Integrand size = 19, antiderivative size = 229 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {b^3}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac {b^2 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {(a-2 b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}-\frac {(a+2 b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}+\frac {b \left (3 a^4+8 a^2 b^2+b^4\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d} \] Output:
-1/2*b^3/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2+b^2*(3*a^2+b^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-2*b)*ln(1-cos(d*x+c))/(a+b)^4/d-1/4*(a+2*b)*ln(1+cos(d *x+c))/(a-b)^4/d+b*(3*a^4+8*a^2*b^2+b^4)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d
Result contains complex when optimal does not.
Time = 6.44 (sec) , antiderivative size = 696, normalized size of antiderivative = 3.04 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {b^3 (b+a \cos (c+d x)) \tan ^3(c+d x)}{2 (-a+b)^2 (a+b)^2 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {b^2 \left (3 a^2+b^2\right ) (b+a \cos (c+d x))^2 \tan ^3(c+d x)}{(-a+b)^3 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {2 i \left (3 a^4 b+8 a^2 b^3+b^5\right ) (c+d x) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{(a-b)^4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (-a-2 b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (a-2 b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {(b+a \cos (c+d x))^3 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(-a-2 b) (b+a \cos (c+d x))^3 \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\left (3 a^4 b+8 a^2 b^3+b^5\right ) (b+a \cos (c+d x))^3 \log (b+a \cos (c+d x)) \tan ^3(c+d x)}{\left (-a^2+b^2\right )^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(a-2 b) (b+a \cos (c+d x))^3 \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {(b+a \cos (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (-a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3} \] Input:
Integrate[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-3),x]
Output:
-1/2*(b^3*(b + a*Cos[c + d*x])*Tan[c + d*x]^3)/((-a + b)^2*(a + b)^2*d*(a* Sin[c + d*x] + b*Tan[c + d*x])^3) - (b^2*(3*a^2 + b^2)*(b + a*Cos[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)*(3*a^4*b + 8*a^2*b^3 + b^5)*(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)*(-a - 2*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)*(a - 2*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) + ((-a - 2*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) + ((3*a^4* b + 8*a^2*b^3 + b^5)*(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) + ((a - 2 *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 = 0.82 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4897, 3042, 25, 3200, 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.
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cot ^3(c+d x)}{(a \cos (c+d x)+b)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan \left (c+d x-\frac {\pi }{2}\right )^3}{\left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}{\left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle -\frac {\int \frac {a^3 \cos ^3(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle -\frac {-\frac {\int \frac {-\frac {\left (a^2+3 b^2\right ) \cos ^3(c+d x) a^7}{\left (a^2-b^2\right )^3}+\frac {b \left (3 a^2-7 b^2\right ) \cos ^2(c+d x) a^6}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (a^2+3 b^2\right ) a^4}{\left (a^2-b^2\right )^3}+\frac {b^2 \left (3 a^4+3 b^2 a^2-2 b^4\right ) \cos (c+d x) a^3}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}-\frac {a^2 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle -\frac {-\frac {\int \left (\frac {2 a^2 b^3}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^3}-\frac {2 a^2 \left (3 a^2+b^2\right ) b^2}{\left (a^2-b^2\right )^3 (b+a \cos (c+d x))^2}+\frac {2 a^2 \left (3 a^4+8 b^2 a^2+b^4\right ) b}{\left (a^2-b^2\right )^4 (b+a \cos (c+d x))}-\frac {a^2 (a-2 b)}{2 (a+b)^4 (a-a \cos (c+d x))}-\frac {a^2 (a+2 b)}{2 (a-b)^4 (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))}{2 a^2}-\frac {a^2 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {a^2 \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {\frac {2 a^2 b^2 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {a^2 b^3}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {a^2 (a-2 b) \log (a-a \cos (c+d x))}{2 (a+b)^4}-\frac {a^2 (a+2 b) \log (a \cos (c+d x)+a)}{2 (a-b)^4}+\frac {2 a^2 b \left (3 a^4+8 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}}{2 a^2}}{d}\) |
