Integrand size = 26, antiderivative size = 231 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {a b^2}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac {2 a b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac {\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {(2 a-b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d} \] Output:
1/2*a*b^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2-2*a*b*(a^2+b^2)/(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-b)*ln(1-cos(d*x+c))/(a+b)^4/d+1/4*(2*a+b)*ln(1+cos( d*x+c))/(a-b)^4/d-a*(a^4+8*a^2*b^2+3*b^4)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d
Result contains complex when optimal does not.
Time = 6.62 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.04 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {a b^2 (b+a \cos (c+d x)) \tan ^3(c+d x)}{2 (-a+b)^2 (a+b)^2 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {2 a b (-i a+b) (i a+b) (b+a \cos (c+d x))^2 \tan ^3(c+d x)}{(-a+b)^3 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {2 i \left (a^5+8 a^3 b^2+3 a b^4\right ) (c+d x) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{(a-b)^4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a-b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a+b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {(b+a \cos (c+d x))^3 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a+b) (b+a \cos (c+d x))^3 \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\left (-a^5-8 a^3 b^2-3 a b^4\right ) (b+a \cos (c+d x))^3 \log (b+a \cos (c+d x)) \tan ^3(c+d x)}{\left (-a^2+b^2\right )^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a-b) (b+a \cos (c+d x))^3 \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(b+a \cos (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (-a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3} \] Input:
Integrate[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
(a*b^2*(b + a*Cos[c + d*x])*Tan[c + d*x]^3)/(2*(-a + b)^2*(a + b)^2*d*(a*S in[c + d*x] + b*Tan[c + d*x])^3) + (2*a*b*((-I)*a + b)*(I*a + b)*(b + a*Co s[c + d*x])^2*Tan[c + d*x]^3)/((-a + b)^3*(a + b)^3*d*(a*Sin[c + d*x] + b* Tan[c + d*x])^3) + ((2*I)*(a^5 + 8*a^3*b^2 + 3*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*Ta n[c + d*x])^3) - ((I/2)*(2*a - 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 + 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*Cos[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 + 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) + ((-a^5 - 8*a^3*b^2 - 3*a*b^4)*(b + a*Cos[c + d*x])^3*Log[b + a*Cos[c + d*x ]]*Tan[c + d*x]^3)/((-a^2 + b^2)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((2*a - 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)
Time = 0.97 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4897, 3042, 3316, 27, 601, 25, 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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a \cos (c+d x)+b)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {a^3 \int \frac {\cos ^2(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {a^2 \cos ^2(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle -\frac {a \left (\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {\int -\frac {-\frac {b \left (3 a^2+b^2\right ) \cos ^3(c+d x) a^5}{\left (a^2-b^2\right )^3}+\frac {\left (2 a^4-3 b^2 a^2-3 b^4\right ) \cos ^2(c+d x) a^4}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (7 a^2-3 b^2\right ) \cos (c+d x) a^3}{\left (a^2-b^2\right )^3}+\frac {b^4 \left (3 a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a \left (\frac {\int \frac {-\frac {b \left (3 a^2+b^2\right ) \cos ^3(c+d x) a^5}{\left (a^2-b^2\right )^3}+\frac {\left (2 a^4-3 b^2 a^2-3 b^4\right ) \cos ^2(c+d x) a^4}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (7 a^2-3 b^2\right ) \cos (c+d x) a^3}{\left (a^2-b^2\right )^3}+\frac {b^4 \left (3 a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}+\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle -\frac {a \left (\frac {\int \left (-\frac {4 b \left (a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3 (b+a \cos (c+d x))^2}+\frac {2 b^2 a^2}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^3}+\frac {(2 a-b) a}{2 (a+b)^4 (a-a \cos (c+d x))}-\frac {(2 a+b) a}{2 (a-b)^4 (\cos (c+d x) a+a)}+\frac {2 \left (a^6+8 b^2 a^4+3 b^4 a^2\right )}{\left (a^2-b^2\right )^4 (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{2 a^2}+\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a \left (\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}+\frac {\frac {4 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {a^2 b^2}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {2 a^2 \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}-\frac {a (2 a-b) \log (a-a \cos (c+d x))}{2 (a+b)^4}-\frac {a (2 a+b) \log (a \cos (c+d x)+a)}{2 (a-b)^4}}{2 a^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((a*((a^2*(a^2 + 3*b^2) - a*b*(3*a^2 + b^2)*Cos[c + d*x])/(2*(a^2 - b^2)^ 3*(a^2 - a^2*Cos[c + d*x]^2)) + (-((a^2*b^2)/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2)) + (4*a^2*b*(a^2 + b^2))/((a^2 - b^2)^3*(b + a*Cos[c + d*x])) - (a*(2*a - b)*Log[a - a*Cos[c + d*x]])/(2*(a + b)^4) - (a*(2*a + b)*Log[a + a*Cos[c + d*x]])/(2*(a - b)^4) + (2*a^2*(a^4 + 8*a^2*b^2 + 3*b^4)*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^4)/(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 = 10.46 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.85
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}-\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} a}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {2 a b \left (a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(196\) |
| default | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}-\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} a}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {2 a b \left (a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(196\) |
| risch | \(\text {Expression too large to display}\) | \(1304\) |
Input:
int(sec(d*x+c)/(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*(2*a-b)/(a+b)^4*ln(-1+cos(d*x+c))-1/4 /(a-b)^3/(1+cos(d*x+c))+1/4*(2*a+b)/(a-b)^4*ln(1+cos(d*x+c))-a*(a^4+8*a^2* b^2+3*b^4)/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))+1/2*b^2/(a+b)^2*a/(a-b)^2/(b +a*cos(d*x+c))^2-2*a*b*(a^2+b^2)/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 1076 vs. \(2 (223) = 446\).
Time = 0.25 (sec) , antiderivative size = 1076, normalized size of antiderivative = 4.66 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/4*(8*a^5*b^2 + 8*a^3*b^4 - 16*a*b^6 - 2*(7*a^6*b - 2*a^4*b^3 - 5*a^2*b^5 )*cos(d*x + c)^3 + 2*(a^7 - 7*a^5*b^2 - a^3*b^4 + 7*a*b^6)*cos(d*x + c)^2 + 2*(6*a^6*b + a^4*b^3 - 8*a^2*b^5 + b^7)*cos(d*x + c) + 4*(a^5*b^2 + 8*a^ 3*b^4 + 3*a*b^6 - (a^7 + 8*a^5*b^2 + 3*a^3*b^4)*cos(d*x + c)^4 - 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^3 + (a^7 + 7*a^5*b^2 - 5*a^3*b^4 - 3 *a*b^6)*cos(d*x + c)^2 + 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c))*l og(a*cos(d*x + c) + b) - (2*a^5*b^2 + 9*a^4*b^3 + 16*a^3*b^4 + 14*a^2*b^5 + 6*a*b^6 + b^7 - (2*a^7 + 9*a^6*b + 16*a^5*b^2 + 14*a^4*b^3 + 6*a^3*b^4 + a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^ 4 + 6*a^2*b^5 + a*b^6)*cos(d*x + c)^3 + (2*a^7 + 9*a^6*b + 14*a^5*b^2 + 5* a^4*b^3 - 10*a^3*b^4 - 13*a^2*b^5 - 6*a*b^6 - b^7)*cos(d*x + c)^2 + 2*(2*a ^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4 + 6*a^2*b^5 + a*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - (2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14 *a^2*b^5 + 6*a*b^6 - b^7 - (2*a^7 - 9*a^6*b + 16*a^5*b^2 - 14*a^4*b^3 + 6* a^3*b^4 - a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)*cos(d*x + c)^3 + (2*a^7 - 9*a^6*b + 14*a^5 *b^2 - 5*a^4*b^3 - 10*a^3*b^4 + 13*a^2*b^5 - 6*a*b^6 + b^7)*cos(d*x + c)^2 + 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)*c os(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*...
\[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
Output:
Integral(sec(c + d*x)/(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. 602 vs. \(2 (223) = 446\).
