Integrand size = 17, antiderivative size = 45 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{b}+\frac {\sec (a+b x) \tan (a+b x)}{2 b} \] Output:
3/2*arctanh(sin(b*x+a))/b-csc(b*x+a)/b+1/2*sec(b*x+a)*tan(b*x+a)/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.60 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\sin ^2(a+b x)\right )}{b} \] Input:
Integrate[Csc[a + b*x]^2*Sec[a + b*x]^3,x]
Output:
-((Csc[a + b*x]*Hypergeometric2F1[-1/2, 2, 1/2, Sin[a + b*x]^2])/b)
Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 3101, 252, 262, 219}
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.
\(\displaystyle \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^2 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^4(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\csc ^2(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\csc ^2(a+b x)}d\csc (a+b x)-\csc (a+b x)\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\csc (a+b x))-\csc (a+b x))}{b}\) |
Input:
Int[Csc[a + b*x]^2*Sec[a + b*x]^3,x]
Output:
-(((-3*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x]))/2 + Csc[a + b*x]^3/(2*(1 - Csc[a + b*x]^2)))/b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 2.00 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {\frac {1}{2 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {3}{2 \sin \left (b x +a \right )}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(50\) |
default | \(\frac {\frac {1}{2 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {3}{2 \sin \left (b x +a \right )}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(50\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{5 i \left (b x +a \right )}+2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}\) | \(104\) |
parallelrisch | \(\frac {\left (-3 \cos \left (2 b x +2 a \right )-3\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (3 \cos \left (2 b x +2 a \right )+3\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (-6 \cos \left (b x +a \right )+6\right ) \cot \left (\frac {b x}{2}+\frac {a}{2}\right )-2 \sec \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b \left (1+\cos \left (2 b x +2 a \right )\right )}\) | \(112\) |
norman | \(\frac {-\frac {1}{2 b}+\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{2 b}+\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{2 b}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{2 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{2}}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{2 b}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{2 b}\) | \(117\) |
Input:
int(csc(b*x+a)^2*sec(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(1/2/sin(b*x+a)/cos(b*x+a)^2-3/2/sin(b*x+a)+3/2*ln(sec(b*x+a)+tan(b*x+ a)))
Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (41) = 82\).
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.89 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 2}{4 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="fricas")
Output:
1/4*(3*cos(b*x + a)^2*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*cos(b*x + a)^ 2*log(-sin(b*x + a) + 1)*sin(b*x + a) - 6*cos(b*x + a)^2 + 2)/(b*cos(b*x + a)^2*sin(b*x + a))
\[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int \csc ^{2}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**2*sec(b*x+a)**3,x)
Output:
Integral(csc(a + b*x)**2*sec(a + b*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.36 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} - 2\right )}}{\sin \left (b x + a\right )^{3} - \sin \left (b x + a\right )} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{4 \, b} \] Input:
integrate(csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="maxima")
Output:
-1/4*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*log(sin (b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.49 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} - \frac {3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} - \frac {3 \, \sin \left (b x + a\right )^{2} - 2}{2 \, {\left (\sin \left (b x + a\right )^{3} - \sin \left (b x + a\right )\right )} b} \] Input:
integrate(csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="giac")
Output:
3/4*log(abs(sin(b*x + a) + 1))/b - 3/4*log(abs(sin(b*x + a) - 1))/b - 1/2* (3*sin(b*x + a)^2 - 2)/((sin(b*x + a)^3 - sin(b*x + a))*b)
Time = 25.77 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.07 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {3\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2\,b}+\frac {\frac {3\,{\sin \left (a+b\,x\right )}^2}{2}-1}{b\,\left (\sin \left (a+b\,x\right )-{\sin \left (a+b\,x\right )}^3\right )} \] Input:
int(1/(cos(a + b*x)^3*sin(a + b*x)^2),x)
Output:
(3*atanh(sin(a + b*x)))/(2*b) + ((3*sin(a + b*x)^2)/2 - 1)/(b*(sin(a + b*x ) - sin(a + b*x)^3))
Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.69 \[ \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )-3 \sin \left (b x +a \right )^{2}+2}{2 \sin \left (b x +a \right ) b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(csc(b*x+a)^2*sec(b*x+a)^3,x)
Output:
( - 3*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**3 + 3*log(tan((a + b*x)/2) - 1)*sin(a + b*x) + 3*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**3 - 3*log(tan ((a + b*x)/2) + 1)*sin(a + b*x) - 3*sin(a + b*x)**2 + 2)/(2*sin(a + b*x)*b *(sin(a + b*x)**2 - 1))