Integrand size = 17, antiderivative size = 44 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {3 \text {arctanh}(\cos (a+b x))}{2 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b}+\frac {\sec (a+b x)}{b} \] Output:
-3/2*arctanh(cos(b*x+a))/b-1/2*cot(b*x+a)*csc(b*x+a)/b+sec(b*x+a)/b
Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(44)=88\).
Time = 0.44 (sec) , antiderivative size = 143, normalized size of antiderivative = 3.25 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {\csc ^4(a+b x) \left (2-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-2-3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{2 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )} \] Input:
Integrate[Csc[a + b*x]^3*Sec[a + b*x]^2,x]
Output:
(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Cos[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))/( 2*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))
Time = 0.36 (sec) , antiderivative size = 51, 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, 3102, 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.
\(\displaystyle \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^3 \sec (a+b x)^2dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int \frac {\sec ^4(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\sec ^2(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sec ^2(a+b x)}d\sec (a+b x)-\sec (a+b x)\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\sec (a+b x))-\sec (a+b x))}{b}\) |
Input:
Int[Csc[a + b*x]^3*Sec[a + b*x]^2,x]
Output:
((-3*(ArcTanh[Sec[a + b*x]] - Sec[a + b*x]))/2 + Sec[a + b*x]^3/(2*(1 - Se c[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_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 1.83 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {3}{2 \cos \left (b x +a \right )}+\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(52\) |
default | \(\frac {-\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {3}{2 \cos \left (b x +a \right )}+\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(52\) |
parallelrisch | \(\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-18}{8 b \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-8 b}\) | \(81\) |
norman | \(\frac {\frac {1}{8 b}+\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{8 b}-\frac {9 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{4 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}\) | \(82\) |
risch | \(\frac {3 \,{\mathrm e}^{5 i \left (b x +a \right )}-2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}\) | \(100\) |
Input:
int(csc(b*x+a)^3*sec(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/2/sin(b*x+a)^2/cos(b*x+a)+3/2/cos(b*x+a)+3/2*ln(csc(b*x+a)-cot(b*x +a)))
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (40) = 80\).
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.18 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {6 \, \cos \left (b x + a\right )^{2} - 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{4 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="fricas")
Output:
1/4*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - cos(b*x + a))*log(-1/2*cos(b*x + a) + 1/ 2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))
\[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**3*sec(b*x+a)**2,x)
Output:
Integral(csc(a + b*x)**3*sec(a + b*x)**2, x)
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{4 \, b} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="maxima")
Output:
1/4*(2*(3*cos(b*x + a)^2 - 2)/(cos(b*x + a)^3 - cos(b*x + a)) - 3*log(cos( b*x + a) + 1) + 3*log(cos(b*x + a) - 1))/b
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.52 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {3 \, \log \left ({\left | \cos \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} + \frac {3 \, \log \left ({\left | \cos \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} + \frac {3 \, \cos \left (b x + a\right )^{2} - 2}{2 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} b} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="giac")
Output:
-3/4*log(abs(cos(b*x + a) + 1))/b + 3/4*log(abs(cos(b*x + a) - 1))/b + 1/2 *(3*cos(b*x + a)^2 - 2)/((cos(b*x + a)^3 - cos(b*x + a))*b)
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {3\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{2\,b}-\frac {\frac {3\,{\cos \left (a+b\,x\right )}^2}{2}-1}{b\,\left (\cos \left (a+b\,x\right )-{\cos \left (a+b\,x\right )}^3\right )} \] Input:
int(1/(cos(a + b*x)^2*sin(a + b*x)^3),x)
Output:
- (3*atanh(cos(a + b*x)))/(2*b) - ((3*cos(a + b*x)^2)/2 - 1)/(b*(cos(a + b *x) - cos(a + b*x)^3))
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.70 \[ \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {12 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}-9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}+12 \sin \left (b x +a \right )^{2}-4}{8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b} \] Input:
int(csc(b*x+a)^3*sec(b*x+a)^2,x)
Output:
(12*cos(a + b*x)*log(tan((a + b*x)/2))*sin(a + b*x)**2 - 9*cos(a + b*x)*si n(a + b*x)**2 + 12*sin(a + b*x)**2 - 4)/(8*cos(a + b*x)*sin(a + b*x)**2*b)