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

Optimal. Leaf size=193 \[ \frac {2 a^3 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d}+\frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{8 b^5}+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

[Out]

-1/8*(8*a^4-4*a^2*b^2-b^4)*x/b^5+1/3*a*(3*a^2-b^2)*sin(d*x+c)/b^4/d-1/8*(4*a^2-b^2)*cos(d*x+c)*sin(d*x+c)/b^3/
d+1/3*a*cos(d*x+c)^2*sin(d*x+c)/b^2/d-1/4*cos(d*x+c)^3*sin(d*x+c)/b/d+2*a^3*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2
*c)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^5/d

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Rubi [A]  time = 0.61, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3050, 3049, 3023, 2735, 2659, 205} \[ \frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {2 a^3 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac {x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^5}+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^4 - 4*a^2*b^2 - b^4)*x)/(8*b^5) + (2*a^3*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/
Sqrt[a + b]])/(b^5*d) + (a*(3*a^2 - b^2)*Sin[c + d*x])/(3*b^4*d) - ((4*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(
8*b^3*d) + (a*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(4*b*d)

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 3023

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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(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)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3050

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])))

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos ^2(c+d x) \left (-3 a+b \cos (c+d x)+4 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b}\\ &=\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos (c+d x) \left (8 a^2-a b \cos (c+d x)-3 \left (4 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2}\\ &=-\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {-3 a \left (4 a^2-b^2\right )+b \left (4 a^2+3 b^2\right ) \cos (c+d x)+8 a \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3}\\ &=\frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {-3 a b \left (4 a^2-b^2\right )-3 \left (8 a^4-4 a^2 b^2-b^4\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\left (a^3 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^5}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\left (2 a^3 \left (a^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 )}{b^5 d}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac {2 a^3 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d}+\frac {a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 0.99, size = 168, normalized size = 0.87 \[ \frac {-96 a^4 c-96 a^4 d x-24 a^2 b^2 \sin (2 (c+d x))+24 a b \left (4 a^2-b^2\right ) \sin (c+d x)+48 a^2 b^2 c+48 a^2 b^2 d x+192 a^3 \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )+8 a b^3 \sin (3 (c+d x))-3 b^4 \sin (4 (c+d x))+12 b^4 c+12 b^4 d x}{96 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-96*a^4*c + 48*a^2*b^2*c + 12*b^4*c - 96*a^4*d*x + 48*a^2*b^2*d*x + 12*b^4*d*x + 192*a^3*Sqrt[-a^2 + b^2]*Arc
Tanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + 24*a*b*(4*a^2 - b^2)*Sin[c + d*x] - 24*a^2*b^2*Sin[2*(c +
d*x)] + 8*a*b^3*Sin[3*(c + d*x)] - 3*b^4*Sin[4*(c + d*x)])/(96*b^5*d)

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fricas [A]  time = 0.49, size = 369, normalized size = 1.91 \[ \left [\frac {12 \, \sqrt {-a^{2} + b^{2}} a^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - {\left (6 \, b^{4} \cos \left (d x + c\right )^{3} - 8 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 8 \, a b^{3} + 3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}, \frac {24 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - {\left (6 \, b^{4} \cos \left (d x + c\right )^{3} - 8 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 8 \, a b^{3} + 3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(12*sqrt(-a^2 + b^2)*a^3*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)) - 3*(8*a^4 - 4*
a^2*b^2 - b^4)*d*x - (6*b^4*cos(d*x + c)^3 - 8*a*b^3*cos(d*x + c)^2 - 24*a^3*b + 8*a*b^3 + 3*(4*a^2*b^2 - b^4)
*cos(d*x + c))*sin(d*x + c))/(b^5*d), 1/24*(24*sqrt(a^2 - b^2)*a^3*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^
2)*sin(d*x + c))) - 3*(8*a^4 - 4*a^2*b^2 - b^4)*d*x - (6*b^4*cos(d*x + c)^3 - 8*a*b^3*cos(d*x + c)^2 - 24*a^3*
b + 8*a*b^3 + 3*(4*a^2*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/(b^5*d)]

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giac [B]  time = 0.44, size = 370, normalized size = 1.92 \[ -\frac {\frac {3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} {\left (d x + c\right )}}{b^{5}} + \frac {48 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} - \frac {2 \, {\left (24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 32 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(8*a^4 - 4*a^2*b^2 - b^4)*(d*x + c)/b^5 + 48*(a^5 - a^3*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*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^5)
- 2*(24*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 3*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*a^3*t
an(1/2*d*x + 1/2*c)^5 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 32*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 21*b^3*tan(1/2*d*x
+ 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 32*a*b^2*tan(1/2*d*x + 1/2*c)^
3 + 21*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*a^3*tan(1/2*d*x + 1/2*c) - 12*a^2*b*tan(1/2*d*x + 1/2*c) - 3*b^3*tan(1/
2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d

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maple [B]  time = 0.10, size = 653, normalized size = 3.38 \[ \frac {2 a^{5} \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^{5} \sqrt {\left (a -b \right ) \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} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{d \,b^{5}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^5/b^5/((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-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^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)^7*a^3+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*a^2+1/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1
/2*d*x+1/2*c)^7+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a^3+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4
*tan(1/2*d*x+1/2*c)^5*a^2-8/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a-7/4/d/b/(1+tan(1/2*d*x+1
/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a^3-1/d/b^3/(1+tan(1/2
*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a^2+7/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-8/3/d/b^2/(1
+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a+2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^3-1/d/
b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^2-1/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-2/
d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4+1/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+1/4/d/b*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)^3*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),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 = 2.59, size = 240, normalized size = 1.24 \[ \frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^3\,d}-\frac {\frac {a\,\sin \left (c+d\,x\right )}{4}-\frac {a\,\sin \left (3\,c+3\,d\,x\right )}{12}}{b^2\,d}+\frac {\frac {\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {\sin \left (4\,c+4\,d\,x\right )}{32}}{b\,d}+\frac {a^3\,\sin \left (c+d\,x\right )}{b^4\,d}-\frac {2\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^5\,d}-\frac {2\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (a^2*sin(2*c + 2*d*x))/4)/(b^3*d) - ((a*sin(c + d*x))/4 - (
a*sin(3*c + 3*d*x))/12)/(b^2*d) + (atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/4 - sin(4*c + 4*d*x)/32)/(b*d)
+ (a^3*sin(c + d*x))/(b^4*d) - (2*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^5*d) - (2*a^3*atanh((sin
(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2))/(b^5*d)

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

[Out]

Timed out

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