3.265 \(\int \frac{\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=232 \[ -\frac{b^5}{2 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac{b^4 \left (5 a^2-b^2\right )}{a^2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac{\csc ^2(c+d x) \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{(a+4 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{(a-4 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

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

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Rubi [A]  time = 0.761775, antiderivative size = 232, 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} \[ -\frac{b^5}{2 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac{b^4 \left (5 a^2-b^2\right )}{a^2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac{\csc ^2(c+d x) \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{(a+4 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{(a-4 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b^5/(2*a^2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) + (b^4*(5*a^2 - b^2))/(a^2*(a^2 - b^2)^3*d*(b + a*Cos[c +
d*x])) + ((b*(3*a^2 + b^2) - a*(a^2 + 3*b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) - ((a + 4*b)*Lo
g[1 - Cos[c + d*x]])/(4*(a + b)^4*d) + ((a - 4*b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) + (2*b^3*(5*a^2 + b^2
)*Log[b + a*Cos[c + d*x]])/((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 ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac{\cos ^2(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^5}{a^5 (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^5}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{a^2 d}\\ &=\frac{\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 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^3 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3}-\frac{a^4 b^2 \left (3 a^4+3 a^2 b^2-2 b^4\right ) x}{\left (a^2-b^2\right )^3}-\frac{a^6 b \left (3 a^2-7 b^2\right ) x^2}{\left (a^2-b^2\right )^3}-\frac{a^2 \left (a^6-9 a^4 b^2+6 a^2 b^4-2 b^6\right ) x^3}{\left (a^2-b^2\right )^3}}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 a^4 d}\\ &=\frac{\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 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^4 (a+4 b)}{2 (a+b)^4 (a-x)}-\frac{a^4 (a-4 b)}{2 (a-b)^4 (a+x)}-\frac{2 a^2 b^5}{\left (a^2-b^2\right )^2 (b+x)^3}+\frac{2 \left (5 a^4 b^4-a^2 b^6\right )}{\left (a^2-b^2\right )^3 (b+x)^2}-\frac{4 a^4 b^3 \left (5 a^2+b^2\right )}{\left (a^2-b^2\right )^4 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 a^4 d}\\ &=-\frac{b^5}{2 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{b^4 \left (5 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{(a+4 b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac{(a-4 b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end{align*}

Mathematica [A]  time = 6.17891, size = 204, normalized size = 0.88 \[ \frac{-\frac{4 b^5}{a^2 (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)^2}+\frac{8 b^4 \left (b^2-5 a^2\right )}{a^2 (b-a)^3 (a+b)^3 (a \cos (c+d x)+b)}+\frac{16 b^3 \left (5 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{(a+b)^3}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{(a-b)^3}-\frac{4 (a+4 b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^4}+\frac{4 (a-4 b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^4}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b^5)/(a^2*(a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2) + (8*b^4*(-5*a^2 + b^2))/(a^2*(-a + b)^3*(a + b)^3*
(b + a*Cos[c + d*x])) - Csc[(c + d*x)/2]^2/(a + b)^3 + (4*(a - 4*b)*Log[Cos[(c + d*x)/2]])/(a - b)^4 + (16*b^3
*(5*a^2 + b^2)*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^4 - (4*(a + 4*b)*Log[Sin[(c + d*x)/2]])/(a + b)^4 + Sec[(c
+ d*x)/2]^2/(a - b)^3)/(8*d)

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Maple [A]  time = 0.174, size = 295, 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 ) }{4\, \left ( a-b \right ) ^{4}d}}-{\frac{b\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{ \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}{4\,d \left ( a+b \right ) ^{4}}}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{d \left ( a+b \right ) ^{4}}}-{\frac{{b}^{5}}{2\,{a}^{2}d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}+10\,{\frac{{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+2\,{\frac{{b}^{5}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+5\,{\frac{{b}^{4}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-{\frac{{b}^{6}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}{a}^{2} \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)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

1/4/d/(a-b)^3/(cos(d*x+c)+1)+1/4*a*ln(cos(d*x+c)+1)/(a-b)^4/d-b*ln(cos(d*x+c)+1)/(a-b)^4/d+1/4/d/(a+b)^3/(-1+c
os(d*x+c))-1/4/d/(a+b)^4*ln(-1+cos(d*x+c))*a-1/d/(a+b)^4*ln(-1+cos(d*x+c))*b-1/2/d*b^5/a^2/(a+b)^2/(a-b)^2/(b+
a*cos(d*x+c))^2+10/d*b^3/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))*a^2+2/d*b^5/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))+5/d
*b^4/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))-1/d*b^6/(a+b)^3/(a-b)^3/a^2/(b+a*cos(d*x+c))

