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

Optimal. Leaf size=155 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3056, 3001, 3770, 2659, 205} \[ -\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(3*a^2 - 2*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(3/2)*(a + b)^(3/2)*d))
+ ArcTanh[Sin[c + d*x]]/(a^3*d) - Sin[c + d*x]/(2*a*d*(a + b*Cos[c + d*x])^2) - ((a^2 - 2*b^2)*Sin[c + d*x])/(
2*a^2*(a^2 - b^2)*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 (c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}+\frac {\int \frac {\left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (2 \left (a^2-b^2\right )^2-a b \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \sec (c+d x) \, dx}{a^3}-\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (b \left (3 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 \left (a^2-b^2\right ) d}\\ &=-\frac {b \left (3 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 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 180, normalized size = 1.16 \[ \frac {\frac {2 b \left (2 b^2-3 a^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac {a \sin (c+d x) \left (2 a^3+b \left (a^2-2 b^2\right ) \cos (c+d x)-3 a b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))^2}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.81, size = 915, normalized size = 5.90 \[ \left [-\frac {{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\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 (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{7} b^{2} - 2 \, a^{5} b^{4} + a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d\right )}}, -\frac {{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\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 (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} b^{2} - 2 \, a^{5} b^{4} + a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 3.71, size = 311, normalized size = 2.01 \[ -\frac {\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 496, normalized size = 3.20 \[ -\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \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 )^{2} \left (a +b \right )}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d a \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 )^{2} \left (a +b \right )}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{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 )^{2} \left (a +b \right )}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \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 )^{2} \left (a -b \right )}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d a \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 )^{2} \left (a -b \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{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 )^{2} \left (a -b \right )}-\frac {3 b \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 \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{3} \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^{3} \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {\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)/(a+b*cos(d*x+c))^3,x)

[Out]

-2/d/(a*tan(1/2*d*x+1/2*c)^2-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/a/(a*tan(1/2*d*x+1/2
*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3*b+2/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1
/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3*b^2-2/d/(a*tan(1/2*d*x+1/2*c)^2-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/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*c)*b+2/
d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*c)*b^2-3/d/a*b/(a^2-b^2)/((a
-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+2/d/a^3*b^3/(a^2-b^2)/((a-b)*(a+b))^(1/2
)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)+1/d/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)/(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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 6.99, size = 3083, normalized size = 19.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((tan(c/2 + (d*x)/2)*(a*b - 2*a^2 + 2*b^2))/(a^2*b - a^3) + (tan(c/2 + (d*x)/2)^3*(a*b + 2*a^2 - 2*b^2))/(a^
2*(a + b)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2
+ b^2)) - (atan((((((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^
6*b^3 - a^7*b^2) - (8*tan(c/2 + (d*x)/2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*
b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6
- 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2))*1i)/a^3 - ((((8*(6*a^10*b - 4*a^11
+ 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (8*tan(c/2 + (d*x)/2)*(
8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^
2)))/a^3 + (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a
^6*b + a^7 - a^4*b^3 - a^5*b^2))*1i)/a^3)/((((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^
9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (8*tan(c/2 + (d*x)/2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b
^4 - 16*a^9*b^3 - 8*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8
*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2))/a^3 - (16*
(6*a^4*b - 2*a*b^4 + 4*b^5 - 10*a^2*b^3 + 3*a^3*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*(6*a^10*b - 4*a
^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (8*tan(c/2 + (d*x)/
2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^
5*b^2)))/a^3 + (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2)
)/(a^6*b + a^7 - a^4*b^3 - a^5*b^2))/a^3))*2i)/(a^3*d) - (b*atan(((b*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1
/2)*((8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b +
a^7 - a^4*b^3 - a^5*b^2) - (b*((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*
b + a^9 - a^6*b^3 - a^7*b^2) - (4*b*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(8*a^11*b
- 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3
*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 -
3*a^7*b^2)))*1i)/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)) + (b*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)
*((8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^
7 - a^4*b^3 - a^5*b^2) + (b*((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b +
a^9 - a^6*b^3 - a^7*b^2) + (4*b*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(8*a^11*b - 8
*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^
6 + 3*a^5*b^4 - 3*a^7*b^2)))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a
^7*b^2)))*1i)/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))/((16*(6*a^4*b - 2*a*b^4 + 4*b^5 - 10*a^2*b^3 + 3*a^
3*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x
)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b
^2) - (b*((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^
7*b^2) - (4*b*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^
5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a
^7*b^2)))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))))/(2*(a^9
- a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)) - (b*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(
4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) +
(b*((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2)
+ (4*b*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16
*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2
)))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))))/(2*(a^9 - a^3*
b^6 + 3*a^5*b^4 - 3*a^7*b^2))))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*1i)/(d*(a^9 - a^3*b^6 + 3*a^5*b^4
- 3*a^7*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sec {\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-sec(c + d*x)/(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)/(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)

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