\(\int \frac {(1-\cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [616]

Optimal result

Integrand size = 33, antiderivative size = 204 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (2 a^4-9 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

-(2*a^4-9*a^2*b^2+6*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(3/2)/(a+b)^(3/2)/d-3*b*
arctanh(sin(d*x+c))/a^4/d+1/2*(5*a^2-6*b^2)*tan(d*x+c)/a^3/(a^2-b^2)/d-1/2*tan(d*x+c)/a/d/(a+b*cos(d*x+c))^2-1
/2*(2*a^2-3*b^2)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3135, 3134, 3080, 3855, 2738, 211} \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\left (2 a^4-9 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

-(((2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^
(3/2)*d)) - (3*b*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((5*a^2 - 6*b^2)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)*d) - Tan[c
+ d*x]/(2*a*d*(a + b*Cos[c + d*x])^2) - ((2*a^2 - 3*b^2)*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Cos[c + d*
x]))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 3080

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 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + 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] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3135

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)),
Int[(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] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}+\frac {\int \frac {\left (3 \left (a^2-b^2\right )-2 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )} \\ & = -\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (5 a^4-11 a^2 b^2+6 b^4-a b \left (a^2-b^2\right ) \cos (c+d x)-\left (2 a^2-3 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}{2 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-6 b \left (a^2-b^2\right )^2-a \left (2 a^4-5 a^2 b^2+3 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {(3 b) \int \sec (c+d x) \, dx}{a^4}-\frac {\left (2 a^4-9 a^2 b^2+6 b^4\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )} \\ & = -\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (2 a^4-9 a^2 b^2+6 b^4\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 \left (a^2-b^2\right ) d} \\ & = -\frac {\left (2 a^4-9 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.24 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^4-9 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+6 b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {a b \left (4 a^3-5 a b^2+b \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+2 a \tan (c+d x)}{2 a^4 d} \]

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

((-2*(2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 6*
b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (a*b*(4*a^3 - 5*a*b^
2 + b*(3*a^2 - 4*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + 2*a*Tan[c + d*x])
/(2*a^4*d)

Maple [A] (verified)

Time = 2.77 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (4 a^{2}+a b -4 b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a +b \right )}-\frac {\left (4 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 a^{4}-9 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\)	\(257\)
default	\(\frac {-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (4 a^{2}+a b -4 b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a +b \right )}-\frac {\left (4 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 a^{4}-9 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\)	\(257\)
risch	\(\frac {i \left (2 a^{3} {\mathrm e}^{5 i \left (d x +c \right )} b -3 b^{3} a \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-3 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+20 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-24 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}+14 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+18 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-21 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a +5 a^{2} b^{2}-6 b^{4}\right )}{a^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2} \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\)	\(803\)

[In]

int((-cos(d*x+c)^2+1)*sec(d*x+c)^2/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/a^3/(tan(1/2*d*x+1/2*c)+1)-3*b/a^4*ln(tan(1/2*d*x+1/2*c)+1)-1/a^3/(tan(1/2*d*x+1/2*c)-1)+3*b/a^4*ln(ta
n(1/2*d*x+1/2*c)-1)-2/a^4*((-1/2*(4*a^2+a*b-4*b^2)*a*b/(a+b)*tan(1/2*d*x+1/2*c)^3-1/2*(4*a^2-a*b-4*b^2)*a*b/(a
-b)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)^2+1/2*(2*a^4-9*a^2*b^2+6*b^4)/(a^2
-b^2)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (189) = 378\).

Time = 0.61 (sec) , antiderivative size = 1121, normalized size of antiderivative = 5.50 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*
a^6 - 9*a^4*b^2 + 6*a^2*b^4)*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)) - 6*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2
+ (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x
+ c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(-sin
(d*x + c) + 1) + 2*(2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + (5*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + (8*a^6
*b - 17*a^4*b^3 + 9*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^3 + 2
*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^2 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)), -1/2*(((2*a^4*
b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 - 9*a^4*b^
2 + 6*a^2*b^4)*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))) + 3*
((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*
b^3 + a^2*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - 3*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^
2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - (
2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + (5*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + (8*a^6*b - 17*a^4*b^3 + 9*
a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^3 + 2*(a^9*b - 2*a^7*b^3
+ a^5*b^5)*d*cos(d*x + c)^2 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c))]

