Integrand size = 17, antiderivative size = 60 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=-\frac {5 \text {arctanh}(\cos (a+b x))}{2 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b}+\frac {2 \sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b} \] Output:
-5/2*arctanh(cos(b*x+a))/b-1/2*cot(b*x+a)*csc(b*x+a)/b+2*sec(b*x+a)/b+1/3* sec(b*x+a)^3/b
Leaf count is larger than twice the leaf count of optimal. \(205\) vs. \(2(60)=120\).
Time = 0.61 (sec) , antiderivative size = 205, normalized size of antiderivative = 3.42 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\frac {2 \csc ^8(a+b x) \left (22-40 \cos (2 (a+b x))+13 \cos (3 (a+b x))-30 \cos (4 (a+b x))+13 \cos (5 (a+b x))+15 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+15 \cos (5 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-15 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )-15 \cos (5 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-26-30 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+30 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{3 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )^3} \] Input:
Integrate[Csc[a + b*x]^3*Sec[a + b*x]^4,x]
Output:
(2*Csc[a + b*x]^8*(22 - 40*Cos[2*(a + b*x)] + 13*Cos[3*(a + b*x)] - 30*Cos [4*(a + b*x)] + 13*Cos[5*(a + b*x)] + 15*Cos[3*(a + b*x)]*Log[Cos[(a + b*x )/2]] + 15*Cos[5*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 15*Cos[3*(a + b*x)]*Lo g[Sin[(a + b*x)/2]] - 15*Cos[5*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-26 - 30*Log[Cos[(a + b*x)/2]] + 30*Log[Sin[(a + b*x)/2]])))/(3*b*(C sc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)
Time = 0.40 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05, 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, 254, 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.
\(\displaystyle \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^3 \sec (a+b x)^4dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sec ^4(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (a+b x))-\frac {1}{3} \sec ^3(a+b x)-\sec (a+b x)\right )}{b}\) |
Input:
Int[Csc[a + b*x]^3*Sec[a + b*x]^4,x]
Output:
(Sec[a + b*x]^5/(2*(1 - Sec[a + b*x]^2)) - (5*(ArcTanh[Sec[a + b*x]] - Sec [a + b*x] - Sec[a + b*x]^3/3))/2)/b
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[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 2.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {\frac {1}{3 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{3}}-\frac {5}{6 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {5}{2 \cos \left (b x +a \right )}+\frac {5 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(70\) |
default | \(\frac {\frac {1}{3 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{3}}-\frac {5}{6 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {5}{2 \cos \left (b x +a \right )}+\frac {5 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(70\) |
norman | \(\frac {\frac {1}{8 b}+\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}}{8 b}+\frac {75 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{8 b}-\frac {65 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{12 b}-\frac {55 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{8 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{3}}+\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}\) | \(114\) |
risch | \(\frac {15 \,{\mathrm e}^{9 i \left (b x +a \right )}+20 \,{\mathrm e}^{7 i \left (b x +a \right )}-22 \,{\mathrm e}^{5 i \left (b x +a \right )}+20 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}\) | \(123\) |
parallelrisch | \(\frac {60 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3}+3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}-165 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+3 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+225 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-130}{24 b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3}}\) | \(124\) |
Input:
int(csc(b*x+a)^3*sec(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*(1/3/sin(b*x+a)^2/cos(b*x+a)^3-5/6/sin(b*x+a)^2/cos(b*x+a)+5/2/cos(b*x +a)+5/2*ln(csc(b*x+a)-cot(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (54) = 108\).
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.87 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\frac {30 \, \cos \left (b x + a\right )^{4} - 20 \, \cos \left (b x + a\right )^{2} - 15 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{12 \, {\left (b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^4,x, algorithm="fricas")
Output:
1/12*(30*cos(b*x + a)^4 - 20*cos(b*x + a)^2 - 15*(cos(b*x + a)^5 - cos(b*x + a)^3)*log(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*x + a)^5 - cos(b*x + a)^3 )*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^5 - b*cos(b*x + a)^3)
\[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \sec ^{4}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**3*sec(b*x+a)**4,x)
Output:
Integral(csc(a + b*x)**3*sec(a + b*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {2 \, {\left (15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{12 \, b} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^4,x, algorithm="maxima")
Output:
1/12*(2*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 - 2)/(cos(b*x + a)^5 - cos( b*x + a)^3) - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x + a) - 1))/b
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=-\frac {5 \, \log \left ({\left | \cos \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} + \frac {5 \, \log \left ({\left | \cos \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} + \frac {\cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} b} + \frac {6 \, \cos \left (b x + a\right )^{2} + 1}{3 \, b \cos \left (b x + a\right )^{3}} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^4,x, algorithm="giac")
Output:
-5/4*log(abs(cos(b*x + a) + 1))/b + 5/4*log(abs(cos(b*x + a) - 1))/b + 1/2 *cos(b*x + a)/((cos(b*x + a)^2 - 1)*b) + 1/3*(6*cos(b*x + a)^2 + 1)/(b*cos (b*x + a)^3)
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\frac {-\frac {5\,{\cos \left (a+b\,x\right )}^4}{2}+\frac {5\,{\cos \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\cos \left (a+b\,x\right )}^3-{\cos \left (a+b\,x\right )}^5\right )}-\frac {5\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{2\,b} \] Input:
int(1/(cos(a + b*x)^4*sin(a + b*x)^3),x)
Output:
((5*cos(a + b*x)^2)/3 - (5*cos(a + b*x)^4)/2 + 1/3)/(b*(cos(a + b*x)^3 - c os(a + b*x)^5)) - (5*atanh(cos(a + b*x)))/(2*b)
Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.32 \[ \int \csc ^3(a+b x) \sec ^4(a+b x) \, dx=\frac {60 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}-60 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}-65 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}+65 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}+60 \sin \left (b x +a \right )^{4}-80 \sin \left (b x +a \right )^{2}+12}{24 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(csc(b*x+a)^3*sec(b*x+a)^4,x)
Output:
(60*cos(a + b*x)*log(tan((a + b*x)/2))*sin(a + b*x)**4 - 60*cos(a + b*x)*l og(tan((a + b*x)/2))*sin(a + b*x)**2 - 65*cos(a + b*x)*sin(a + b*x)**4 + 6 5*cos(a + b*x)*sin(a + b*x)**2 + 60*sin(a + b*x)**4 - 80*sin(a + b*x)**2 + 12)/(24*cos(a + b*x)*sin(a + b*x)**2*b*(sin(a + b*x)**2 - 1))