Integrand size = 28, antiderivative size = 232 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {b^5}{2 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac {b^4 \left (5 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac {(a+4 b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac {(a-4 b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}+\frac {2 b^3 \left (5 a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d} \] Output:
-1/2*b^5/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2+b^4*(5*a^2-b^2)/a^2/(a^2-b^2 )^3/d/(b+a*cos(d*x+c))+1/2*(b*(3*a^2+b^2)-a*(a^2+3*b^2)*cos(d*x+c))*csc(d* x+c)^2/(a^2-b^2)^3/d-1/4*(a+4*b)*ln(1-cos(d*x+c))/(a+b)^4/d+1/4*(a-4*b)*ln (1+cos(d*x+c))/(a-b)^4/d+2*b^3*(5*a^2+b^2)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/ d
Time = 5.19 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {-\frac {4 b^5}{a^2 (a-b)^2 (a+b)^2 (b+a \cos (c+d x))^2}+\frac {8 b^4 \left (-5 a^2+b^2\right )}{a^2 (-a+b)^3 (a+b)^3 (b+a \cos (c+d x))}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b)^3}+\frac {4 (a-4 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^4}+\frac {16 b^3 \left (5 a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4}-\frac {4 (a+4 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^4}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{(a-b)^3}}{8 d} \] Input:
Integrate[Cos[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
((-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)
Time = 1.04 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4897, 3042, 25, 3316, 25, 27, 601, 2160, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2}{(a \sin (c+d x)+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a \cos (c+d x)+b)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^5}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^5}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {a^3 \int -\frac {\cos ^5(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^3 \int \frac {\cos ^5(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a^5 \cos ^5(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle -\frac {-\frac {\int \frac {\frac {b \left (3 a^2-7 b^2\right ) \cos ^2(c+d x) a^8}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (a^2+3 b^2\right ) a^6}{\left (a^2-b^2\right )^3}+\frac {\left (a^6-9 b^2 a^4+6 b^4 a^2-2 b^6\right ) \cos ^3(c+d x) a^5}{\left (a^2-b^2\right )^3}+\frac {b^2 \left (3 a^4+3 b^2 a^2-2 b^4\right ) \cos (c+d x) a^5}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}-\frac {a^4 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}}{a^2 d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle -\frac {-\frac {\int \left (\frac {2 a^2 b^5}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^3}-\frac {2 a^2 \left (5 a^2-b^2\right ) b^4}{\left (a^2-b^2\right )^3 (b+a \cos (c+d x))^2}+\frac {4 a^4 \left (5 a^2+b^2\right ) b^3}{\left (a^2-b^2\right )^4 (b+a \cos (c+d x))}+\frac {a^4 (a+4 b)}{2 (a+b)^4 (a-a \cos (c+d x))}+\frac {a^4 (a-4 b)}{2 (a-b)^4 (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))}{2 a^2}-\frac {a^4 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}}{a^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {a^4 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {-\frac {a^4 (a+4 b) \log (a-a \cos (c+d x))}{2 (a+b)^4}+\frac {a^4 (a-4 b) \log (a \cos (c+d x)+a)}{2 (a-b)^4}-\frac {a^2 b^5}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {2 a^2 b^4 \left (5 a^2-b^2\right )}{\left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac {4 a^4 b^3 \left (5 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}}{2 a^2}}{a^2 d}\) |
Input:
Int[Cos[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((-1/2*(a^4*(b*(3*a^2 + b^2) - a*(a^2 + 3*b^2)*Cos[c + d*x]))/((a^2 - b^2 )^3*(a^2 - a^2*Cos[c + d*x]^2)) - (-((a^2*b^5)/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2)) + (2*a^2*b^4*(5*a^2 - b^2))/((a^2 - b^2)^3*(b + a*Cos[c + d*x ])) - (a^4*(a + 4*b)*Log[a - a*Cos[c + d*x]])/(2*(a + b)^4) + (a^4*(a - 4* b)*Log[a + a*Cos[c + d*x]])/(2*(a - b)^4) + (4*a^4*b^3*(5*a^2 + b^2)*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^4)/(2*a^2))/(a^2*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^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]
Time = 2.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (a -4 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}-\frac {b^{5}}{2 a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {2 b^{3} \left (5 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{4} \left (5 a^{2}-b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(199\) |
| default | \(\frac {\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (a -4 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}-\frac {b^{5}}{2 a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {2 b^{3} \left (5 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{4} \left (5 a^{2}-b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(199\) |
| risch | \(\text {Expression too large to display}\) | \(1300\) |
Input:
int(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/4/(a-b)^3/(1+cos(d*x+c))+1/4*(a-4*b)/(a-b)^4*ln(1+cos(d*x+c))+1/4/( a+b)^3/(-1+cos(d*x+c))+1/4/(a+b)^4*(-a-4*b)*ln(-1+cos(d*x+c))-1/2*b^5/a^2/ (a+b)^2/(a-b)^2/(b+a*cos(d*x+c))^2+2*b^3*(5*a^2+b^2)/(a+b)^4/(a-b)^4*ln(b+ a*cos(d*x+c))+b^4*(5*a^2-b^2)/(a+b)^3/(a-b)^3/a^2/(b+a*cos(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 1045 vs. \(2 (224) = 448\).
Time = 0.26 (sec) , antiderivative size = 1045, normalized size of antiderivative = 4.50 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas" )
Output:
-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 - 1 1*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)*cos(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^1...
\[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(cos(d*x+c)**2/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
Output:
Integral(cos(c + d*x)**2/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 589 vs. \(2 (224) = 448\).
