\(\int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1450]

Optimal result

Integrand size = 29, antiderivative size = 59 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 a b \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}+\frac {\left (a^2+b^2\right ) \tan (c+d x)}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.73 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\left (2 a^2+b^2\right ) \cos (2 (c+d x))-b \left (b+4 a \sin (c+d x)-2 a \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (2 (c+d x))\right )\right )}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3390, 3042, 3102, 25, 262, 219, 3679, 244, 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.

\(\displaystyle \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sin (c+d x)^2 \cos (c+d x)^2}dx\)

\(\Big \downarrow \) 3390

\(\displaystyle \int \csc ^2(c+d x) \sec ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )dx+2 a b \int \csc (c+d x) \sec ^2(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^2}dx+2 a b \int \csc (c+d x) \sec (c+d x)^2dx\)

\(\Big \downarrow \) 3102

\(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^2}dx+\frac {2 a b \int -\frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^2}dx-\frac {2 a b \int \frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 262

\(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^2}dx+\frac {2 a b \left (\sec (c+d x)-\int \frac {1}{1-\sec ^2(c+d x)}d\sec (c+d x)\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^2}dx+\frac {2 a b (\sec (c+d x)-\text {arctanh}(\sec (c+d x)))}{d}\)

\(\Big \downarrow \) 3679

\(\displaystyle \frac {\int \cot ^2(c+d x) \left (a^2+\left (a^2+b^2\right ) \tan ^2(c+d x)\right )d\tan (c+d x)}{d}+\frac {2 a b (\sec (c+d x)-\text {arctanh}(\sec (c+d x)))}{d}\)

\(\Big \downarrow \) 244

\(\displaystyle \frac {\int \left (\cot ^2(c+d x) a^2+\left (\frac {b^2}{a^2}+1\right ) a^2\right )d\tan (c+d x)}{d}+\frac {2 a b (\sec (c+d x)-\text {arctanh}(\sec (c+d x)))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (a^2+b^2\right ) \tan (c+d x)-a^2 \cot (c+d x)}{d}+\frac {2 a b (\sec (c+d x)-\text {arctanh}(\sec (c+d x)))}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

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

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3102

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

rule 3390

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[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 3679

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_)*((a_) + (b_.) 
*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[ 
e + f*x], x]}, Simp[ff^(n + 1)/f   Subst[Int[x^n*((a + (a + b)*ff^2*x^2)^p/ 
(1 + ff^2*x^2)^((m + n)/2 + p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[ 
{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+2 a b \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+\tan \left (d x +c \right ) b^{2}}{d}\)	\(75\)
default	\(\frac {a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+2 a b \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+\tan \left (d x +c \right ) b^{2}}{d}\)	\(75\)
parallelrisch	\(\frac {4 b a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-3 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right ) a -8 b \right )}{2 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(103\)
risch	\(\frac {2 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-4 i a^{2}-2 i b^{2}-4 a b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(125\)
norman	\(\frac {\frac {a^{2}}{2 d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 d}-\frac {2 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {2 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {\left (5 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(214\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.92 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - a b \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}{d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - a^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + b^{2} \tan \left (d x + c\right )}{d} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (59) = 118\).

Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.17 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {12 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{6 \, d} \] Input:

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

Output:

1/6*(12*a*b*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^2*tan(1/2*d*x + 1/2*c) - 
(4*a*b*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan 
(1/2*d*x + 1/2*c)^2 + 20*a*b*tan(1/2*d*x + 1/2*c) - 3*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 = 19.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.83 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^2+4\,b^2\right )-a^2+8\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)^2*(5*a^2 + 4*b^2) - a^2 + 8*a*b*tan(c/2 + (d*x)/2))/(d 
*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3)) + (a^2*tan(c/2 + (d*x)/2 
))/(2*d) + (2*a*b*log(tan(c/2 + (d*x)/2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.75 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +2 \sin \left (d x +c \right )^{2} a^{2}+\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b -a^{2}}{\cos \left (d x +c \right ) \sin \left (d x +c \right ) d} \] Input:

int(csc(d*x+c)^2*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x)
 

Output:

(2*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a*b - 2*cos(c + d*x)*si 
n(c + d*x)*a*b + 2*sin(c + d*x)**2*a**2 + sin(c + d*x)**2*b**2 + 2*sin(c + 
 d*x)*a*b - a**2)/(cos(c + d*x)*sin(c + d*x)*d)