Integrand size = 39, antiderivative size = 54 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=\frac {\log (\cos (a+b x))}{b}-\frac {\log (1+\tan (a+b x))}{3 b}+\frac {2 \log \left (1-\tan (a+b x)+\tan ^2(a+b x)\right )}{3 b} \] Output:
ln(cos(b*x+a))/b-1/3*ln(1+tan(b*x+a))/b+2/3*ln(1-tan(b*x+a)+tan(b*x+a)^2)/ b
Time = 11.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=-\frac {\log (\cos (a+b x)+\sin (a+b x))}{3 b}+\frac {2 \log (2-\sin (2 (a+b x)))}{3 b} \] Input:
Integrate[(-Csc[a + b*x]^3 + Sec[a + b*x]^3)/(Csc[a + b*x]^3 + Sec[a + b*x ]^3),x]
Output:
-1/3*Log[Cos[a + b*x] + Sin[a + b*x]]/b + (2*Log[2 - Sin[2*(a + b*x)]])/(3 *b)
Time = 0.68 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {3042, 4889, 25, 7276, 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.
\(\displaystyle \int \frac {\sec ^3(a+b x)-\csc ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (a+b x)^3-\csc (a+b x)^3}{\csc (a+b x)^3+\sec (a+b x)^3}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \frac {\int -\frac {1-\tan ^3(a+b x)}{\left (\tan ^2(a+b x)+1\right ) \left (\tan ^3(a+b x)+1\right )}d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1-\tan ^3(a+b x)}{\left (\tan ^2(a+b x)+1\right ) \left (\tan ^3(a+b x)+1\right )}d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {\int \left (\frac {\tan (a+b x)}{\tan ^2(a+b x)+1}+\frac {1}{3 (\tan (a+b x)+1)}-\frac {2 (2 \tan (a+b x)-1)}{3 \left (\tan ^2(a+b x)-\tan (a+b x)+1\right )}\right )d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} \log \left (\tan ^2(a+b x)+1\right )+\frac {2}{3} \log \left (\tan ^2(a+b x)-\tan (a+b x)+1\right )-\frac {1}{3} \log (\tan (a+b x)+1)}{b}\) |
Input:
Int[(-Csc[a + b*x]^3 + Sec[a + b*x]^3)/(Csc[a + b*x]^3 + Sec[a + b*x]^3),x ]
Output:
(-1/3*Log[1 + Tan[a + b*x]] - Log[1 + Tan[a + b*x]^2]/2 + (2*Log[1 - Tan[a + b*x] + Tan[a + b*x]^2])/3)/b
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.78 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (1+\tan \left (b x +a \right )^{2}\right )}{2}-\frac {\ln \left (1+\tan \left (b x +a \right )\right )}{3}+\frac {2 \ln \left (1-\tan \left (b x +a \right )+\tan \left (b x +a \right )^{2}\right )}{3}}{b}\) | \(51\) |
default | \(\frac {-\frac {\ln \left (1+\tan \left (b x +a \right )^{2}\right )}{2}-\frac {\ln \left (1+\tan \left (b x +a \right )\right )}{3}+\frac {2 \ln \left (1-\tan \left (b x +a \right )+\tan \left (b x +a \right )^{2}\right )}{3}}{b}\) | \(51\) |
risch | \(-i x -\frac {2 i a}{b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+i\right )}{3 b}+\frac {2 \ln \left ({\mathrm e}^{4 i \left (b x +a \right )}-4 i {\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{3 b}\) | \(60\) |
parallelrisch | \(\frac {-3 \ln \left (\sec \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )-\ln \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-1\right )+2 \ln \left (\sec \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}-4 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )}{3 b}\) | \(96\) |
norman | \(-\frac {\ln \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )}{b}-\frac {\ln \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-1\right )}{3 b}+\frac {2 \ln \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}{3 b}\) | \(107\) |
Input:
int((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x,method=_RET URNVERBOSE)
Output:
1/b*(-1/2*ln(1+tan(b*x+a)^2)-1/3*ln(1+tan(b*x+a))+2/3*ln(1-tan(b*x+a)+tan( b*x+a)^2))
