3.112 \(\int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac {b \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )}+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )}+\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {a x}{2 \left (a^2+b^2\right )}+\frac {b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3100, 2635, 8, 3098, 3133} \[ \frac {b \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )}+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )}+\frac {b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {a x}{2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3100

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(b*Cos[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

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

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Mathematica [C]  time = 0.23, size = 143, normalized size = 1.20 \[ \frac {a^3 \sin (2 (c+d x))+2 a^3 c+2 a^3 d x+b \left (a^2+b^2\right ) \cos (2 (c+d x))+2 b^3 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x-4 i b^3 \tan ^{-1}(\tan (c+d x))+4 i b^3 c+4 i b^3 d x}{4 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*c + 6*a*b^2*c + (4*I)*b^3*c + 2*a^3*d*x + 6*a*b^2*d*x + (4*I)*b^3*d*x - (4*I)*b^3*ArcTan[Tan[c + d*x]]
+ b*(a^2 + b^2)*Cos[2*(c + d*x)] + 2*b^3*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2] + a^3*Sin[2*(c + d*x)] + a*b
^2*Sin[2*(c + d*x)])/(4*(a^2 + b^2)^2*d)

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fricas [A]  time = 0.59, size = 119, normalized size = 1.00 \[ \frac {b^{3} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + {\left (a^{3} + 3 \, a b^{2}\right )} d x + {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + (a^3 + 3*a*b^2)*d*x + (a^2*
b + b^3)*cos(d*x + c)^2 + (a^3 + a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d)

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giac [A]  time = 0.28, size = 182, normalized size = 1.53 \[ \frac {\frac {2 \, b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} + 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {b^{3} \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right ) + a b^{2} \tan \left (d x + c\right ) + a^{2} b + 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.16, size = 236, normalized size = 1.98 \[ \frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\tan \left (d x +c \right ) a^{3}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {\tan \left (d x +c \right ) a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {b^{3}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )}-\frac {b^{3} \ln \left (\tan ^{2}\left (d x +c \right )+1\right )}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{2 d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

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

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maxima [B]  time = 2.01, size = 284, normalized size = 2.39 \[ \frac {\frac {b^{3} \log \left (-a - \frac {2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {b^{3} \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2} + b^{2} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.16, size = 3572, normalized size = 30.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a*cos(c + d*x) + b*sin(c + d*x)),x)

[Out]

