\(\int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx\) [237]

Optimal result

Integrand size = 20, antiderivative size = 131 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {2 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}+\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2} \] Output:

-2*I*d*x*arctan(exp(I*(b*x+a)))/b-d*arctanh(cos(b*x+a))/b^2-d*x*arctanh(si 
n(b*x+a))/b+(d*x+c)*arctanh(sin(b*x+a))/b-(d*x+c)*csc(b*x+a)/b+I*d*polylog 
(2,-I*exp(I*(b*x+a)))/b^2-I*d*polylog(2,I*exp(I*(b*x+a)))/b^2
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.28 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.95 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

(d*(a*Cos[(a + b*x)/2] - (a + b*x)*Cos[(a + b*x)/2])*Csc[(a + b*x)/2])/(2* 
b^2) - (c*Csc[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[a + b*x]^2])/b 
- (d*Log[Cos[(a + b*x)/2]])/b^2 + (d*Log[Sin[(a + b*x)/2]])/b^2 - (d*x*(a* 
Log[1 - Tan[(a + b*x)/2]] - a*Log[1 + Tan[(a + b*x)/2]] - I*(Log[1 + I*Tan 
[(a + b*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(a + b*x)/2])] + PolyLog[2, ((1 + 
I) - (1 - I)*Tan[(a + b*x)/2])/2]) + I*(Log[1 - I*Tan[(a + b*x)/2]]*Log[(1 
/2 + I/2)*(1 + Tan[(a + b*x)/2])] + PolyLog[2, (-1/2 - I/2)*(I + Tan[(a + 
b*x)/2])]) - I*(Log[1 - I*Tan[(a + b*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(a 
+ b*x)/2])] + PolyLog[2, ((1 + I) + (1 - I)*Tan[(a + b*x)/2])/2]) + I*(Log 
[1 + I*Tan[(a + b*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(a + b*x)/2])] + PolyL 
og[2, ((1 - I) + (1 + I)*Tan[(a + b*x)/2])/2])))/(b*(a - I*Log[1 - I*Tan[( 
a + b*x)/2]] + I*Log[1 + I*Tan[(a + b*x)/2]])) + (d*Sec[(a + b*x)/2]*(a*Si 
n[(a + b*x)/2] - (a + b*x)*Sin[(a + b*x)/2]))/(2*b^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4920, 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.

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

\(\Big \downarrow \) 4920

\(\displaystyle -d \int \left (\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b}\right )dx+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -d \left (\frac {2 i x \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {\text {arctanh}(\cos (a+b x))}{b^2}+\frac {x \text {arctanh}(\sin (a+b x))}{b}-\frac {i \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {i \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}\right )+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}\)

Input:

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

Output:

((c + d*x)*ArcTanh[Sin[a + b*x]])/b - ((c + d*x)*Csc[a + b*x])/b - d*(((2* 
I)*x*ArcTan[E^(I*(a + b*x))])/b + ArcTanh[Cos[a + b*x]]/b^2 + (x*ArcTanh[S 
in[a + b*x]])/b - (I*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 + (I*PolyLog[2, 
 I*E^(I*(a + b*x))])/b^2)
 

Defintions of rubi rules used

rule 2009

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

rule 4920

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Module[{u = IntHide[Csc[a + b*x]^n*Sec[a + b* 
x]^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, x 
], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, 
p]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.79

method	result	size
risch	\(-\frac {2 i \left (d x +c \right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {2 i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}-\frac {d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 i d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {i d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(235\)

Input:

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

Output:

-2*I*(d*x+c)*exp(I*(b*x+a))/b/(exp(2*I*(b*x+a))-1)+d/b^2*ln(exp(I*(b*x+a)) 
-1)-d/b^2*ln(exp(I*(b*x+a))+1)-2*I/b*c*arctan(exp(I*(b*x+a)))-1/b^2*d*ln(I 
*exp(I*(b*x+a))+1)*a-1/b*d*ln(I*exp(I*(b*x+a))+1)*x+1/b*d*ln(1-I*exp(I*(b* 
x+a)))*x+1/b^2*d*ln(1-I*exp(I*(b*x+a)))*a+2*I/b^2*d*a*arctan(exp(I*(b*x+a) 
))+I/b^2*d*dilog(I*exp(I*(b*x+a))+1)-I/b^2*d*dilog(1-I*exp(I*(b*x+a)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (113) = 226\).

Time = 0.10 (sec) , antiderivative size = 434, normalized size of antiderivative = 3.31 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {2 \, b d x + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + d \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - d \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + 2 \, b c}{2 \, b^{2} \sin \left (b x + a\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/2*(4*(b*d*x + b*c)*cos(b*x + a)*sin(2*b*x + 2*a) - 4*(b*d*x + b*c)*cos( 
2*b*x + 2*a)*sin(b*x + a) - 4*(b^2*d*cos(2*b*x + 2*a)^2 + b^2*d*sin(2*b*x 
+ 2*a)^2 - 2*b^2*d*cos(2*b*x + 2*a) + b^2*d)*integrate((x*cos(2*b*x + 2*a) 
*cos(b*x + a) + x*sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x + a))/(cos(2*b 
*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) - (b*c*cos( 
2*b*x + 2*a)^2 + b*c*sin(2*b*x + 2*a)^2 - 2*b*c*cos(2*b*x + 2*a) + b*c)*lo 
g(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + (b*c*cos(2*b*x + 
 2*a)^2 + b*c*sin(2*b*x + 2*a)^2 - 2*b*c*cos(2*b*x + 2*a) + b*c)*log(cos(b 
*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) + (d*cos(2*b*x + 2*a)^2 + 
 d*sin(2*b*x + 2*a)^2 - 2*d*cos(2*b*x + 2*a) + d)*log(cos(b*x)^2 + 2*cos(b 
*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - (d*co 
s(2*b*x + 2*a)^2 + d*sin(2*b*x + 2*a)^2 - 2*d*cos(2*b*x + 2*a) + d)*log(co 
s(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + 
 sin(a)^2) + 4*(b*d*x + b*c)*sin(b*x + a))/(b^2*cos(2*b*x + 2*a)^2 + b^2*s 
in(2*b*x + 2*a)^2 - 2*b^2*cos(2*b*x + 2*a) + b^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\frac {8 \cos \left (b x +a \right ) b d x -4 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}d x \right ) \sin \left (b x +a \right ) b^{2} d +8 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right ) d -2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right ) b c +2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right ) b c -14 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) d +\sin \left (b x +a \right ) b^{2} d \,x^{2}-2 b c +6 b d x}{2 \sin \left (b x +a \right ) b^{2}} \] Input:

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

Output:

(8*cos(a + b*x)*b*d*x - 4*int(x/(tan((a + b*x)/2)**4 - tan((a + b*x)/2)**2 
),x)*sin(a + b*x)*b**2*d + 8*log(tan((a + b*x)/2)**2 + 1)*sin(a + b*x)*d - 
 2*log(tan((a + b*x)/2) - 1)*sin(a + b*x)*b*c + 2*log(tan((a + b*x)/2) + 1 
)*sin(a + b*x)*b*c - 14*log(tan((a + b*x)/2))*sin(a + b*x)*d + sin(a + b*x 
)*b**2*d*x**2 - 2*b*c + 6*b*d*x)/(2*sin(a + b*x)*b**2)