\(\int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx\) [33]

Optimal result

Integrand size = 21, antiderivative size = 57 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d} \]

[Out]

2*a^2*arctanh(sin(d*x+c))/d-2*a^2*cot(d*x+c)/d-2*a^2*csc(d*x+c)/d+a^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3957, 2952, 3852, 8, 2701, 327, 213, 2700, 14} \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]^2*(a + a*Sec[c + d*x])^2,x]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Cot[c + d*x])/d - (2*a^2*Csc[c + d*x])/d + (a^2*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int (-a-a \cos (c+d x))^2 \csc ^2(c+d x) \sec ^2(c+d x) \, dx \\ & = \int \left (a^2 \csc ^2(c+d x)+2 a^2 \csc ^2(c+d x) \sec (c+d x)+a^2 \csc ^2(c+d x) \sec ^2(c+d x)\right ) \, dx \\ & = a^2 \int \csc ^2(c+d x) \, dx+a^2 \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^2(c+d x) \sec (c+d x) \, dx \\ & = -\frac {a^2 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {a^2 \text {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d} \\ & = -\frac {a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \text {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d} \\ & = \frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(401\) vs. \(2(57)=114\).

Time = 6.65 (sec) , antiderivative size = 401, normalized size of antiderivative = 7.04 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {\cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2}{2 d}+\frac {\cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2}{2 d}+\frac {\cos ^2(c+d x) \csc \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \sin \left (\frac {d x}{2}\right )}{2 d}+\frac {\cos ^2(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \sin \left (\frac {d x}{2}\right )}{4 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\cos ^2(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \sin \left (\frac {d x}{2}\right )}{4 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]

[In]

Integrate[Csc[c + d*x]^2*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/2*(Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)
/d + (Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)
/(2*d) + (Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])
/(2*d) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(4*d*(Cos[c/2] - Sin[c/2])*
(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(
d*x)/2])/(4*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+2 a^{2} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{2} \cot \left (d x +c \right )}{d}\)	\(77\)
default	\(\frac {a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+2 a^{2} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{2} \cot \left (d x +c \right )}{d}\)	\(77\)
parallelrisch	\(-\frac {a^{2} \left (3 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \cos \left (d x +c \right )}\)	\(86\)
norman	\(\frac {\frac {2 a^{2}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(97\)
risch	\(-\frac {2 i a^{2} \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+3\right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(103\)

[In]

int(csc(d*x+c)^2*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(1/sin(d*x+c)/cos(d*x+c)-2*cot(d*x+c))+2*a^2*(-1/sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))-a^2*cot(d*x+c)
)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.77 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}{d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

(a^2*cos(d*x + c)*log(sin(d*x + c) + 1)*sin(d*x + c) - a^2*cos(d*x + c)*log(-sin(d*x + c) + 1)*sin(d*x + c) -
3*a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)/(d*cos(d*x + c)*sin(d*x + c))

Sympy [F]

\[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int 2 \csc ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(csc(d*x+c)**2*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*csc(c + d*x)**2*sec(c + d*x), x) + Integral(csc(c + d*x)**2*sec(c + d*x)**2, x) + Integral(cs
c(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.30 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^{2} {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + a^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac {a^{2}}{\tan \left (d x + c\right )}}{d} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-(a^2*(2/sin(d*x + c) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + a^2*(1/tan(d*x + c) - tan(d*x + c)) +
a^2/tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 \, {\left (a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )}}{d} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

2*(a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (2*a^2*tan(1/2*d*x + 1/2*
c)^2 - a^2)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)))/d

Mupad [B] (verification not implemented)

Time = 13.50 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,a^2}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int((a + a/cos(c + d*x))^2/sin(c + d*x)^2,x)

[Out]

(4*a^2*tan(c/2 + (d*x)/2)^2 - 2*a^2)/(d*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^3)) + (4*a^2*atanh(tan(c/2 +
(d*x)/2)))/d