Integrand size = 19, antiderivative size = 74 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {(a+b)^2 \log (1-\cos (c+d x))}{2 d}-\frac {2 a b \log (\cos (c+d x))}{d}-\frac {(a-b)^2 \log (1+\cos (c+d x))}{2 d}+\frac {b^2 \sec (c+d x)}{d} \]
1/2*(a+b)^2*ln(1-cos(d*x+c))/d-2*a*b*ln(cos(d*x+c))/d-1/2*(a-b)^2*ln(1+cos (d*x+c))/d+b^2*sec(d*x+c)/d
Time = 0.83 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {-(a-b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (\cos (c+d x))+a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 \sec (c+d x)}{d} \]
(-((a - b)^2*Log[Cos[(c + d*x)/2]]) - 2*a*b*Log[Cos[c + d*x]] + a^2*Log[Si n[(c + d*x)/2]] + 2*a*b*Log[Sin[(c + d*x)/2]] + b^2*Log[Sin[(c + d*x)/2]] + b^2*Sec[c + d*x])/d
Time = 0.43 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 4360, 3042, 3316, 27, 526, 25, 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.
\(\displaystyle \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \csc (c+d x) \sec ^2(c+d x) (-a \cos (c+d x)-b)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {a \int \frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(-a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x)}{a^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 526 |
\(\displaystyle \frac {a^3 \left (\frac {\int \frac {\left (2 b a^2+\left (a^2+b^2\right ) \cos (c+d x) a\right ) \sec (c+d x)}{a \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{a^2}+\frac {b^2 \sec (c+d x)}{a^3}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a^3 \left (\frac {b^2 \sec (c+d x)}{a^3}-\frac {\int -\frac {\left (2 b a^2+\left (a^2+b^2\right ) \cos (c+d x) a\right ) \sec (c+d x)}{a \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{a^2}\right )}{d}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {a^3 \left (\frac {b^2 \sec (c+d x)}{a^3}-\frac {\int \left (-\frac {(a-b)^2}{2 a (\cos (c+d x) a+a)}-\frac {2 b \sec (c+d x)}{a}-\frac {(a+b)^2}{2 a (a-a \cos (c+d x))}\right )d(-a \cos (c+d x))}{a^2}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\frac {b^2 \sec (c+d x)}{a^3}-\frac {\frac {(a-b)^2 \log (a \cos (c+d x)+a)}{2 a}+2 b \log (-a \cos (c+d x))-\frac {(a+b)^2 \log (a-a \cos (c+d x))}{2 a}}{a^2}\right )}{d}\) |
(a^3*(-((2*b*Log[-(a*Cos[c + d*x])] - ((a + b)^2*Log[a - a*Cos[c + d*x]])/ (2*a) + ((a - b)^2*Log[a + a*Cos[c + d*x]])/(2*a))/a^2) + (b^2*Sec[c + d*x ])/a^3))/d
3.2.77.3.1 Defintions of rubi rules used
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), x_Symbol] : > With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemainder [(c + d*x)^n, x, x]}, Simp[R*(x^(m + 1)/(a*(m + 1))), x] + Simp[1/a Int[x ^(m + 1)*(ExpandToSum[a*Qx - b*R*x, x]/(a + b*x^2)), x], x]] /; FreeQ[{a, b , c, d}, x] && IGtQ[n, 1] && ILtQ[m, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.51 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {a^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+2 a b \ln \left (\tan \left (d x +c \right )\right )+b^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{d}\) | \(66\) |
default | \(\frac {a^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+2 a b \ln \left (\tan \left (d x +c \right )\right )+b^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{d}\) | \(66\) |
norman | \(-\frac {2 b^{2}}{d \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(88\) |
parallelrisch | \(\frac {-2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\cos \left (d x +c \right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\cos \left (d x +c \right )+1\right ) b^{2}}{d \cos \left (d x +c \right )}\) | \(92\) |
risch | \(\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(167\) |
1/d*(a^2*ln(-cot(d*x+c)+csc(d*x+c))+2*a*b*ln(tan(d*x+c))+b^2*(1/cos(d*x+c) +ln(-cot(d*x+c)+csc(d*x+c))))
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.31 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {4 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, b^{2}}{2 \, d \cos \left (d x + c\right )} \]
-1/2*(4*a*b*cos(d*x + c)*log(-cos(d*x + c)) + (a^2 - 2*a*b + b^2)*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2) - (a^2 + 2*a*b + b^2)*cos(d*x + c)*log(-1 /2*cos(d*x + c) + 1/2) - 2*b^2)/(d*cos(d*x + c))
\[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \csc {\left (c + d x \right )}\, dx \]
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {4 \, a b \log \left (\cos \left (d x + c\right )\right ) + {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, b^{2}}{\cos \left (d x + c\right )}}{2 \, d} \]
-1/2*(4*a*b*log(cos(d*x + c)) + (a^2 - 2*a*b + b^2)*log(cos(d*x + c) + 1) - (a^2 + 2*a*b + b^2)*log(cos(d*x + c) - 1) - 2*b^2/cos(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.68 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {4 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac {4 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1}}{2 \, d} \]
-1/2*(4*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) - (a^2 + 2*a*b + b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 4*(a*b + b^2 + a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos( d*x + c) + 1) + 1))/d
Time = 14.60 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \csc (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2}{2}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2}{2}+\frac {b^2}{\cos \left (c+d\,x\right )}-2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]