\(\int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx\) [319]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=-\frac {3 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}+\frac {3 c \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {3 i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b} \] Output:

-3*I*d*x*arctan(exp(I*(b*x+a)))/b-d*arctanh(cos(b*x+a))/b^2+3/2*c*arctanh( 
sin(b*x+a))/b-3/2*(d*x+c)*csc(b*x+a)/b+3/2*I*d*polylog(2,-I*exp(I*(b*x+a)) 
)/b^2-3/2*I*d*polylog(2,I*exp(I*(b*x+a)))/b^2-1/2*d*sec(b*x+a)/b^2+1/2*(d* 
x+c)*csc(b*x+a)*sec(b*x+a)^2/b
 

Mathematica [C] (warning: unable to verify)

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

Time = 6.75 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.13 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx =\text {Too large to display} \] Input:

Integrate[(c + d*x)*Csc[a + b*x]^2*Sec[a + b*x]^3,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, 2, 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 - (3*d*x*( 
a*Log[1 - Tan[(a + b*x)/2]] - a*Log[1 + Tan[(a + b*x)/2]] - I*(Log[1 + I*T 
an[(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*(L 
og[1 + I*Tan[(a + b*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(a + b*x)/2])] + Pol 
yLog[2, ((1 - I) + (1 + I)*Tan[(a + b*x)/2])/2])))/(2*b*(a - I*Log[1 - I*T 
an[(a + b*x)/2]] + I*Log[1 + I*Tan[(a + b*x)/2]])) + (d*x)/(4*b*(Cos[(a + 
b*x)/2] - Sin[(a + b*x)/2])^2) - (d*Sin[(a + b*x)/2])/(2*b^2*(Cos[(a + b*x 
)/2] - Sin[(a + b*x)/2])) - (d*x)/(4*b*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2 
])^2) + (d*Sin[(a + b*x)/2])/(2*b^2*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) 
 + (d*Sec[(a + b*x)/2]*(a*Sin[(a + b*x)/2] - (a + b*x)*Sin[(a + b*x)/2]))/ 
(2*b^2)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

\(\displaystyle \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx\)

\(\Big \downarrow \) 4920

\(\displaystyle -d \int \left (\frac {\csc (a+b x) \sec ^2(a+b x)}{2 b}+\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 \csc (a+b x)}{2 b}\right )dx+\frac {3 (c+d x) \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -d \left (\frac {3 i x \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {\text {arctanh}(\cos (a+b x))}{b^2}+\frac {3 x \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 i \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 i \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {\sec (a+b x)}{2 b^2}\right )+\frac {3 (c+d x) \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}\)

Input:

Int[(c + d*x)*Csc[a + b*x]^2*Sec[a + b*x]^3,x]
 

Output:

(3*(c + d*x)*ArcTanh[Sin[a + b*x]])/(2*b) - (3*(c + d*x)*Csc[a + b*x])/(2* 
b) + ((c + d*x)*Csc[a + b*x]*Sec[a + b*x]^2)/(2*b) - d*(((3*I)*x*ArcTan[E^ 
(I*(a + b*x))])/b + ArcTanh[Cos[a + b*x]]/b^2 + (3*x*ArcTanh[Sin[a + b*x]] 
)/(2*b) - (((3*I)/2)*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 + (((3*I)/2)*Po 
lyLog[2, I*E^(I*(a + b*x))])/b^2 + Sec[a + b*x]/(2*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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (139 ) = 278\).

Time = 0.57 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.12

method	result	size
risch	\(-\frac {i \left (3 b d x \,{\mathrm e}^{5 i \left (b x +a \right )}+3 c b \,{\mathrm e}^{5 i \left (b x +a \right )}+2 b d x \,{\mathrm e}^{3 i \left (b x +a \right )}+2 c b \,{\mathrm e}^{3 i \left (b x +a \right )}-i d \,{\mathrm e}^{5 i \left (b x +a \right )}+3 b d x \,{\mathrm e}^{i \left (b x +a \right )}+3 b c \,{\mathrm e}^{i \left (b x +a \right )}+i d \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {3 d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{2 b}+\frac {3 d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}-\frac {3 d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{2 b^{2}}+\frac {3 d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}+\frac {3 i d \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b^{2}}-\frac {3 i d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {3 i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {3 i a d \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}\)	\(344\)