Time = 0.06 (sec) , antiderivative size = 602, normalized size of antiderivative = 2.61 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {\frac {8 \, {\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\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 (2 \, 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} - 20 \, a^{5} b - 11 \, a^{4} b^{2} - 24 \, a^{3} b^{3} - 29 \, 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 (a^{6} - 38 \, a^{5} b + 31 \, a^{4} b^{2} - 52 \, a^{3} b^{3} + 63 \, 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)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/8*(8*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(a + b - (a - b)*sin(d*x + c)^2/(co s(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*(2* a - 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 - 20*a^5*b - 11*a^4*b^2 - 24*a^3*b^3 - 29*a^2*b^4 + 4*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 38*a^5*b + 31*a^4*b^2 - 52*a^3*b^3 + 63*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 = 0.34 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.56 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {{\left (a^{6} + 8 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{9} d - 4 \, a^{7} b^{2} d + 6 \, a^{5} b^{4} d - 4 \, a^{3} b^{6} d + a b^{8} d} + \frac {{\left (2 \, a - b\right )} \log \left ({\left | -\cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} + \frac {{\left (2 \, a + b\right )} \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} + \frac {4 \, a^{5} b^{2} + 4 \, a^{3} b^{4} - 8 \, a b^{6} - {\left (7 \, a^{6} b - 2 \, a^{4} b^{3} - 5 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} - 7 \, a^{5} b^{2} - a^{3} b^{4} + 7 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, a^{6} b + a^{4} b^{3} - 8 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + b\right )}^{2} {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
-(a^6 + 8*a^4*b^2 + 3*a^2*b^4)*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*(2*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*(2*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*(4*a^5*b^2 + 4*a^3*b^4 - 8*a*b^6 - (7*a^6*b - 2* a^4*b^3 - 5*a^2*b^5)*cos(d*x + c)^3 + (a^7 - 7*a^5*b^2 - a^3*b^4 + 7*a*b^6 )*cos(d*x + c)^2 + (6*a^6*b + a^4*b^3 - 8*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 = 16.33 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.15 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-a^5+37\,a^4\,b+6\,a^3\,b^2+58\,a^2\,b^3-5\,a\,b^4+b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-21\,a^4\,b+10\,a^3\,b^2-34\,a^2\,b^3+5\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a-b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (a^5+8\,a^3\,b^2+3\,a\,b^4\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \] Input:
int(1/(cos(c + d*x)*(a*sin(c + d*x) + b*tan(c + d*x))^3),x)
Output:
((tan(c/2 + (d*x)/2)^4*(37*a^4*b - 5*a*b^4 - a^5 + b^5 + 58*a^2*b^3 + 6*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 - 21*a^4*b + a^5 - b^5 - 34*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))*(2*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)*(3*a*b^4 + a^5 + 8*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 = 0.17 (sec) , antiderivative size = 1654, normalized size of antiderivative = 7.16 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
( - 8*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 - 64*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - t an((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**4*b**3 - 24*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**2*b**5 + 8*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**6*b - 36*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**5*b**2 + 64*cos( c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4*b**3 - 56*cos(c + d*x) *log(tan((c + d*x)/2))*sin(c + d*x)**2*a**3*b**4 + 24*cos(c + d*x)*log(tan ((c + d*x)/2))*sin(c + d*x)**2*a**2*b**5 - 4*cos(c + d*x)*log(tan((c + d*x )/2))*sin(c + d*x)**2*a*b**6 - 14*cos(c + d*x)*sin(c + d*x)**2*a**6*b + 26 *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 + 44*cos(c + d*x)*sin(c + d*x)**2*a**3*b**4 - 26*cos(c + d*x)*si n(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*cos(c + d*x)*a**4*b**3 + 6*cos(c + d*x)*a**2*b**5 - 2*co s(c + d*x)*b**7 + 4*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**4*a**7 + 32*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 + 12*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**4*a**3*b**4 - 4*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**7 - 36*l og(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)*...