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

Optimal. Leaf size=248 \[ \frac{b^6}{2 a^3 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{2 b^5 \left (3 a^2-b^2\right )}{a^3 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{b^4 \left (-4 a^2 b^2+15 a^4+b^4\right ) \log (a \cos (c+d x)+b)}{a^3 d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{(2 a+5 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac{(2 a-5 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

b^6/(2*a^3*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (2*b^5*(3*a^2 - b^2))/(a^3*(a^2 - b^2)^3*d*(b + a*Cos[c +
d*x])) - ((a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) - ((2*a + 5*b)
*Log[1 - Cos[c + d*x]])/(4*(a + b)^4*d) - ((2*a - 5*b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (b^4*(15*a^4 -
4*a^2*b^2 + b^4)*Log[b + a*Cos[c + d*x]])/(a^3*(a^2 - b^2)^4*d)

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Rubi [A]  time = 0.901639, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4397, 2837, 12, 1647, 1629} \[ \frac{b^6}{2 a^3 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{2 b^5 \left (3 a^2-b^2\right )}{a^3 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{b^4 \left (-4 a^2 b^2+15 a^4+b^4\right ) \log (a \cos (c+d x)+b)}{a^3 d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{(2 a+5 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac{(2 a-5 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^6/(2*a^3*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (2*b^5*(3*a^2 - b^2))/(a^3*(a^2 - b^2)^3*d*(b + a*Cos[c +
d*x])) - ((a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) - ((2*a + 5*b)
*Log[1 - Cos[c + d*x]])/(4*(a + b)^4*d) - ((2*a - 5*b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (b^4*(15*a^4 -
4*a^2*b^2 + b^4)*Log[b + a*Cos[c + d*x]])/(a^3*(a^2 - b^2)^4*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{x^6}{a^6 (b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^6 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac{a^6 b^3 \left (7 a^2-3 b^2\right ) x}{\left (a^2-b^2\right )^3}+\frac{a^4 b^2 \left (3 a^4-9 a^2 b^2+2 b^4\right ) x^2}{\left (a^2-b^2\right )^3}-\frac{a^6 b \left (3 a^2+b^2\right ) x^3}{\left (a^2-b^2\right )^3}-2 a^2 x^4}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 a^5 d}\\ &=-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{a^5 (2 a+5 b)}{2 (a+b)^4 (a-x)}+\frac{a^5 (2 a-5 b)}{2 (a-b)^4 (a+x)}+\frac{2 a^2 b^6}{(a-b)^2 (a+b)^2 (b+x)^3}-\frac{4 a^2 b^5 \left (3 a^2-b^2\right )}{(a-b)^3 (a+b)^3 (b+x)^2}+\frac{2 a^2 b^4 \left (15 a^4-4 a^2 b^2+b^4\right )}{(a-b)^4 (a+b)^4 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 a^5 d}\\ &=\frac{b^6}{2 a^3 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac{2 b^5 \left (3 a^2-b^2\right )}{a^3 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{(2 a+5 b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}-\frac{(2 a-5 b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac{b^4 \left (15 a^4-4 a^2 b^2+b^4\right ) \log (b+a \cos (c+d x))}{a^3 \left (a^2-b^2\right )^4 d}\\ \end{align*}

