Integrand size = 21, antiderivative size = 92 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {a^2 \log (\cos (c+d x))}{d}-\frac {a (a+b) \log (1-\sec (c+d x))}{2 d}-\frac {a (a-b) \log (1+\sec (c+d x))}{2 d}-\frac {\cot ^2(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{2 d} \] Output:
-a^2*ln(cos(d*x+c))/d-1/2*a*(a+b)*ln(1-sec(d*x+c))/d-1/2*a*(a-b)*ln(1+sec( d*x+c))/d-1/2*cot(d*x+c)^2*(a^2+b^2+2*a*b*sec(d*x+c))/d
Time = 0.51 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {4 a^2 \log (\cos (c+d x))+2 a (a+b) \log (1-\sec (c+d x))+2 a (a-b) \log (1+\sec (c+d x))+\frac {(a+b)^2}{-1+\sec (c+d x)}-\frac {(a-b)^2}{1+\sec (c+d x)}}{4 d} \] Input:
Integrate[Cot[c + d*x]^3*(a + b*Sec[c + d*x])^2,x]
Output:
-1/4*(4*a^2*Log[Cos[c + d*x]] + 2*a*(a + b)*Log[1 - Sec[c + d*x]] + 2*a*(a - b)*Log[1 + Sec[c + d*x]] + (a + b)^2/(-1 + Sec[c + d*x]) - (a - b)^2/(1 + Sec[c + d*x]))/d
Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 25, 4373, 532, 27, 523, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\cot \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle \frac {b^4 \int \frac {\cos (c+d x) (a+b \sec (c+d x))^2}{b \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {b^4 \left (\frac {a^2+2 a b \sec (c+d x)+b^2}{2 b^2 \left (b^2-b^2 \sec ^2(c+d x)\right )}-\frac {\int -\frac {2 a \cos (c+d x) (a+b \sec (c+d x))}{b \left (b^2-b^2 \sec ^2(c+d x)\right )}d(b \sec (c+d x))}{2 b^2}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^4 \left (\frac {a \int \frac {\cos (c+d x) (a+b \sec (c+d x))}{b \left (b^2-b^2 \sec ^2(c+d x)\right )}d(b \sec (c+d x))}{b^2}+\frac {a^2+2 a b \sec (c+d x)+b^2}{2 b^2 \left (b^2-b^2 \sec ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {b^4 \left (\frac {a \int \left (\frac {b-a}{2 b^2 (\sec (c+d x) b+b)}+\frac {a \cos (c+d x)}{b^3}+\frac {a+b}{2 b^2 (b-b \sec (c+d x))}\right )d(b \sec (c+d x))}{b^2}+\frac {a^2+2 a b \sec (c+d x)+b^2}{2 b^2 \left (b^2-b^2 \sec ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^4 \left (\frac {a^2+2 a b \sec (c+d x)+b^2}{2 b^2 \left (b^2-b^2 \sec ^2(c+d x)\right )}+\frac {a \left (\frac {a \log (b \sec (c+d x))}{b^2}-\frac {(a+b) \log (b-b \sec (c+d x))}{2 b^2}-\frac {(a-b) \log (b \sec (c+d x)+b)}{2 b^2}\right )}{b^2}\right )}{d}\) |
Input:
Int[Cot[c + d*x]^3*(a + b*Sec[c + d*x])^2,x]
Output:
(b^4*((a*((a*Log[b*Sec[c + d*x]])/b^2 - ((a + b)*Log[b - b*Sec[c + d*x]])/ (2*b^2) - ((a - b)*Log[b + b*Sec[c + d*x]])/(2*b^2)))/b^2 + (a^2 + b^2 + 2 *a*b*Sec[c + d*x])/(2*b^2*(b^2 - b^2*Sec[c + d*x]^2))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
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[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )-\frac {b^{2}}{2 \sin \left (d x +c \right )^{2}}}{d}\) | \(92\) |
default | \(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )-\frac {b^{2}}{2 \sin \left (d x +c \right )^{2}}}{d}\) | \(92\) |
risch | \(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {2 a b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a b}{d}\) | \(165\) |
Input:
int(cot(d*x+c)^3*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+2*a*b*(-1/2/sin(d*x+c)^2*cos(d *x+c)^3-1/2*cos(d*x+c)-1/2*ln(csc(d*x+c)-cot(d*x+c)))-1/2*b^2/sin(d*x+c)^2 )
Time = 0.14 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 \, a b \cos \left (d x + c\right ) + a^{2} + b^{2} - {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} - a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{2} - a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/2*(2*a*b*cos(d*x + c) + a^2 + b^2 - ((a^2 - a*b)*cos(d*x + c)^2 - a^2 + a*b)*log(1/2*cos(d*x + c) + 1/2) - ((a^2 + a*b)*cos(d*x + c)^2 - a^2 - a*b )*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^2 - d)
\[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cot ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**3*(a+b*sec(d*x+c))**2,x)
Output:
Integral((a + b*sec(c + d*x))**2*cot(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.78 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {{\left (a^{2} - a b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (a^{2} + a b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, a b \cos \left (d x + c\right ) + a^{2} + b^{2}}{\cos \left (d x + c\right )^{2} - 1}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*((a^2 - a*b)*log(cos(d*x + c) + 1) + (a^2 + a*b)*log(cos(d*x + c) - 1 ) - (2*a*b*cos(d*x + c) + a^2 + b^2)/(cos(d*x + c)^2 - 1))/d
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {{\left (a^{2} - a b\right )} \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, d} - \frac {{\left (a^{2} + a b\right )} \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} + \frac {2 \, a b \cos \left (d x + c\right ) + a^{2} + b^{2}}{2 \, d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
-1/2*(a^2 - a*b)*log(abs(cos(d*x + c) + 1))/d - 1/2*(a^2 + a*b)*log(abs(co s(d*x + c) - 1))/d + 1/2*(2*a*b*cos(d*x + c) + a^2 + b^2)/(d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1))
Time = 10.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\left (a-b\right )}^2}{8\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+b\,a\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+\frac {a\,b}{4}+\frac {b^2}{8}\right )}{d} \] Input:
int(cot(c + d*x)^3*(a + b/cos(c + d*x))^2,x)
Output:
(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (tan(c/2 + (d*x)/2)^2*(a - b)^2)/( 8*d) - (log(tan(c/2 + (d*x)/2))*(a*b + a^2))/d - (cot(c/2 + (d*x)/2)^2*((a *b)/4 + a^2/8 + b^2/8))/d
Time = 0.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {-4 \cos \left (d x +c \right ) a b +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a b +\sin \left (d x +c \right )^{2} a^{2}+\sin \left (d x +c \right )^{2} b^{2}-2 a^{2}-2 b^{2}}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+b*sec(d*x+c))^2,x)
Output:
( - 4*cos(c + d*x)*a*b + 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a* *2 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2 - 4*log(tan((c + d*x)/2) )*sin(c + d*x)**2*a*b + sin(c + d*x)**2*a**2 + sin(c + d*x)**2*b**2 - 2*a* *2 - 2*b**2)/(4*sin(c + d*x)**2*d)