Input:

int((d*x+c)*csc(b*x+a)^2*sec(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-I/b^2/(exp(2*I*(b*x+a))+1)^2/(exp(2*I*(b*x+a))-1)*(3*b*d*x*exp(5*I*(b*x+a 
))+3*c*b*exp(5*I*(b*x+a))+2*b*d*x*exp(3*I*(b*x+a))+2*c*b*exp(3*I*(b*x+a))- 
I*d*exp(5*I*(b*x+a))+3*b*d*x*exp(I*(b*x+a))+3*b*c*exp(I*(b*x+a))+I*d*exp(I 
*(b*x+a)))-3/2/b*d*ln(I*exp(I*(b*x+a))+1)*x+3/2/b*d*ln(1-I*exp(I*(b*x+a))) 
*x-3/2/b^2*d*ln(I*exp(I*(b*x+a))+1)*a+3/2/b^2*d*ln(1-I*exp(I*(b*x+a)))*a+3 
/2*I/b^2*d*dilog(I*exp(I*(b*x+a))+1)-3/2*I/b^2*d*dilog(1-I*exp(I*(b*x+a))) 
-3*I/b*c*arctan(exp(I*(b*x+a)))+3*I/b^2*a*d*arctan(exp(I*(b*x+a)))+d/b^2*l 
n(exp(I*(b*x+a))-1)-d/b^2*ln(exp(I*(b*x+a))+1)
 

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. 592 vs. \(2 (132) = 264\).

Time = 0.11 (sec) , antiderivative size = 592, normalized size of antiderivative = 3.65 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)*csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="fricas")
 

Output:

1/4*(-3*I*d*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a))*sin(b*x + 
a) - 3*I*d*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a 
) + 3*I*d*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a 
) + 3*I*d*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a 
) + 3*(b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) + I*sin(b*x + a) + I)*si 
n(b*x + a) - 3*(b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) - I*sin(b*x + a 
) + I)*sin(b*x + a) - 2*d*cos(b*x + a)^2*log(1/2*cos(b*x + a) + 1/2)*sin(b 
*x + a) + 3*(b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) + sin(b*x + a) 
 + 1)*sin(b*x + a) - 3*(b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) - s 
in(b*x + a) + 1)*sin(b*x + a) + 3*(b*d*x + a*d)*cos(b*x + a)^2*log(-I*cos( 
b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) - 3*(b*d*x + a*d)*cos(b*x + a)^2 
*log(-I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) + 2*d*cos(b*x + a)^2 
*log(-1/2*cos(b*x + a) + 1/2)*sin(b*x + a) + 3*(b*c - a*d)*cos(b*x + a)^2* 
log(-cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) - 3*(b*c - a*d)*cos(b 
*x + a)^2*log(-cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + 2*b*d*x - 
 6*(b*d*x + b*c)*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x + a) + 2*b*c)/( 
b^2*cos(b*x + a)^2*sin(b*x + a))
 

Sympy [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int \left (c + d x\right ) \csc ^{2}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)*csc(b*x+a)**2*sec(b*x+a)**3,x)
 

Output:

Integral((c + d*x)*csc(a + b*x)**2*sec(a + b*x)**3, x)
 

Maxima [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)*csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="maxima")
 

Output:

1/4*(8*(b*d*x + b*c)*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) - 4*(d*cos(5*b*x + 
5*a) - d*cos(b*x + a) - 3*(b*d*x + b*c)*sin(5*b*x + 5*a) - 2*(b*d*x + b*c) 
*sin(3*b*x + 3*a) - 3*(b*d*x + b*c)*sin(b*x + a))*cos(6*b*x + 6*a) - 4*(d* 
cos(4*b*x + 4*a) - d*cos(2*b*x + 2*a) + 3*(b*d*x + b*c)*sin(4*b*x + 4*a) - 
 3*(b*d*x + b*c)*sin(2*b*x + 2*a) - d)*cos(5*b*x + 5*a) + 4*(d*cos(b*x + a 
) + 2*(b*d*x + b*c)*sin(3*b*x + 3*a) + 3*(b*d*x + b*c)*sin(b*x + a))*cos(4 
*b*x + 4*a) - 4*(d*cos(b*x + a) + 3*(b*d*x + b*c)*sin(b*x + a))*cos(2*b*x 
+ 2*a) - 4*d*cos(b*x + a) + 12*(b^2*d*cos(6*b*x + 6*a)^2 + b^2*d*cos(4*b*x 
 + 4*a)^2 + b^2*d*cos(2*b*x + 2*a)^2 + b^2*d*sin(6*b*x + 6*a)^2 + b^2*d*si 
n(4*b*x + 4*a)^2 - 2*b^2*d*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + b^2*d*sin(2 
*b*x + 2*a)^2 + 2*b^2*d*cos(2*b*x + 2*a) + b^2*d + 2*(b^2*d*cos(4*b*x + 4* 
a) - b^2*d*cos(2*b*x + 2*a) - b^2*d)*cos(6*b*x + 6*a) - 2*(b^2*d*cos(2*b*x 
 + 2*a) + b^2*d)*cos(4*b*x + 4*a) + 2*(b^2*d*sin(4*b*x + 4*a) - b^2*d*sin( 
2*b*x + 2*a))*sin(6*b*x + 6*a))*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) + 3*(b*c*cos(6*b*x + 6*a 
)^2 + b*c*cos(4*b*x + 4*a)^2 + b*c*cos(2*b*x + 2*a)^2 + b*c*sin(6*b*x + 6* 
a)^2 + b*c*sin(4*b*x + 4*a)^2 - 2*b*c*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 
b*c*sin(2*b*x + 2*a)^2 + 2*b*c*cos(2*b*x + 2*a) + b*c + 2*(b*c*cos(4*b*x + 
 4*a) - b*c*cos(2*b*x + 2*a) - b*c)*cos(6*b*x + 6*a) - 2*(b*c*cos(2*b*x...
 

Giac [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)*csc(b*x+a)^2*sec(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)*csc(b*x + a)^2*sec(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Hanged} \] Input:

int((c + d*x)/(cos(a + b*x)^3*sin(a + b*x)^2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx =\text {Too large to display} \] Input:

int((d*x+c)*csc(b*x+a)^2*sec(b*x+a)^3,x)
 

Output:

(126*cos(a + b*x)*sin(a + b*x)**2*b*d*x + 20*cos(a + b*x)*sin(a + b*x)*d - 
 96*cos(a + b*x)*b*d*x - 48*int(x/(tan((a + b*x)/2)**8 - 3*tan((a + b*x)/2 
)**6 + 3*tan((a + b*x)/2)**4 - tan((a + b*x)/2)**2),x)*sin(a + b*x)**3*b** 
2*d + 48*int(x/(tan((a + b*x)/2)**8 - 3*tan((a + b*x)/2)**6 + 3*tan((a + b 
*x)/2)**4 - tan((a + b*x)/2)**2),x)*sin(a + b*x)*b**2*d + 126*log(tan((a + 
 b*x)/2)**2 + 1)*sin(a + b*x)**3*d - 126*log(tan((a + b*x)/2)**2 + 1)*sin( 
a + b*x)*d - 69*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**3*b*c - 30*log(tan 
((a + b*x)/2) - 1)*sin(a + b*x)**3*d + 69*log(tan((a + b*x)/2) - 1)*sin(a 
+ b*x)*b*c + 30*log(tan((a + b*x)/2) - 1)*sin(a + b*x)*d + 69*log(tan((a + 
 b*x)/2) + 1)*sin(a + b*x)**3*b*c - 30*log(tan((a + b*x)/2) + 1)*sin(a + b 
*x)**3*d - 69*log(tan((a + b*x)/2) + 1)*sin(a + b*x)*b*c + 30*log(tan((a + 
 b*x)/2) + 1)*sin(a + b*x)*d - 146*log(tan((a + b*x)/2))*sin(a + b*x)**3*d 
 + 146*log(tan((a + b*x)/2))*sin(a + b*x)*d + 3*sin(a + b*x)**3*b**2*d*x** 
2 - 69*sin(a + b*x)**2*b*c + 30*sin(a + b*x)**2*b*d*x - 3*sin(a + b*x)*b** 
2*d*x**2 + 46*b*c - 50*b*d*x)/(46*sin(a + b*x)*b**2*(sin(a + b*x)**2 - 1))