Integrand size = 21, antiderivative size = 87 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {10 a^2 \tan (c+d x)}{3 d}-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2} \]
2*a^2*arctanh(sin(d*x+c))/d+10/3*a^2*tan(d*x+c)/d-2*a^2*tan(d*x+c)/d/(1-co s(d*x+c))-1/3*a^4*tan(d*x+c)/d/(a-a*cos(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(228\) vs. \(2(87)=174\).
Time = 1.97 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.62 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\cot \left (\frac {c}{2}\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )-(-8+7 \cos (c+d x)) \csc \left (\frac {c}{2}\right ) \csc ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )+6 \left (-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{24 d} \]
(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-(Cot[c/2]*Csc[(c + d*x)/2]^ 2) - (-8 + 7*Cos[c + d*x])*Csc[c/2]*Csc[(c + d*x)/2]^3*Sin[(d*x)/2] + 6*(- 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*Log[Cos[(c + d*x)/2] + Sin[ (c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos [(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))) )/(24*d)
Time = 0.88 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4360, 3042, 3348, 3042, 3245, 27, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \csc ^4(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \csc ^4(c+d x) \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3348 |
| \(\displaystyle a^4 \int \frac {\sec ^2(c+d x)}{(a-a \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle a^4 \left (\frac {\int \frac {2 (\cos (c+d x) a+2 a) \sec ^2(c+d x)}{a-a \cos (c+d x)}dx}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a^4 \left (\frac {2 \int \frac {(\cos (c+d x) a+2 a) \sec ^2(c+d x)}{a-a \cos (c+d x)}dx}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) a+2 a}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {\int \left (3 \cos (c+d x) a^2+5 a^2\right ) \sec ^2(c+d x)dx}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {\int \frac {3 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+5 a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {5 a^2 \int \sec ^2(c+d x)dx+3 a^2 \int \sec (c+d x)dx}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^2 \int 1d(-\tan (c+d x))}{d}}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^2 \tan (c+d x)}{d}}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a^4 \left (\frac {2 \left (\frac {\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^2 \tan (c+d x)}{d}}{a^2}-\frac {3 \tan (c+d x)}{d (1-\cos (c+d x))}\right )}{3 a^2}-\frac {\tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\right )\) |
a^4*(-1/3*Tan[c + d*x]/(d*(a - a*Cos[c + d*x])^2) + (2*((-3*Tan[c + d*x])/ (d*(1 - Cos[c + d*x])) + ((3*a^2*ArcTanh[Sin[c + d*x]])/d + (5*a^2*Tan[c + d*x])/d)/a^2))/(3*a^2))
3.1.34.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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 = 1.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13
| method | result | size |
| parallelrisch | \(-\frac {2 \left (\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\frac {7 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\cos \left (d x +c \right )-\frac {5 \cos \left (2 d x +2 c \right )}{14}-\frac {4}{7}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{6}\right ) a^{2}}{d \cos \left (d x +c \right )}\) | \(98\) |
| norman | \(\frac {\frac {a^{2}}{6 d}+\frac {7 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(116\) |
| derivativedivides | \(\frac {a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+2 a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\) | \(117\) |
| default | \(\frac {a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+2 a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\) | \(117\) |
| risch | \(-\frac {4 i a^{2} \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}-9 \,{\mathrm e}^{3 i \left (d x +c \right )}+11 \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(125\) |
-2*(ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)-ln(tan(1/2*d*x+1/2*c)+1)*cos(d*x+c )+7/6*csc(1/2*d*x+1/2*c)^2*(cos(d*x+c)-5/14*cos(2*d*x+2*c)-4/7)*cot(1/2*d* x+1/2*c))*a^2/d/cos(d*x+c)
Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.83 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {10 \, a^{2} \cos \left (d x + c\right )^{3} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, a^{2}}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]
-1/3*(10*a^2*cos(d*x + c)^3 - 4*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c) - 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(sin(d*x + c) + 1)*sin(d*x + c) + 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(-sin(d*x + c) + 1)*sin (d*x + c) + 3*a^2)/((d*cos(d*x + c)^2 - d*cos(d*x + c))*sin(d*x + c))
\[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int 2 \csc ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(2*csc(c + d*x)**4*sec(c + d*x), x) + Integral(csc(c + d*x)* *4*sec(c + d*x)**2, x) + Integral(csc(c + d*x)**4, x))
Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.30 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + a^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]
-1/3*(a^2*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + a^2*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + (3*tan(d*x + c)^2 + 1)*a^2/tan(d*x + c)^3)/d
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{6 \, d} \]
1/6*(12*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*a^2*log(abs(tan(1/2*d* x + 1/2*c) - 1)) - 12*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1 ) - (15*a^2*tan(1/2*d*x + 1/2*c)^2 + a^2)/tan(1/2*d*x + 1/2*c)^3)/d
Time = 15.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {a^2}{3}}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )} \]