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Maxima [B]  time = 1.2071, size = 795, normalized size = 3.43 \begin{align*} \frac{\frac{16 \,{\left (5 \, a^{2} b^{3} + b^{5}\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^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac{4 \,{\left (a + 4 \, 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^{6} - 2 \, a^{5} b - a^{4} b^{2} + 4 \, a^{3} b^{3} - a^{2} b^{4} - 2 \, a b^{5} + b^{6} - \frac{2 \,{\left (a^{6} - 4 \, a^{5} b + 5 \, a^{4} b^{2} + 35 \, a^{2} b^{4} + 44 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{{\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 95 \, a^{2} b^{4} - 70 \, a b^{5} - 15 \, b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac{{\left (a^{9} + a^{8} b - 4 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} - 4 \, a^{2} b^{7} + a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{2 \,{\left (a^{9} - a^{8} b - 4 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} + 4 \, a^{2} b^{7} + a b^{8} - b^{9}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{{\left (a^{9} - 3 \, a^{8} b + 8 \, a^{6} b^{3} - 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} + 8 \, a^{3} b^{6} - 3 \, a b^{8} + 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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/8*(16*(5*a^2*b^3 + b^5)*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^8 - 4*a^6*b^2 + 6*a^4*b^
4 - 4*a^2*b^6 + b^8) - 4*(a + 4*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^6 - 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 - 2*(a^6 - 4*a^5*b + 5*a^4*b^2 + 35*a^2
*b^4 + 44*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 95*a^2
*b^4 - 70*a*b^5 - 15*b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/((a^9 + a^8*b - 4*a^7*b^2 - 4*a^6*b^3 + 6*a^5*b
^4 + 6*a^4*b^5 - 4*a^3*b^6 - 4*a^2*b^7 + a*b^8 + b^9)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(a^9 - a^8*b - 4
*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 6*a^4*b^5 - 4*a^3*b^6 + 4*a^2*b^7 + a*b^8 - b^9)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + (a^9 - 3*a^8*b + 8*a^6*b^3 - 6*a^5*b^4 - 6*a^4*b^5 + 8*a^3*b^6 - 3*a*b^8 + b^9)*sin(d*x + c)^6/(co
s(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))/d

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Fricas [B]  time = 1.19362, size = 2279, normalized size = 9.82 \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)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(6*a^6*b^3 + 14*a^4*b^5 - 22*a^2*b^7 + 2*b^9 - 2*(a^9 + 2*a^7*b^2 + 7*a^5*b^4 - 12*a^3*b^6 + 2*a*b^8)*cos
(d*x + c)^3 + 2*(a^8*b - 6*a^6*b^3 - 4*a^4*b^5 + 10*a^2*b^7 - b^9)*cos(d*x + c)^2 + 2*(5*a^7*b^2 + 4*a^5*b^4 -
11*a^3*b^6 + 2*a*b^8)*cos(d*x + c) + 8*(5*a^4*b^5 + a^2*b^7 - (5*a^6*b^3 + a^4*b^5)*cos(d*x + c)^4 - 2*(5*a^5
*b^4 + a^3*b^6)*cos(d*x + c)^3 + (5*a^6*b^3 - 4*a^4*b^5 - a^2*b^7)*cos(d*x + c)^2 + 2*(5*a^5*b^4 + a^3*b^6)*co
s(d*x + c))*log(a*cos(d*x + c) + b) + (a^7*b^2 - 10*a^5*b^4 - 20*a^4*b^5 - 15*a^3*b^6 - 4*a^2*b^7 - (a^9 - 10*
a^7*b^2 - 20*a^6*b^3 - 15*a^5*b^4 - 4*a^4*b^5)*cos(d*x + c)^4 - 2*(a^8*b - 10*a^6*b^3 - 20*a^5*b^4 - 15*a^4*b^
5 - 4*a^3*b^6)*cos(d*x + c)^3 + (a^9 - 11*a^7*b^2 - 20*a^6*b^3 - 5*a^5*b^4 + 16*a^4*b^5 + 15*a^3*b^6 + 4*a^2*b
^7)*cos(d*x + c)^2 + 2*(a^8*b - 10*a^6*b^3 - 20*a^5*b^4 - 15*a^4*b^5 - 4*a^3*b^6)*cos(d*x + c))*log(1/2*cos(d*
x + c) + 1/2) - (a^7*b^2 - 10*a^5*b^4 + 20*a^4*b^5 - 15*a^3*b^6 + 4*a^2*b^7 - (a^9 - 10*a^7*b^2 + 20*a^6*b^3 -
15*a^5*b^4 + 4*a^4*b^5)*cos(d*x + c)^4 - 2*(a^8*b - 10*a^6*b^3 + 20*a^5*b^4 - 15*a^4*b^5 + 4*a^3*b^6)*cos(d*x
+ c)^3 + (a^9 - 11*a^7*b^2 + 20*a^6*b^3 - 5*a^5*b^4 - 16*a^4*b^5 + 15*a^3*b^6 - 4*a^2*b^7)*cos(d*x + c)^2 + 2
*(a^8*b - 10*a^6*b^3 + 20*a^5*b^4 - 15*a^4*b^5 + 4*a^3*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^12
- 4*a^10*b^2 + 6*a^8*b^4 - 4*a^6*b^6 + a^4*b^8)*d*cos(d*x + c)^4 + 2*(a^11*b - 4*a^9*b^3 + 6*a^7*b^5 - 4*a^5*
b^7 + a^3*b^9)*d*cos(d*x + c)^3 - (a^12 - 5*a^10*b^2 + 10*a^8*b^4 - 10*a^6*b^6 + 5*a^4*b^8 - a^2*b^10)*d*cos(d
*x + c)^2 - 2*(a^11*b - 4*a^9*b^3 + 6*a^7*b^5 - 4*a^5*b^7 + a^3*b^9)*d*cos(d*x + c) - (a^10*b^2 - 4*a^8*b^4 +
6*a^6*b^6 - 4*a^4*b^8 + a^2*b^10)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )}}{\left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Integral(cos(c + d*x)**2/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)

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Giac [B]  time = 1.37794, size = 913, normalized size = 3.94 \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)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(2*(a + 4*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
) - 16*(5*a^2*b^3 + b^5)*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^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) - (a + b + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c
) + 1) + 8*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)) + 8*(15*a^
4*b^3 + 20*a^3*b^4 - 2*a^2*b^5 - 4*a*b^6 + 3*b^7 + 30*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*a^3*b
^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 26*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 10*a*b^6*(cos(d*
x + c) - 1)/(cos(d*x + c) + 1) - 4*b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 15*a^4*b^3*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2 - 30*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*a^2*b^5*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2 - 6*a*b^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*b^7*(cos(d*x + c) - 1)^2/(cos(d*
x + c) + 1)^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + 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))/d