*a^3*b^4 + 19*a^4*b^3 + 7*a^5*b^2))/a^6 - (b*((a + b)*(a - b))^(1/2)*((8*(2*a^12 - 10*a^11*b -
12*a^8*b^4 + 18*a^9*b^3 + 2*a^10*b^2))/a^9 + (8*b*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*(
8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2))
)*((a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*2i)/(d*(a^6 - a^4*b^2))

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