Integrand size = 28, antiderivative size = 211 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {1}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac {1}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac {4 a^3 b}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {(4 a+b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac {(4 a-b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac {2 a^3 \left (a^2+5 b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d} \] Output:
-1/4/(a+b)^3/d/(1-cos(d*x+c))-1/4/(a-b)^3/d/(1+cos(d*x+c))+1/2*a^3/(a^2-b^ 2)^2/d/(b+a*cos(d*x+c))^2-4*a^3*b/(a^2-b^2)^3/d/(b+a*cos(d*x+c))+1/4*(4*a+ b)*ln(1-cos(d*x+c))/(a+b)^4/d+1/4*(4*a-b)*ln(1+cos(d*x+c))/(a-b)^4/d-2*a^3 *(a^2+5*b^2)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d
Time = 6.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {a^3}{2 (-a+b)^2 (a+b)^2 d (b+a \cos (c+d x))^2}+\frac {4 a^3 b}{(-a+b)^3 (a+b)^3 d (b+a \cos (c+d x))}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 (a+b)^3 d}+\frac {(4 a-b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 (-a+b)^4 d}-\frac {2 \left (a^5+5 a^3 b^2\right ) \log (b+a \cos (c+d x))}{\left (-a^2+b^2\right )^4 d}+\frac {(4 a+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 (a+b)^4 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 (-a+b)^3 d} \] Input:
Integrate[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
a^3/(2*(-a + b)^2*(a + b)^2*d*(b + a*Cos[c + d*x])^2) + (4*a^3*b)/((-a + b )^3*(a + b)^3*d*(b + a*Cos[c + d*x])) - Csc[(c + d*x)/2]^2/(8*(a + b)^3*d) + ((4*a - b)*Log[Cos[(c + d*x)/2]])/(2*(-a + b)^4*d) - (2*(a^5 + 5*a^3*b^ 2)*Log[b + a*Cos[c + d*x]])/((-a^2 + b^2)^4*d) + ((4*a + b)*Log[Sin[(c + d *x)/2]])/(2*(a + b)^4*d) + Sec[(c + d*x)/2]^2/(8*(-a + b)^3*d)
Time = 0.64 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4897, 3042, 3147, 477, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^3}{(a \sin (c+d x)+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\csc ^3(c+d x)}{(a \cos (c+d x)+b)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle -\frac {a^3 \int \frac {1}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle -\frac {\int \left (\frac {2 \left (a^2+5 b^2\right ) a^4}{\left (a^2-b^2\right )^4 (b+a \cos (c+d x))}-\frac {4 b a^4}{\left (a^2-b^2\right )^3 (b+a \cos (c+d x))^2}+\frac {a^4}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^3}+\frac {a^2}{4 (a+b)^3 (a-a \cos (c+d x))^2}-\frac {a^2}{4 (a-b)^3 (\cos (c+d x) a+a)^2}+\frac {(4 a+b) a}{4 (a+b)^4 (a-a \cos (c+d x))}-\frac {(4 a-b) a}{4 (a-b)^4 (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {a^2}{4 (a+b)^3 (a-a \cos (c+d x))}+\frac {a^2}{4 (a-b)^3 (a \cos (c+d x)+a)}+\frac {4 a^4 b}{\left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {a^4}{2 \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {2 a^4 \left (a^2+5 b^2\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}-\frac {a (4 a+b) \log (a-a \cos (c+d x))}{4 (a+b)^4}-\frac {a (4 a-b) \log (a \cos (c+d x)+a)}{4 (a-b)^4}}{a d}\) |
Input:
Int[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((a^2/(4*(a + b)^3*(a - a*Cos[c + d*x])) + a^2/(4*(a - b)^3*(a + a*Cos[c + d*x])) - a^4/(2*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (4*a^4*b)/((a^2 - b^2)^3*(b + a*Cos[c + d*x])) - (a*(4*a + b)*Log[a - a*Cos[c + d*x]])/(4* (a + b)^4) - (a*(4*a - b)*Log[a + a*Cos[c + d*x]])/(4*(a - b)^4) + (2*a^4* (a^2 + 5*b^2)*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^4)/(a*d))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 57.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (4 a -b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (4 a +b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}+\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {4 a^{3} b}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}-\frac {2 a^{3} \left (a^{2}+5 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}}{d}\) | \(184\) |
| default | \(\frac {-\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (4 a -b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (4 a +b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}+\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {4 a^{3} b}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}-\frac {2 a^{3} \left (a^{2}+5 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}}{d}\) | \(184\) |
| risch | \(\text {Expression too large to display}\) | \(1138\) |
Input:
int(sec(d*x+c)^3/(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*(4*a-b)/(a-b)^4*ln(1+cos(d*x+c))+1/4/ (a+b)^3/(-1+cos(d*x+c))+1/4*(4*a+b)/(a+b)^4*ln(-1+cos(d*x+c))+1/2*a^3/(a+b )^2/(a-b)^2/(b+a*cos(d*x+c))^2-4*a^3*b/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))-2* a^3*(a^2+5*b^2)/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 971 vs. \(2 (199) = 398\).