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=-\frac {\log \left (2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) - 4 \, \log \left (-\cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{6 \, b} \] Input:
integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x, algo rithm="fricas")
Output:
-1/6*(log(2*cos(b*x + a)*sin(b*x + a) + 1) - 4*log(-cos(b*x + a)*sin(b*x + a) + 1))/b
\[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=- \int \frac {\csc ^{3}{\left (a + b x \right )}}{\csc ^{3}{\left (a + b x \right )} + \sec ^{3}{\left (a + b x \right )}}\, dx - \int \left (- \frac {\sec ^{3}{\left (a + b x \right )}}{\csc ^{3}{\left (a + b x \right )} + \sec ^{3}{\left (a + b x \right )}}\right )\, dx \] Input:
integrate((-csc(b*x+a)**3+sec(b*x+a)**3)/(csc(b*x+a)**3+sec(b*x+a)**3),x)
Output:
-Integral(csc(a + b*x)**3/(csc(a + b*x)**3 + sec(a + b*x)**3), x) - Integr al(-sec(a + b*x)**3/(csc(a + b*x)**3 + sec(a + b*x)**3), x)
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (50) = 100\).
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.85 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=\frac {2 \, \log \left (-\frac {2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac {2 \, \sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (b x + a\right )^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {\sin \left (b x + a\right )^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right ) - \log \left (-\frac {2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right ) - 3 \, \log \left (\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}{3 \, b} \] Input:
integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x, algo rithm="maxima")
Output:
1/3*(2*log(-2*sin(b*x + a)/(cos(b*x + a) + 1) + 2*sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 2*sin(b*x + a)^3/(cos(b*x + a) + 1)^3 + sin(b*x + a)^4/(cos( b*x + a) + 1)^4 + 1) - log(-2*sin(b*x + a)/(cos(b*x + a) + 1) + sin(b*x + a)^2/(cos(b*x + a) + 1)^2 - 1) - 3*log(sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 1))/b
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=\frac {4 \, \log \left (\tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (b x + a\right ) + 1 \right |}\right )}{6 \, b} \] Input:
integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x, algo rithm="giac")
Output:
1/6*(4*log(tan(b*x + a)^2 - tan(b*x + a) + 1) - 3*log(tan(b*x + a)^2 + 1) - 2*log(abs(tan(b*x + a) + 1)))/b
Time = 16.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.96 \[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=\frac {2\,\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )+1\right )}{3\,b}-\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )-1\right )}{3\,b}-\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}{b} \] Input:
int((1/cos(a + b*x)^3 - 1/sin(a + b*x)^3)/(1/cos(a + b*x)^3 + 1/sin(a + b* x)^3),x)
Output:
(2*log(2*tan(a/2 + (b*x)/2)^2 - 2*tan(a/2 + (b*x)/2) + 2*tan(a/2 + (b*x)/2 )^3 + tan(a/2 + (b*x)/2)^4 + 1))/(3*b) - log(tan(a/2 + (b*x)/2)^2 - 2*tan( a/2 + (b*x)/2) - 1)/(3*b) - log(tan(a/2 + (b*x)/2)^2 + 1)/b
\[ \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx=-\left (\int \frac {\csc \left (b x +a \right )^{3}}{\csc \left (b x +a \right )^{3}+\sec \left (b x +a \right )^{3}}d x \right )+\int \frac {\sec \left (b x +a \right )^{3}}{\csc \left (b x +a \right )^{3}+\sec \left (b x +a \right )^{3}}d x \] Input:
int((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x)
Output:
- int(csc(a + b*x)**3/(csc(a + b*x)**3 + sec(a + b*x)**3),x) + int(sec(a + b*x)**3/(csc(a + b*x)**3 + sec(a + b*x)**3),x)