(b^3*log(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^2))/(d*(a^4 + b^4 + 2*a^2*b^2)) - (4*b^3*log(1/(cos
(c + d*x) + 1)))/(d*(4*a^4 + 4*b^4 + 8*a^2*b^2)) - ((a*tan(c/2 + (d*x)/2)^3)/(a^2 + b^2) + (2*b*tan(c/2 + (d*x
)/2)^2)/(a^2 + b^2) - (a*tan(c/2 + (d*x)/2))/(a^2 + b^2))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 +
1)) - (a*atan((tan(c/2 + (d*x)/2)*((((4*b^3*((a*((8*(4*a*b^9 + 4*a^9*b + 28*a^3*b^7 + 48*a^5*b^5 + 28*a^7*b^3)
)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2
))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^2 + 3*b^2))/(2*(a^4 + b^4 + 2*a^2*b^2
)) - (16*a*b^3*(a^2 + 3*b^2)*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4
+ 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2) - (a*(
(8*(a^9 - 12*a*b^8 - 6*a^3*b^6 + 13*a^5*b^4 + 8*a^7*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*b^3*((8*(4*
a*b^9 + 4*a^9*b + 28*a^3*b^7 + 48*a^5*b^5 + 28*a^7*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a*b
^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4
+ 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2))*(a^2 + 3*b^2))/(2*(a^4 + b^4 + 2*a^2*b^2)) + (a^3*(a^2 + 3*b^2)^3
*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((a^4 + b^4 + 2*a^2*b^2)^3*(a^6 + b^6 + 3*a^
2*b^4 + 3*a^4*b^2)))*(a^8 + 16*b^8 - 73*a^2*b^6 - 13*a^4*b^4 + 5*a^6*b^2))/(a^8 + 16*b^8 + 25*a^2*b^6 + 15*a^4
*b^4 + 7*a^6*b^2)^2 - (2*a*b*(a^6 - 28*b^6 + 17*a^2*b^4 + 10*a^4*b^2)*((8*(4*a*b^7 + 6*a^3*b^5 + a^5*b^3))/(a^
6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (4*b^3*((8*(a^9 - 12*a*b^8 - 6*a^3*b^6 + 13*a^5*b^4 + 8*a^7*b^2))/(a^6 + b^
6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*b^3*((8*(4*a*b^9 + 4*a^9*b + 28*a^3*b^7 + 48*a^5*b^5 + 28*a^7*b^3))/(a^6 + b^6
+ 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 +
4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2)))/(4*a^4 + 4*b^4 + 8*a^
2*b^2) + (a*(a^2 + 3*b^2)*((a*((8*(4*a*b^9 + 4*a^9*b + 28*a^3*b^7 + 48*a^5*b^5 + 28*a^7*b^3))/(a^6 + b^6 + 3*a
^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4
+ 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^2 + 3*b^2))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (16*a*b^3*(a^
2 + 3*b^2)*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4
+ b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (8*a^2*b^3*(a^2 + 3*b^
2)^2*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4
+ 2*a^2*b^2)^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(a^8 + 16*b^8 + 25*a^2*b^6 + 15*a^4*b^4 + 7*a^6*b^2)^2)*
(a^10 + b^10 + 5*a^2*b^8 + 10*a^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/(4*a^4 + 12*a^2*b^2) + (((4*b^3*((a*(a^2 + 3*
b^2)*((8*(2*a^10 - 10*a^2*b^8 - 16*a^4*b^6 + 8*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a^1
0*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4
+ 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (16*a*b^3*(a^2 + 3*b^2)*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 7
2*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b
^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2) - (a*(a^2 + 3*b^2)*((8*(a^8*b + a^4*b^5 + 2*a^6*b^3))/(a^6 + b^6 + 3*a^2*b^
4 + 3*a^4*b^2) - (4*b^3*((8*(2*a^10 - 10*a^2*b^8 - 16*a^4*b^6 + 8*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2
) - (32*b^3*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6
+ b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2)))/(2*(a^4 + b^4 + 2*a^2*b^2)) + (a^3*(a^2 + 3*b
^2)^3*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((a^4 + b^4 + 2*a^2*b^2)^3*(a^6 + b^6 +
3*a^2*b^4 + 3*a^4*b^2)))*(a^8 + 16*b^8 - 73*a^2*b^6 - 13*a^4*b^4 + 5*a^6*b^2)*(a^10 + b^10 + 5*a^2*b^8 + 10*a
^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/((4*a^4 + 12*a^2*b^2)*(a^8 + 16*b^8 + 25*a^2*b^6 + 15*a^4*b^4 + 7*a^6*b^2)^2
) - (2*a*b*(a^6 - 28*b^6 + 17*a^2*b^4 + 10*a^4*b^2)*((8*(2*a^2*b^6 + a^4*b^4))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*
b^2) + (4*b^3*((8*(a^8*b + a^4*b^5 + 2*a^6*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*b^3*((8*(2*a^10 - 10
*a^2*b^8 - 16*a^4*b^6 + 8*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a^10*b + 12*a^2*b^9 + 48
*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^
4 + 4*b^4 + 8*a^2*b^2)))/(4*a^4 + 4*b^4 + 8*a^2*b^2) + (a*((a*(a^2 + 3*b^2)*((8*(2*a^10 - 10*a^2*b^8 - 16*a^4*
b^6 + 8*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*b^3*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*
b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^2*b
^2)) - (16*a*b^3*(a^2 + 3*b^2)*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^
4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^2 + 3*b^2))/(2*(a^4 + b^4 + 2*
a^2*b^2)) - (8*a^2*b^3*(a^2 + 3*b^2)^2*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^
4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^10 + b^10 + 5*a^2*b^
8 + 10*a^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/((4*a^4 + 12*a^2*b^2)*(a^8 + 16*b^8 + 25*a^2*b^6 + 15*a^4*b^4 + 7*a^
6*b^2)^2))*(a^2 + 3*b^2))/(d*(a^4 + b^4 + 2*a^2*b^2))

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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/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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