\(\int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\) [182]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{d}-\frac {2 a b \csc (c+d x)}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3957, 2990, 2701, 327, 213, 14} \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{d}+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \csc (c+d x)}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[In]

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

[Out]

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

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 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 2990

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[2*a*(b/d), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^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 (-b-a \cos (c+d x))^2 \csc ^2(c+d x) \sec ^2(c+d x) \, dx \\ & = (2 a b) \int \csc ^2(c+d x) \sec (c+d x) \, dx+\int \left (b^2+a^2 \cos ^2(c+d x)\right ) \csc ^2(c+d x) \sec ^2(c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {a^2+b^2+b^2 x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d} \\ & = -\frac {2 a b \csc (c+d x)}{d}+\frac {\text {Subst}\left (\int \left (b^2+\frac {a^2+b^2}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d} \\ & = \frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{d}-\frac {2 a b \csc (c+d x)}{d}+\frac {b^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(138\) vs. \(2(59)=118\).

Time = 1.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.34 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {\csc ^3\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (4 a b \cos (c+d x)+\left (a^2+2 b^2\right ) \cos (2 (c+d x))+a \left (a+2 b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (2 (c+d x))\right )\right )}{4 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

-1/4*(Csc[(c + d*x)/2]^3*Sec[(c + d*x)/2]*(4*a*b*Cos[c + d*x] + (a^2 + 2*b^2)*Cos[2*(c + d*x)] + a*(a + 2*b*(L
og[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Sin[2*(c + d*x)])))/(d*(-1
+ Cot[(c + d*x)/2]^2))

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(\frac {-a^{2} \cot \left (d x +c \right )+2 a b \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\)	\(76\)
default	\(\frac {-a^{2} \cot \left (d x +c \right )+2 a b \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\)	\(76\)
parallelrisch	\(\frac {-8 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+8 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-\left (\left (a^{2}+2 b^{2}\right ) \cos \left (2 d x +2 c \right )+a \left (a -4 b \right )\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-8 a b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \cos \left (d x +c \right )}\)	\(119\)
risch	\(-\frac {2 i \left (2 a b \,{\mathrm e}^{3 i \left (d x +c \right )}+a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a b \,{\mathrm e}^{i \left (d x +c \right )}+a^{2}+2 b^{2}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(122\)
norman	\(\frac {\frac {a^{2}+2 a b +b^{2}}{2 d}-\frac {\left (a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.76 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}{d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {a b {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + b^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac {a^{2}}{\tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (59) = 118\).

Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.83 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*(4*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + a^2*tan(1/2*d*x + 1
/2*c) - 2*a*b*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c) - (a^2*tan(1/2*d*x + 1/2*c)^2 + 2*a*b*tan(1/2*d*
x + 1/2*c)^2 + 5*b^2*tan(1/2*d*x + 1/2*c)^2 - a^2 - 2*a*b - b^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 = 14.58 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.83 \[ \int \csc ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a-b\right )}^2}{2\,d}-\frac {2\,a\,b+a^2+b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+2\,a\,b+5\,b^2\right )}{d\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {4\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(a - b)^2)/(2*d) - (2*a*b + a^2 + b^2 - tan(c/2 + (d*x)/2)^2*(2*a*b + a^2 + 5*b^2))/(d*(2*
tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3)) + (4*a*b*atanh(tan(c/2 + (d*x)/2)))/d