Mathematica [C]  time = 6.34652, size = 713, normalized size = 2.88 \[ \frac{b^6 \tan ^3(c+d x) (a \cos (c+d x)+b)}{2 a^3 d (b-a)^2 (a+b)^2 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac{2 b^5 \left (b^2-3 a^2\right ) \tan ^3(c+d x) (a \cos (c+d x)+b)^2}{a^3 d (b-a)^3 (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac{2 i \left (-4 a^3 b^2+a^5-9 a b^4\right ) (c+d x) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{d (a-b)^4 (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac{\left (4 a^2 b^6-15 a^4 b^4-b^8\right ) \tan ^3(c+d x) (a \cos (c+d x)+b)^3 \log (a \cos (c+d x)+b)}{a^3 d \left (b^2-a^2\right )^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac{i (5 b-2 a) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac{i (-2 a-5 b) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac{(5 b-2 a) \tan ^3(c+d x) \log \left (\cos ^2\left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac{(-2 a-5 b) \tan ^3(c+d x) \log \left (\sin ^2\left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac{\tan ^3(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac{\tan ^3(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (b-a)^3 (a \sin (c+d x)+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^6*(b + a*Cos[c + d*x])*Tan[c + d*x]^3)/(2*a^3*(-a + b)^2*(a + b)^2*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) -
(2*b^5*(-3*a^2 + b^2)*(b + a*Cos[c + d*x])^2*Tan[c + d*x]^3)/(a^3*(-a + b)^3*(a + b)^3*d*(a*Sin[c + d*x] + b*
Tan[c + d*x])^3) - ((2*I)*(a^5 - 4*a^3*b^2 - 9*a*b^4)*(c + d*x)*(b + a*Cos[c + d*x])^3*Tan[c + d*x]^3)/((a - b
)^4*(a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((I/2)*(-2*a - 5*b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c +
d*x])^3*Tan[c + d*x]^3)/((a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((I/2)*(-2*a + 5*b)*ArcTan[Tan[c
+ d*x]]*(b + a*Cos[c + d*x])^3*Tan[c + d*x]^3)/((-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((b + a*Co
s[c + d*x])^3*Csc[(c + d*x)/2]^2*Tan[c + d*x]^3)/(8*(a + b)^3*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((-2*a
+ 5*b)*(b + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2]^2]*Tan[c + d*x]^3)/(4*(-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[
c + d*x])^3) + ((-15*a^4*b^4 + 4*a^2*b^6 - b^8)*(b + a*Cos[c + d*x])^3*Log[b + a*Cos[c + d*x]]*Tan[c + d*x]^3)
/(a^3*(-a^2 + b^2)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((-2*a - 5*b)*(b + a*Cos[c + d*x])^3*Log[Sin[(c
+ d*x)/2]^2]*Tan[c + d*x]^3)/(4*(a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((b + a*Cos[c + d*x])^3*Sec
[(c + d*x)/2]^2*Tan[c + d*x]^3)/(8*(-a + b)^3*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3)

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Maple [A]  time = 0.18, size = 333, normalized size = 1.3 \begin{align*} -{\frac{1}{4\,d \left ( a-b \right ) ^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{2\, \left ( a-b \right ) ^{4}d}}+{\frac{5\,b\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{4\, \left ( a-b \right ) ^{4}d}}+{\frac{1}{4\,d \left ( a+b \right ) ^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) a}{2\,d \left ( a+b \right ) ^{4}}}-{\frac{5\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,d \left ( a+b \right ) ^{4}}}+{\frac{{b}^{6}}{2\,{a}^{3}d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}-15\,{\frac{a{b}^{4}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+4\,{\frac{{b}^{6}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}a}}-{\frac{{b}^{8}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}{a}^{3}}}-6\,{\frac{{b}^{5}}{ad \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }}+2\,{\frac{{b}^{7}}{{a}^{3}d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/(a-b)^3/(cos(d*x+c)+1)-1/2*a*ln(cos(d*x+c)+1)/(a-b)^4/d+5/4*b*ln(cos(d*x+c)+1)/(a-b)^4/d+1/4/d/(a+b)^3/
(-1+cos(d*x+c))-1/2/d/(a+b)^4*ln(-1+cos(d*x+c))*a-5/4/d/(a+b)^4*ln(-1+cos(d*x+c))*b+1/2/d*b^6/a^3/(a+b)^2/(a-b
)^2/(b+a*cos(d*x+c))^2-15/d*b^4/(a+b)^4/(a-b)^4*a*ln(b+a*cos(d*x+c))+4/d*b^6/(a+b)^4/(a-b)^4/a*ln(b+a*cos(d*x+
c))-1/d*b^8/(a+b)^4/(a-b)^4/a^3*ln(b+a*cos(d*x+c))-6/d*b^5/a/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))+2/d*b^7/a^3/(a+b
)^3/(a-b)^3/(b+a*cos(d*x+c))