Sympy [F]

\[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=- \int \left (- \frac {\sec ^{2}{\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\, dx \]

[In]

integrate((1-cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)

[Out]

-Integral(-sec(c + d*x)**2/(a**3 + 3*a**2*b*cos(c + d*x) + 3*a*b**2*cos(c + d*x)**2 + b**3*cos(c + d*x)**3), x
) - Integral(cos(c + d*x)**2*sec(c + d*x)**2/(a**3 + 3*a**2*b*cos(c + d*x) + 3*a*b**2*cos(c + d*x)**2 + b**3*c
os(c + d*x)**3), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,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 de

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.75 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{4} - 9 \, a^{2} b^{2} + 6 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} - \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} + \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}}}{d} \]

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

((2*a^4 - 9*a^2*b^2 + 6*b^4)*(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^6 - a^4*b^2)*sqrt(a^2 - b^2)) + (4*a^3*b*tan(1/2*d*x + 1/2*
c)^3 - 3*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^
3*b*tan(1/2*d*x + 1/2*c) + 3*a^2*b^2*tan(1/2*d*x + 1/2*c) - 5*a*b^3*tan(1/2*d*x + 1/2*c) - 4*b^4*tan(1/2*d*x +
1/2*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) - 3*b*log(abs(tan(1
/2*d*x + 1/2*c) + 1))/a^4 + 3*b*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x
+ 1/2*c)^2 - 1)*a^3))/d

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 3376, normalized size of antiderivative = 16.55 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

int(-(cos(c + d*x)^2 - 1)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^3),x)

[Out]

((tan(c/2 + (d*x)/2)*(3*a*b^2 - 6*a^2*b - 2*a^3 + 6*b^3))/(a^3*b - a^4) - (tan(c/2 + (d*x)/2)^5*(3*a*b^2 + 6*a
^2*b - 2*a^3 - 6*b^3))/(a^3*(a + b)) + (2*tan(c/2 + (d*x)/2)^3*(2*a^4 + 6*b^4 - 7*a^2*b^2))/(a^3*(a + b)*(a -
b)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a*b - a^2 + 3*b^2) - tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 +
b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b + a^2 - 3*b^2))) - (b*atan(((b*((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 7
2*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (3*b*((8*(8*
a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3
- a^10*b^2) - (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*
b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/a^4)*3i)/a^4 + (b*((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 7
2*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (3*b*((8*(8*
a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3
- a^10*b^2) + (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*
b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/a^4)*3i)/a^4)/((16*(54*a*b^7 - 12*a^7*b - 108*b^8 + 270*a^2*b^
6 - 117*a^3*b^5 - 198*a^4*b^4 + 72*a^5*b^3 + 36*a^6*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (3*b*((8*tan(
c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9
- a^6*b^3 - a^7*b^2) - (3*b*((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*
a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 +
16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/a^4))/a^4 - (3*b*((8*tan(c
/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9
- a^6*b^3 - a^7*b^2) + (3*b*((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a
^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 +
16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/a^4))/a^4))*6i)/(a^4*d) - (
atan(((((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3
))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^1
1*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2
)*(a^4 + 3*b^4 - (9*a^2*b^2)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(
(a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4
+ 3*b^4 - (9*a^2*b^2)/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4
- (9*a^2*b^2)/2)*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 7
2*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (((8*(8*a^13
*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^
10*b^2) - (8*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2)*(8*a^13*b - 8*a^8*b
^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 +
3*a^6*b^4 - 3*a^8*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2))/(a^10 - a^4*b^6 + 3*a^6*
b^4 - 3*a^8*b^2))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2)*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 -
3*a^8*b^2))/((16*(54*a*b^7 - 12*a^7*b - 108*b^8 + 270*a^2*b^6 - 117*a^3*b^5 - 198*a^4*b^4 + 72*a^5*b^3 + 36*a
^6*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144*a^2*b
^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*(8*a^13*b + 4*a^14 - 12*
a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*tan(
c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 +
16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^
8*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))
*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (((8*t
an(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b +
a^9 - a^6*b^3 - a^7*b^2) - (((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30*a^10*b^4 - 14*a^11*b^3 - 22*a
^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b
^4 - (9*a^2*b^2)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9
- a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (
9*a^2*b^2)/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^
2)/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(a^4 + 3*b^4 - (9*a^2*b^2)/2)*2
i)/(d*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))