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

[Out]

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

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Rubi [A]  time = 0.90, antiderivative size = 248, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 4397

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

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*}

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Mathematica [C]  time = 6.36, size = 713, normalized size = 2.88 \[ \frac {b^6 \tan ^3(c+d x) (a \cos (c+d x)+b)}{2 a^3 d (b-a)^2 (a+b)^2 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {2 i \left (a^5-4 a^3 b^2-9 a b^4\right ) (c+d x) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{d (a-b)^4 (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {2 b^5 \left (b^2-3 a^2\right ) \tan ^3(c+d x) (a \cos (c+d x)+b)^2}{a^3 d (b-a)^3 (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\left (-15 a^4 b^4+4 a^2 b^6-b^8\right ) \tan ^3(c+d x) (a \cos (c+d x)+b)^3 \log (a \cos (c+d x)+b)}{a^3 d \left (b^2-a^2\right )^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (5 b-2 a) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (-2 a-5 b) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(5 b-2 a) \tan ^3(c+d x) \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(-2 a-5 b) \tan ^3(c+d x) \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {\tan ^3(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\tan ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (b-a)^3 (a \sin (c+d x)+b \tan (c+d x))^3} \]

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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fricas [B]  time = 1.08, size = 1180, normalized size = 4.76 \[ \text {result too large to display} \]

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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giac [B]  time = 1.64, size = 848, normalized size = 3.42 \[ \text {result too large to display} \]

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

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maple [A]  time = 0.23, size = 333, normalized size = 1.34 \[ \frac {b^{6}}{2 d \,a^{3} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {15 b^{4} a \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {4 b^{6} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4} a}-\frac {b^{8} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4} a^{3}}-\frac {6 b^{5}}{d a \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 b^{7}}{d \,a^{3} \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {1}{4 d \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) a}{2 d \left (a +b \right )^{4}}-\frac {5 \ln \left (\cos \left (d x +c \right )-1\right ) b}{4 d \left (a +b \right )^{4}}-\frac {1}{4 d \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}-\frac {a \ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{4} d}+\frac {5 b \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4} d} \]

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]

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

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maxima [B]  time = 0.48, size = 684, normalized size = 2.76 \[ -\frac {\frac {8 \, {\left (15 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}} + \frac {4 \, {\left (2 \, a + 5 \, b\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {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} - \frac {2 \, {\left (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}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (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}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac {{\left (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}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (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}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (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}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\sin \left (d x + c\right )^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{3}}}{8 \, d} \]

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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mupad [B]  time = 2.05, size = 527, normalized size = 2.12 \[ \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-a^7+5\,a^6\,b-10\,a^5\,b^2+10\,a^4\,b^3-5\,a^3\,b^4+97\,a^2\,b^5+16\,a\,b^6-16\,b^7\right )}{2\,a^2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^7-5\,a^6\,b+10\,a^5\,b^2-10\,a^4\,b^3+5\,a^3\,b^4-49\,a^2\,b^5+8\,b^7\right )}{a^2\,{\left (a+b\right )}^2\,\left (a-b\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a+5\,b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (15\,a^4-4\,a^2\,b^2+b^4\right )}{a^3\,d\,{\left (a^2-b^2\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)^4*(16*a*b^6 + 5*a^6*b - a^7 - 16*b^7 + 97*a^2*b^5 - 5*a^3*b^4 + 10*a^4*b^3 - 10*a^5*b^2))
/(2*a^2*(a + b)*(2*a*b + a^2 + b^2)) - (3*a*b^2 - 3*a^2*b + a^3 - b^3)/(2*(a + b)) + (tan(c/2 + (d*x)/2)^2*(a^
7 - 5*a^6*b + 8*b^7 - 49*a^2*b^5 + 5*a^3*b^4 - 10*a^4*b^3 + 10*a^5*b^2))/(a^2*(a + b)^2*(a - b)))/(d*(tan(c/2
+ (d*x)/2)^2*(4*a*b^4 - 4*a^4*b + 4*a^5 - 4*b^5 + 8*a^2*b^3 - 8*a^3*b^2) - tan(c/2 + (d*x)/2)^4*(8*a^5 - 24*a^
4*b - 24*a*b^4 + 8*b^5 + 16*a^2*b^3 + 16*a^3*b^2) + tan(c/2 + (d*x)/2)^6*(20*a*b^4 - 20*a^4*b + 4*a^5 - 4*b^5
- 40*a^2*b^3 + 40*a^3*b^2))) - tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^3) + log(tan(c/2 + (d*x)/2)^2 + 1)/(a^3*d) -
(log(tan(c/2 + (d*x)/2))*(2*a + 5*b))/(d*(8*a*b^3 + 8*a^3*b + 2*a^4 + 2*b^4 + 12*a^2*b^2)) - (b^4*log(a + b -
a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(15*a^4 + b^4 - 4*a^2*b^2))/(a^3*d*(a^2 - b^2)^4)

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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*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Timed out

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