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Maxima [B]  time = 1.89728, size = 923, normalized size = 3.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*(15*a^4*b^4 - 4*a^2*b^6 + b^8)*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^11 - 4*a^9*
b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8) + 4*(2*a + 5*b)*log(sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6
*a^2*b^2 + 4*a*b^3 + b^4) + (a^8 - 2*a^7*b - a^6*b^2 + 4*a^5*b^3 - a^4*b^4 - 2*a^3*b^5 + a^2*b^6 - 2*(a^8 - 4*
a^7*b + 5*a^6*b^2 - 5*a^4*b^4 - 44*a^3*b^5 - 49*a^2*b^6 + 8*a*b^7 + 8*b^8)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ (a^8 - 6*a^7*b + 15*a^6*b^2 - 20*a^5*b^3 + 15*a^4*b^4 - 102*a^3*b^5 + 81*a^2*b^6 + 32*a*b^7 - 16*b^8)*sin(d
*x + c)^4/(cos(d*x + c) + 1)^4)/((a^11 + a^10*b - 4*a^9*b^2 - 4*a^8*b^3 + 6*a^7*b^4 + 6*a^6*b^5 - 4*a^5*b^6 -
4*a^4*b^7 + a^3*b^8 + a^2*b^9)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(a^11 - a^10*b - 4*a^9*b^2 + 4*a^8*b^3
+ 6*a^7*b^4 - 6*a^6*b^5 - 4*a^5*b^6 + 4*a^4*b^7 + a^3*b^8 - a^2*b^9)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (a^
11 - 3*a^10*b + 8*a^8*b^3 - 6*a^7*b^4 - 6*a^6*b^5 + 8*a^5*b^6 - 3*a^3*b^8 + a^2*b^9)*sin(d*x + c)^6/(cos(d*x +
c) + 1)^6) + sin(d*x + c)^2/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)^2) - 8*log(sin(d*x + c)^2/(co
s(d*x + c) + 1)^2 + 1)/a^3)/d

