3.607 \(\int \frac {(1-\cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=118 \[ -\frac {2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 \tan (c+d x)}{a^2 d}-\frac {2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))} \]

[Out]

-2*b*arctanh(sin(d*x+c))/a^3/d-2*(a^2-2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/d/(a-b)^(1
/2)/(a+b)^(1/2)+2*tan(d*x+c)/a^2/d-tan(d*x+c)/a/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.40, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3056, 3001, 3770, 2659, 205} \[ -\frac {2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 \tan (c+d x)}{a^2 d}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

(-2*(a^2 - 2*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b]*Sqrt[a + b]*d) - (2*b*A
rcTanh[Sin[c + d*x]])/(a^3*d) + (2*Tan[c + d*x])/(a^2*d) - Tan[c + d*x]/(a*d*(a + b*Cos[c + d*x]))

Rule 205

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

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 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3056

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[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac {\int \frac {\left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 \tan (c+d x)}{a^2 d}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-2 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {2 \tan (c+d x)}{a^2 d}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac {(2 b) \int \sec (c+d x) \, dx}{a^3}-\frac {\left (a^2-2 b^2\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^3}\\ &=-\frac {2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 \tan (c+d x)}{a^2 d}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac {\left (2 \left (a^2-2 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^3 d}\\ &=-\frac {2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 \tan (c+d x)}{a^2 d}-\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 143, normalized size = 1.21 \[ \frac {\frac {2 \left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {a b \sin (c+d x)}{a+b \cos (c+d x)}+a \tan (c+d x)+2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

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

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fricas [B]  time = 0.63, size = 628, normalized size = 5.32 \[ \left [\frac {{\left ({\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 2 \, a b^{2}\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 (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} - a^{2} b^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b - a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {{\left ({\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} - a^{2} b^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{5} b - a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(((a^2*b - 2*b^3)*cos(d*x + c)^2 + (a^3 - 2*a*b^2)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c)
+ (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos
(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*((a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*cos(d*x + c))*l
og(sin(d*x + c) + 1) + 2*((a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*cos(d*x + c))*log(-sin(d*x + c) + 1
) + 2*(a^4 - a^2*b^2 + 2*(a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5*b - a^3*b^3)*d*cos(d*x + c)^2 + (a^
6 - a^4*b^2)*d*cos(d*x + c)), -(((a^2*b - 2*b^3)*cos(d*x + c)^2 + (a^3 - 2*a*b^2)*cos(d*x + c))*sqrt(a^2 - b^2
)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + ((a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*
b^3)*cos(d*x + c))*log(sin(d*x + c) + 1) - ((a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*cos(d*x + c))*log
(-sin(d*x + c) + 1) - (a^4 - a^2*b^2 + 2*(a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5*b - a^3*b^3)*d*cos(
d*x + c)^2 + (a^6 - a^4*b^2)*d*cos(d*x + c))]

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giac [B]  time = 0.51, size = 235, normalized size = 1.99 \[ -\frac {2 \, {\left (\frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} a^{2}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.18, size = 231, normalized size = 1.96 \[ \frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{2}}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {1}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.49, size = 1091, normalized size = 9.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(cos(c + d*x)^2 - 1)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^2),x)

[Out]

((2*tan(c/2 + (d*x)/2)*(a + 2*b))/a^2 + (2*tan(c/2 + (d*x)/2)^3*(a - 2*b))/a^2)/(d*(a + b - tan(c/2 + (d*x)/2)
^4*(a - b) - 2*b*tan(c/2 + (d*x)/2)^2)) - (4*b*atanh((128*b*tan(c/2 + (d*x)/2))/(128*b - (128*b^2)/a) - (128*b
^2*tan(c/2 + (d*x)/2))/(128*a*b - 128*b^2)))/(a^3*d) - (atan((((-(a + b)*(a - b))^(1/2)*((32*tan(c/2 + (d*x)/2
)*(a^4*b - 16*a*b^4 - a^5 + 8*b^5 + 8*a^2*b^3))/a^4 + ((-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*((32*(a^9 + 2*a^
6*b^3 - 3*a^7*b^2))/a^6 - (32*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*(2*a^8*b + 2*a^6*b^3 -
4*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(a^2 - 2*b^2)*1i)/(a^5 - a^3*b^2) + ((-(a + b)*(a - b))^
(1/2)*((32*tan(c/2 + (d*x)/2)*(a^4*b - 16*a*b^4 - a^5 + 8*b^5 + 8*a^2*b^3))/a^4 - ((-(a + b)*(a - b))^(1/2)*(a
^2 - 2*b^2)*((32*(a^9 + 2*a^6*b^3 - 3*a^7*b^2))/a^6 + (32*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(a^2 - 2
*b^2)*(2*a^8*b + 2*a^6*b^3 - 4*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(a^2 - 2*b^2)*1i)/(a^5 - a^3
*b^2))/((64*(12*a*b^4 + 2*a^4*b - 8*b^5 - 6*a^3*b^2))/a^6 + ((-(a + b)*(a - b))^(1/2)*((32*tan(c/2 + (d*x)/2)*
(a^4*b - 16*a*b^4 - a^5 + 8*b^5 + 8*a^2*b^3))/a^4 + ((-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*((32*(a^9 + 2*a^6*
b^3 - 3*a^7*b^2))/a^6 - (32*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*(2*a^8*b + 2*a^6*b^3 - 4
*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(a^2 - 2*b^2))/(a^5 - a^3*b^2) - ((-(a + b)*(a - b))^(1/2)
*((32*tan(c/2 + (d*x)/2)*(a^4*b - 16*a*b^4 - a^5 + 8*b^5 + 8*a^2*b^3))/a^4 - ((-(a + b)*(a - b))^(1/2)*(a^2 -
2*b^2)*((32*(a^9 + 2*a^6*b^3 - 3*a^7*b^2))/a^6 + (32*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)
*(2*a^8*b + 2*a^6*b^3 - 4*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(a^2 - 2*b^2))/(a^5 - a^3*b^2)))*
(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*2i)/(d*(a^5 - a^3*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sec ^{2}{\left (c + d x \right )}}{a^{2} + 2 a b \cos {\left (c + d x \right )} + b^{2} \cos ^{2}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a^{2} + 2 a b \cos {\left (c + d x \right )} + b^{2} \cos ^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**2,x)

[Out]

-Integral(-sec(c + d*x)**2/(a**2 + 2*a*b*cos(c + d*x) + b**2*cos(c + d*x)**2), x) - Integral(cos(c + d*x)**2*s
ec(c + d*x)**2/(a**2 + 2*a*b*cos(c + d*x) + b**2*cos(c + d*x)**2), x)

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