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

Optimal. Leaf size=149 \[ \frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}-\frac {x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3032, 3021, 2735, 2659, 205} \[ \frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}-\frac {x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/b^3) + (a*(2*a^2 - 3*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^3*(a + b)^(
3/2)*d) - (a*Sin[c + d*x])/(2*b^2*d*(a + b*Cos[c + d*x])^2) + ((3*a^2 - 2*b^2)*Sin[c + d*x])/(2*b^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 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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1
))/(b^2*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b
*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f
*x] + b*C*d*(m + 1)*(a^2 - b^2)*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] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 1.59, size = 291, normalized size = 1.95 \[ \frac {\frac {\frac {\sin (c+d x) \left (b \left (a^2+2 b^2\right ) \cos (c+d x)+a \left (2 a^2+b^2\right )\right )}{(a+b \cos (c+d x))^2}+\frac {6 a b \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{(a-b)^2 (a+b)^2}-\frac {\frac {a b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-\frac {3 b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac {2 a \left (8 a^4-20 a^2 b^2+15 b^4\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+8 (c+d x)}{b^3}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a
^2 + b^2)^(5/2) + (a*b*(4*a^2 - 3*b^2)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) - (3*b*(4*a^4 -
7*a^2*b^2 + 2*b^4)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/b^3) + ((6*a*b*ArcTanh[((a - b)*T
an[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + ((a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Cos[c + d*x])*Sin[c
+ d*x])/(a + b*Cos[c + d*x])^2)/((a - b)^2*(a + b)^2))/(8*d)

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fricas [B]  time = 0.51, size = 740, normalized size = 4.97 \[ \left [-\frac {4 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 8 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b - 3 \, a^{2} 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 (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 (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 2 \, a^{4} b^{5} + a^{2} b^{7}\right )} d\right )}}, -\frac {2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x - {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\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 (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 2 \, a^{4} b^{5} + a^{2} b^{7}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.66, size = 290, normalized size = 1.95 \[ -\frac {\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {d x + c}{b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.09, size = 475, normalized size = 3.19 \[ \frac {2 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{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 {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \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 )}{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 {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{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 {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \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 {2 a^{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 \,b^{3} \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {3 a \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 b \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^2/b^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)^3-1/d/b/(a*tan(1/2*
d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a+b)*tan(1/2*d*x+1/2*c)^3-2/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+2/d*a^2/b^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)+1/d/b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a-b)*tan(1/2*d*x+1/
2*c)-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)+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))-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/b^3*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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(1-cos(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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.03, size = 3095, normalized size = 20.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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