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Fricas [B]  time = 1.80159, size = 2583, normalized size = 10.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^8*b^2 + 4*a^6*b^4 + 16*a^4*b^6 - 28*a^2*b^8 + 6*b^10 - 2*(3*a^9*b - 2*a^7*b^3 + 11*a^5*b^5 - 16*a^3*b
^7 + 4*a*b^9)*cos(d*x + c)^3 + 2*(a^10 - 4*a^8*b^2 + a^6*b^4 - 9*a^4*b^6 + 14*a^2*b^8 - 3*b^10)*cos(d*x + c)^2
+ 2*(2*a^9*b + a^7*b^3 + 8*a^5*b^5 - 15*a^3*b^7 + 4*a*b^9)*cos(d*x + c) + 4*(15*a^4*b^6 - 4*a^2*b^8 + b^10 -
(15*a^6*b^4 - 4*a^4*b^6 + a^2*b^8)*cos(d*x + c)^4 - 2*(15*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)^3 + (15*a^
6*b^4 - 19*a^4*b^6 + 5*a^2*b^8 - b^10)*cos(d*x + c)^2 + 2*(15*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c))*log(a
*cos(d*x + c) + b) + (2*a^8*b^2 + 3*a^7*b^3 - 8*a^6*b^4 - 22*a^5*b^5 - 18*a^4*b^6 - 5*a^3*b^7 - (2*a^10 + 3*a^
9*b - 8*a^8*b^2 - 22*a^7*b^3 - 18*a^6*b^4 - 5*a^5*b^5)*cos(d*x + c)^4 - 2*(2*a^9*b + 3*a^8*b^2 - 8*a^7*b^3 - 2
2*a^6*b^4 - 18*a^5*b^5 - 5*a^4*b^6)*cos(d*x + c)^3 + (2*a^10 + 3*a^9*b - 10*a^8*b^2 - 25*a^7*b^3 - 10*a^6*b^4
+ 17*a^5*b^5 + 18*a^4*b^6 + 5*a^3*b^7)*cos(d*x + c)^2 + 2*(2*a^9*b + 3*a^8*b^2 - 8*a^7*b^3 - 22*a^6*b^4 - 18*a
^5*b^5 - 5*a^4*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (2*a^8*b^2 - 3*a^7*b^3 - 8*a^6*b^4 + 22*a^5*b^
5 - 18*a^4*b^6 + 5*a^3*b^7 - (2*a^10 - 3*a^9*b - 8*a^8*b^2 + 22*a^7*b^3 - 18*a^6*b^4 + 5*a^5*b^5)*cos(d*x + c)
^4 - 2*(2*a^9*b - 3*a^8*b^2 - 8*a^7*b^3 + 22*a^6*b^4 - 18*a^5*b^5 + 5*a^4*b^6)*cos(d*x + c)^3 + (2*a^10 - 3*a^
9*b - 10*a^8*b^2 + 25*a^7*b^3 - 10*a^6*b^4 - 17*a^5*b^5 + 18*a^4*b^6 - 5*a^3*b^7)*cos(d*x + c)^2 + 2*(2*a^9*b
- 3*a^8*b^2 - 8*a^7*b^3 + 22*a^6*b^4 - 18*a^5*b^5 + 5*a^4*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a
^13 - 4*a^11*b^2 + 6*a^9*b^4 - 4*a^7*b^6 + a^5*b^8)*d*cos(d*x + c)^4 + 2*(a^12*b - 4*a^10*b^3 + 6*a^8*b^5 - 4*
a^6*b^7 + a^4*b^9)*d*cos(d*x + c)^3 - (a^13 - 5*a^11*b^2 + 10*a^9*b^4 - 10*a^7*b^6 + 5*a^5*b^8 - a^3*b^10)*d*c
os(d*x + c)^2 - 2*(a^12*b - 4*a^10*b^3 + 6*a^8*b^5 - 4*a^6*b^7 + a^4*b^9)*d*cos(d*x + c) - (a^11*b^2 - 4*a^9*b
^4 + 6*a^7*b^6 - 4*a^5*b^8 + a^3*b^10)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.48901, size = 1145, normalized size = 4.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(2*a + 5*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b
^4) + 8*(15*a^4*b^4 - 4*a^2*b^6 + b^8)*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x +
c) - 1)/(cos(d*x + c) + 1)))/(a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8) - (a + b + 4*a*(cos(d*x + c
) - 1)/(cos(d*x + c) + 1) + 10*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^4 + 4*a^3*b + 6
*a^2*b^2 + 4*a*b^3 + b^4)*(cos(d*x + c) - 1)) - (cos(d*x + c) - 1)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x +
c) + 1)) - 4*(45*a^6*b^4 + 66*a^5*b^5 - 15*a^4*b^6 - 44*a^3*b^7 - a^2*b^8 + 10*a*b^9 + 3*b^10 + 90*a^6*b^4*(c
os(d*x + c) - 1)/(cos(d*x + c) + 1) - 24*a^5*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 118*a^4*b^6*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1) + 28*a^3*b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 34*a^2*b^8*(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - 4*a*b^9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*b^10*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 45*a^6*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 90*a^5*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 + 33*a^4*b^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 24*a^3*b^7*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 - 9*a^2*b^8*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*a*b^9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^
2 + 3*b^10*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*(a
+ b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2) - 8*log(abs(-(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3)/d