Time = 0.05 (sec) , antiderivative size = 589, normalized size of antiderivative = 2.54 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\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} \] Input:
integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima" )
Output:
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/(c os(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/(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))/d
Time = 0.39 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {2 \, {\left (5 \, a^{3} b^{3} + a b^{5}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{9} d - 4 \, a^{7} b^{2} d + 6 \, a^{5} b^{4} d - 4 \, a^{3} b^{6} d + a b^{8} d} - \frac {{\left (a + 4 \, b\right )} \log \left ({\left | -\cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} + \frac {{\left (a - 4 \, b\right )} \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} - \frac {3 \, a^{6} b^{3} + 7 \, a^{4} b^{5} - 11 \, a^{2} b^{7} + b^{9} - {\left (a^{9} + 2 \, a^{7} b^{2} + 7 \, a^{5} b^{4} - 12 \, a^{3} b^{6} + 2 \, a b^{8}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{8} b - 6 \, a^{6} b^{3} - 4 \, a^{4} b^{5} + 10 \, a^{2} b^{7} - b^{9}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{7} b^{2} + 4 \, a^{5} b^{4} - 11 \, a^{3} b^{6} + 2 \, a b^{8}\right )} \cos \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + b\right )}^{2} {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
2*(5*a^3*b^3 + a*b^5)*log(abs(a*cos(d*x + c) + b))/(a^9*d - 4*a^7*b^2*d + 6*a^5*b^4*d - 4*a^3*b^6*d + a*b^8*d) - 1/4*(a + 4*b)*log(abs(-cos(d*x + c) + 1))/(a^4*d + 4*a^3*b*d + 6*a^2*b^2*d + 4*a*b^3*d + b^4*d) + 1/4*(a - 4* b)*log(abs(-cos(d*x + c) - 1))/(a^4*d - 4*a^3*b*d + 6*a^2*b^2*d - 4*a*b^3* d + b^4*d) - 1/2*(3*a^6*b^3 + 7*a^4*b^5 - 11*a^2*b^7 + b^9 - (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 + (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 + (5*a^7*b^2 + 4*a^5*b^4 - 11*a^3*b^6 + 2*a*b^8)*cos(d*x + c))/((a*cos(d*x + c) + b)^2*(a + b)^4*(a - b)^4*a^2*d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1))
Time = 16.89 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.12 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^5-5\,a^4\,b+10\,a^3\,b^2-10\,a^2\,b^3+85\,a\,b^4+15\,b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-5\,a^4\,b+10\,a^3\,b^2-10\,a^2\,b^3+45\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+4\,b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}+\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (10\,a^2\,b^3+2\,b^5\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \] Input:
int(cos(c + d*x)^2/(a*sin(c + d*x) + b*tan(c + d*x))^3,x)
Output:
tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^3) - ((3*a*b^2 - 3*a^2*b + a^3 - b^3)/(2 *(a + b)) + (tan(c/2 + (d*x)/2)^4*(85*a*b^4 - 5*a^4*b + a^5 + 15*b^5 - 10* a^2*b^3 + 10*a^3*b^2))/(2*(a + b)*(2*a*b + a^2 + b^2)) - (tan(c/2 + (d*x)/ 2)^2*(45*a*b^4 - 5*a^4*b + a^5 - b^5 - 10*a^2*b^3 + 10*a^3*b^2))/((a - b)* (2*a*b + a^2 + b^2)))/(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))) - (log(tan(c/ 2 + (d*x)/2))*(a + 4*b))/(d*(8*a*b^3 + 8*a^3*b + 2*a^4 + 2*b^4 + 12*a^2*b^ 2)) + (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(2*b^5 + 10*a^2*b^3))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))
Time = 0.19 (sec) , antiderivative size = 1451, normalized size of antiderivative = 6.25 \[ \int \frac {\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
(80*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b )*sin(c + d*x)**2*a**3*b**4 + 16*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a*b**6 - 4*cos(c + d*x)*log (tan((c + d*x)/2))*sin(c + d*x)**2*a**6*b + 40*cos(c + d*x)*log(tan((c + d *x)/2))*sin(c + d*x)**2*a**4*b**3 - 80*cos(c + d*x)*log(tan((c + d*x)/2))* sin(c + d*x)**2*a**3*b**4 + 60*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**5 - 16*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2* a*b**6 + 2*cos(c + d*x)*sin(c + d*x)**2*a**7 - 2*cos(c + d*x)*sin(c + d*x) **2*a**6*b + 8*cos(c + d*x)*sin(c + d*x)**2*a**5*b**2 - 20*cos(c + d*x)*si n(c + d*x)**2*a**4*b**3 + 38*cos(c + d*x)*sin(c + d*x)**2*a**3*b**4 - 50*c os(c + d*x)*sin(c + d*x)**2*a**2*b**5 + 24*cos(c + d*x)*sin(c + d*x)**2*a* b**6 - 2*cos(c + d*x)*a**7 + 6*cos(c + d*x)*a**5*b**2 - 6*cos(c + d*x)*a** 3*b**4 + 2*cos(c + d*x)*a*b**6 - 40*log(tan((c + d*x)/2)**2*a - tan((c + d *x)/2)**2*b - a - b)*sin(c + d*x)**4*a**4*b**3 - 8*log(tan((c + d*x)/2)**2 *a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**4*a**2*b**5 + 40*log(tan ((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**4*b **3 + 48*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**2*b**5 + 8*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*b**7 + 2*log(tan((c + d*x)/2))*sin(c + d*x)**4*a* *7 - 20*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**5*b**2 + 40*log(tan((c...