Time = 0.25 (sec) , antiderivative size = 971, normalized size of antiderivative = 4.60 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas" )
Output:
-1/4*(2*a^7 - 22*a^5*b^2 + 14*a^3*b^4 + 6*a*b^6 + 2*(11*a^6*b - 10*a^4*b^3 - a^2*b^5)*cos(d*x + c)^3 - 4*(a^7 - 7*a^5*b^2 + 5*a^3*b^4 + a*b^6)*cos(d *x + c)^2 - 2*(10*a^6*b - 7*a^4*b^3 - 4*a^2*b^5 + b^7)*cos(d*x + c) - 8*(a ^5*b^2 + 5*a^3*b^4 - (a^7 + 5*a^5*b^2)*cos(d*x + c)^4 - 2*(a^6*b + 5*a^4*b ^3)*cos(d*x + c)^3 + (a^7 + 4*a^5*b^2 - 5*a^3*b^4)*cos(d*x + c)^2 + 2*(a^6 *b + 5*a^4*b^3)*cos(d*x + c))*log(a*cos(d*x + c) + b) + (4*a^5*b^2 + 15*a^ 4*b^3 + 20*a^3*b^4 + 10*a^2*b^5 - b^7 - (4*a^7 + 15*a^6*b + 20*a^5*b^2 + 1 0*a^4*b^3 - a^2*b^5)*cos(d*x + c)^4 - 2*(4*a^6*b + 15*a^5*b^2 + 20*a^4*b^3 + 10*a^3*b^4 - a*b^6)*cos(d*x + c)^3 + (4*a^7 + 15*a^6*b + 16*a^5*b^2 - 5 *a^4*b^3 - 20*a^3*b^4 - 11*a^2*b^5 + b^7)*cos(d*x + c)^2 + 2*(4*a^6*b + 15 *a^5*b^2 + 20*a^4*b^3 + 10*a^3*b^4 - a*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (4*a^5*b^2 - 15*a^4*b^3 + 20*a^3*b^4 - 10*a^2*b^5 + b^7 - (4 *a^7 - 15*a^6*b + 20*a^5*b^2 - 10*a^4*b^3 + a^2*b^5)*cos(d*x + c)^4 - 2*(4 *a^6*b - 15*a^5*b^2 + 20*a^4*b^3 - 10*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (4 *a^7 - 15*a^6*b + 16*a^5*b^2 + 5*a^4*b^3 - 20*a^3*b^4 + 11*a^2*b^5 - b^7)* cos(d*x + c)^2 + 2*(4*a^6*b - 15*a^5*b^2 + 20*a^4*b^3 - 10*a^3*b^4 + a*b^6 )*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 4*a^8*b^2 + 6*a^6*b ^4 - 4*a^4*b^6 + a^2*b^8)*d*cos(d*x + c)^4 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5* b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^3 - (a^10 - 5*a^8*b^2 + 10*a^6*b^4 - 10*a^4*b^6 + 5*a^2*b^8 - b^10)*d*cos(d*x + c)^2 - 2*(a^9*b - 4*a^7*b...
\[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
Output:
Integral(sec(c + d*x)**3/(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. 591 vs. \(2 (199) = 398\).