Input:
Int[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-3),x]
Output:
-((-1/2*(a^2*(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^3)/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2)) + (2*a^2*b^2*(3*a^2 + b^2))/((a^2 - b^2)^3*(b + a*Cos[c + d*x ])) + (a^2*(a - 2*b)*Log[a - a*Cos[c + d*x]])/(2*(a + b)^4) - (a^2*(a + 2* b)*Log[a + a*Cos[c + d*x]])/(2*(a - b)^4) + (2*a^2*b*(3*a^4 + 8*a^2*b^2 + b^4)*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^4)/(2*a^2))/d)
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(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] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 1.93 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (a -2 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}-\frac {b^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{4}+8 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} \left (3 a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(196\) |
| default | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (a -2 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}-\frac {b^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{4}+8 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} \left (3 a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(196\) |
| risch | \(\text {Expression too large to display}\) | \(1314\) |
Input:
int(1/(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-2*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-2*b)*ln(1+cos(d*x+c))-1/2*b^3/(a+b )^2/(a-b)^2/(b+a*cos(d*x+c))^2+b*(3*a^4+8*a^2*b^2+b^4)/(a+b)^4/(a-b)^4*ln( b+a*cos(d*x+c))+b^2*(3*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. 1071 vs. \(2 (221) = 442\).
Time = 0.26 (sec) , antiderivative size = 1071, normalized size of antiderivative = 4.68 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-1/4*(16*a^4*b^3 - 8*a^2*b^5 - 8*b^7 - 2*(a^7 + 8*a^5*b^2 - 7*a^3*b^4 - 2* a*b^6)*cos(d*x + c)^3 + 2*(a^6*b - 11*a^4*b^3 + 7*a^2*b^5 + 3*b^7)*cos(d*x + c)^2 + 2*(11*a^5*b^2 - 10*a^3*b^4 - a*b^6)*cos(d*x + c) + 4*(3*a^4*b^3 + 8*a^2*b^5 + b^7 - (3*a^6*b + 8*a^4*b^3 + a^2*b^5)*cos(d*x + c)^4 - 2*(3* a^5*b^2 + 8*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (3*a^6*b + 5*a^4*b^3 - 7*a^2 *b^5 - b^7)*cos(d*x + c)^2 + 2*(3*a^5*b^2 + 8*a^3*b^4 + a*b^6)*cos(d*x + c ))*log(a*cos(d*x + c) + b) - (a^5*b^2 + 6*a^4*b^3 + 14*a^3*b^4 + 16*a^2*b^ 5 + 9*a*b^6 + 2*b^7 - (a^7 + 6*a^6*b + 14*a^5*b^2 + 16*a^4*b^3 + 9*a^3*b^4 + 2*a^2*b^5)*cos(d*x + c)^4 - 2*(a^6*b + 6*a^5*b^2 + 14*a^4*b^3 + 16*a^3* b^4 + 9*a^2*b^5 + 2*a*b^6)*cos(d*x + c)^3 + (a^7 + 6*a^6*b + 13*a^5*b^2 + 10*a^4*b^3 - 5*a^3*b^4 - 14*a^2*b^5 - 9*a*b^6 - 2*b^7)*cos(d*x + c)^2 + 2* (a^6*b + 6*a^5*b^2 + 14*a^4*b^3 + 16*a^3*b^4 + 9*a^2*b^5 + 2*a*b^6)*cos(d* x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^5*b^2 - 6*a^4*b^3 + 14*a^3*b^4 - 16*a^2*b^5 + 9*a*b^6 - 2*b^7 - (a^7 - 6*a^6*b + 14*a^5*b^2 - 16*a^4*b^3 + 9*a^3*b^4 - 2*a^2*b^5)*cos(d*x + c)^4 - 2*(a^6*b - 6*a^5*b^2 + 14*a^4*b^3 - 16*a^3*b^4 + 9*a^2*b^5 - 2*a*b^6)*cos(d*x + c)^3 + (a^7 - 6*a^6*b + 13*a ^5*b^2 - 10*a^4*b^3 - 5*a^3*b^4 + 14*a^2*b^5 - 9*a*b^6 + 2*b^7)*cos(d*x + c)^2 + 2*(a^6*b - 6*a^5*b^2 + 14*a^4*b^3 - 16*a^3*b^4 + 9*a^2*b^5 - 2*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*...
\[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\int \frac {1}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
Output:
Integral((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. 601 vs. \(2 (221) = 442\).