Time = 0.06 (sec) , antiderivative size = 591, normalized size of antiderivative = 2.80 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {\frac {16 \, {\left (a^{5} + 5 \, a^{3} b^{2}\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 (4 \, a + 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} - 44 \, a^{5} b - 35 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 4 \, 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 (15 \, a^{6} + 70 \, a^{5} b - 95 \, a^{4} b^{2} + 20 \, a^{3} b^{3} - 15 \, a^{2} b^{4} + 6 \, a b^{5} - 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(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima" )
Output:
-1/8*(16*(a^5 + 5*a^3*b^2)*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*(4*a + b)*lo g(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 - 44*a^5*b - 35*a^4*b^2 - 5*a^2*b^4 + 4*a*b^5 - b^6)*sin(d*x + c)^2/( cos(d*x + c) + 1)^2 - (15*a^6 + 70*a^5*b - 95*a^4*b^2 + 20*a^3*b^3 - 15*a^ 2*b^4 + 6*a*b^5 - 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 = 0.33 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (a^{6} + 5 \, a^{4} b^{2}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{9} d - 4 \, a^{7} b^{2} d + 6 \, a^{5} b^{4} d - 4 \, a^{3} b^{6} d + a b^{8} d} + \frac {{\left (4 \, a + b\right )} \log \left ({\left | -\cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} + \frac {{\left (4 \, a - b\right )} \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} - \frac {a^{7} - 11 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 3 \, a b^{6} + {\left (11 \, a^{6} b - 10 \, a^{4} b^{3} - a^{2} b^{5}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{7} - 7 \, a^{5} b^{2} + 5 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left (10 \, a^{6} b - 7 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + b\right )}^{2} {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
-2*(a^6 + 5*a^4*b^2)*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*(4*a + 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*(4*a - 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*(a^7 - 11*a^5*b^2 + 7*a^3*b^4 + 3*a*b^6 + (11*a^6*b - 10*a ^4*b^3 - a^2*b^5)*cos(d*x + c)^3 - 2*(a^7 - 7*a^5*b^2 + 5*a^3*b^4 + a*b^6) *cos(d*x + c)^2 - (10*a^6*b - 7*a^4*b^3 - 4*a^2*b^5 + b^7)*cos(d*x + c))/( (a*cos(d*x + c) + b)^2*(a + b)^4*(a - b)^4*d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1))
Time = 16.58 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.32 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (15\,a^5+85\,a^4\,b-10\,a^3\,b^2+10\,a^2\,b^3-5\,a\,b^4+b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-45\,a^4\,b+10\,a^3\,b^2-10\,a^2\,b^3+5\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,a+b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (2\,a^5+10\,a^3\,b^2\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \] Input:
int(1/(cos(c + d*x)^3*(a*sin(c + d*x) + b*tan(c + d*x))^3),x)
Output:
((tan(c/2 + (d*x)/2)^4*(85*a^4*b - 5*a*b^4 + 15*a^5 + b^5 + 10*a^2*b^3 - 1 0*a^3*b^2))/(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*(5*a*b^4 - 45*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))) - tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^3) + (log(tan(c/ 2 + (d*x)/2))*(4*a + 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*a^5 + 10*a^3*b^2))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))
Time = 0.22 (sec) , antiderivative size = 1428, normalized size of antiderivative = 6.77 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
( - 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**6*b - 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**4*b**3 + 16*cos(c + d*x) *log(tan((c + d*x)/2))*sin(c + d*x)**2*a**6*b - 60*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**5*b**2 + 80*cos(c + d*x)*log(tan((c + d*x)/ 2))*sin(c + d*x)**2*a**4*b**3 - 40*cos(c + d*x)*log(tan((c + d*x)/2))*sin( c + d*x)**2*a**3*b**4 + 4*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)* *2*a*b**6 - 26*cos(c + d*x)*sin(c + d*x)**2*a**6*b + 50*cos(c + d*x)*sin(c + d*x)**2*a**5*b**2 - 32*cos(c + d*x)*sin(c + d*x)**2*a**4*b**3 + 20*cos( c + d*x)*sin(c + d*x)**2*a**3*b**4 - 14*cos(c + d*x)*sin(c + d*x)**2*a**2* b**5 + 2*cos(c + d*x)*sin(c + d*x)**2*a*b**6 + 2*cos(c + d*x)*a**6*b - 6*c os(c + d*x)*a**4*b**3 + 6*cos(c + d*x)*a**2*b**5 - 2*cos(c + d*x)*b**7 + 8 *log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)** 4*a**7 + 40*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin (c + d*x)**4*a**5*b**2 - 8*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2 *b - a - b)*sin(c + d*x)**2*a**7 - 48*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**5*b**2 - 40*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 - 8*log(t an((c + d*x)/2))*sin(c + d*x)**4*a**7 + 30*log(tan((c + d*x)/2))*sin(c + d *x)**4*a**6*b - 40*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**5*b**2 + 20...