Time = 0.07 (sec) , antiderivative size = 601, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {\frac {8 \, {\left (3 \, a^{4} b + 8 \, 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 - 2 \, 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 + 29 \, a^{4} b^{2} + 24 \, a^{3} b^{3} + 11 \, a^{2} b^{4} + 20 \, 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 + 63 \, a^{4} b^{2} - 52 \, a^{3} b^{3} + 31 \, a^{2} b^{4} - 38 \, 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(1/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/8*(8*(3*a^4*b + 8*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 - 2*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 + 29*a^4*b^2 + 24*a^3*b^3 + 11*a^2*b^4 + 20*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 6*a^5*b + 63*a^4*b^2 - 52*a^3*b^3 + 31*a^2*b^4 - 38*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
Leaf count of result is larger than twice the leaf count of optimal. 800 vs. \(2 (221) = 442\).
Time = 0.26 (sec) , antiderivative size = 800, normalized size of antiderivative = 3.49 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/8*(2*(a - 2*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*(3*a^4*b + 8*a^2*b^3 + b^5)*log(a bs(-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) + 4*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*(9*a^6*b + 6*a^5*b^2 + 9*a^4*b^3 + 28*a^ 3*b^4 + 11*a^2*b^5 - 2*a*b^6 + 3*b^7 + 18*a^6*b*(cos(d*x + c) - 1)/(cos(d* x + c) + 1) - 12*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 26*a^4*b^ 3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4*a^3*b^4*(cos(d*x + c) - 1)/(co s(d*x + c) + 1) - 38*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*a*b ^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*b^7*(cos(d*x + c) - 1)/(cos(d *x + c) + 1) + 9*a^6*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 18*a^5* b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 33*a^4*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 48*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 27*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...
Time = 17.21 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^5-5\,a^4\,b+58\,a^3\,b^2+6\,a^2\,b^3+37\,a\,b^4-b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-5\,a^4\,b+34\,a^3\,b^2-10\,a^2\,b^3+21\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a-2\,b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}+\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (3\,a^4\,b+8\,a^2\,b^3+b^5\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \] Input:
int(1/(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*(37*a*b^4 - 5*a^4*b + a^5 - b^5 + 6*a^2* b^3 + 58*a^3*b^2))/(2*(a + b)*(2*a*b + a^2 + b^2)) - (tan(c/2 + (d*x)/2)^2 *(21*a*b^4 - 5*a^4*b + a^5 - b^5 - 10*a^2*b^3 + 34*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 - 2*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^4*b + b^5 + 8*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 = 0.17 (sec) , antiderivative size = 1673, normalized size of antiderivative = 7.31 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(1/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
(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**5*b**2 + 64*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 + 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*b**6 + 4*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**6*b - 24 *cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**5*b**2 + 56*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4*b**3 - 64*cos(c + d*x)*lo g(tan((c + d*x)/2))*sin(c + d*x)**2*a**3*b**4 + 36*cos(c + d*x)*log(tan((c + d*x)/2))*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*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 + 20*cos(c + d*x)*sin(c + d*x)**2*a**5*b**2 - 44*cos(c + d*x)*sin(c + d*x)**2*a**4*b**3 + 38*cos(c + d*x)*sin(c + d*x) **2*a**3*b**4 - 26*cos(c + d*x)*sin(c + d*x)**2*a**2*b**5 + 12*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 - 12*log(tan((c + d*x)/ 2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**4*a**6*b - 32*log(t an((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**4*a**4 *b**3 - 4*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 + 12*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 + 44*log(tan((c + d*x)/2)**2*a - tan(...