3.243 \(\int (c+d x) \csc ^3(a+b x) \sec (a+b x) \, dx\)
Optimal. Leaf size=141 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}+\frac{(c+d x) \log (\tan (a+b x))}{b}-\frac{2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d x \log (\tan (a+b x))}{b}-\frac{d x}{2 b} \]
[Out]
-(d*x)/(2*b) - (2*d*x*ArcTanh[E^((2*I)*(a + b*x))])/b - (d*Cot[a + b*x])/(2*b^2) - ((c + d*x)*Cot[a + b*x]^2)/
(2*b) - (d*x*Log[Tan[a + b*x]])/b + ((c + d*x)*Log[Tan[a + b*x]])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))
])/b^2 - ((I/2)*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2
________________________________________________________________________________________
Rubi [A] time = 0.138571, antiderivative size = 141, normalized size of antiderivative = 1.,
number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used =
{2620, 14, 4420, 3473, 8, 2548, 12, 4183, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}+\frac{(c+d x) \log (\tan (a+b x))}{b}-\frac{2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d x \log (\tan (a+b x))}{b}-\frac{d x}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x],x]
[Out]
-(d*x)/(2*b) - (2*d*x*ArcTanh[E^((2*I)*(a + b*x))])/b - (d*Cot[a + b*x])/(2*b^2) - ((c + d*x)*Cot[a + b*x]^2)/
(2*b) - (d*x*Log[Tan[a + b*x]])/b + ((c + d*x)*Log[Tan[a + b*x]])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))
])/b^2 - ((I/2)*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2
Rule 2620
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 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 4420
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Modul
e[{u = IntHide[Csc[a + b*x]^n*Sec[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[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]
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2548
Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 4183
Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]
Rule 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2391
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rubi steps
\begin{align*} \int (c+d x) \csc ^3(a+b x) \sec (a+b x) \, dx &=-\frac{(c+d x) \cot ^2(a+b x)}{2 b}+\frac{(c+d x) \log (\tan (a+b x))}{b}-d \int \left (-\frac{\cot ^2(a+b x)}{2 b}+\frac{\log (\tan (a+b x))}{b}\right ) \, dx\\ &=-\frac{(c+d x) \cot ^2(a+b x)}{2 b}+\frac{(c+d x) \log (\tan (a+b x))}{b}+\frac{d \int \cot ^2(a+b x) \, dx}{2 b}-\frac{d \int \log (\tan (a+b x)) \, dx}{b}\\ &=-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}-\frac{d x \log (\tan (a+b x))}{b}+\frac{(c+d x) \log (\tan (a+b x))}{b}-\frac{d \int 1 \, dx}{2 b}+\frac{d \int 2 b x \csc (2 a+2 b x) \, dx}{b}\\ &=-\frac{d x}{2 b}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}-\frac{d x \log (\tan (a+b x))}{b}+\frac{(c+d x) \log (\tan (a+b x))}{b}+(2 d) \int x \csc (2 a+2 b x) \, dx\\ &=-\frac{d x}{2 b}-\frac{2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}-\frac{d x \log (\tan (a+b x))}{b}+\frac{(c+d x) \log (\tan (a+b x))}{b}-\frac{d \int \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac{d \int \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=-\frac{d x}{2 b}-\frac{2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}-\frac{d x \log (\tan (a+b x))}{b}+\frac{(c+d x) \log (\tan (a+b x))}{b}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^2}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^2}\\ &=-\frac{d x}{2 b}-\frac{2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \cot (a+b x)}{2 b^2}-\frac{(c+d x) \cot ^2(a+b x)}{2 b}-\frac{d x \log (\tan (a+b x))}{b}+\frac{(c+d x) \log (\tan (a+b x))}{b}+\frac{i d \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{i d \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.957523, size = 210, normalized size = 1.49 \[ \frac{d \left (\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )+\frac{1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )\right )}{b^2}+\frac{d \left ((a+b x) \log \left (1-e^{2 i (a+b x)}\right )-\frac{1}{2} i \left ((a+b x)^2+\text{PolyLog}\left (2,e^{2 i (a+b x)}\right )\right )\right )}{b^2}-\frac{d \cot (a+b x)}{2 b^2}-\frac{a d \log (\sin (a+b x))}{b^2}+\frac{a d \log (\cos (a+b x))}{b^2}-\frac{c \left (\csc ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))\right )}{2 b}-\frac{d x \csc ^2(a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x],x]
[Out]
-(d*Cot[a + b*x])/(2*b^2) - (d*x*Csc[a + b*x]^2)/(2*b) + (a*d*Log[Cos[a + b*x]])/b^2 - (c*(Csc[a + b*x]^2 + 2*
Log[Cos[a + b*x]] - 2*Log[Sin[a + b*x]]))/(2*b) - (a*d*Log[Sin[a + b*x]])/b^2 + (d*((I/2)*(a + b*x)^2 - (a + b
*x)*Log[1 + E^((2*I)*(a + b*x))] + (I/2)*PolyLog[2, -E^((2*I)*(a + b*x))]))/b^2 + (d*((a + b*x)*Log[1 - E^((2*
I)*(a + b*x))] - (I/2)*((a + b*x)^2 + PolyLog[2, E^((2*I)*(a + b*x))])))/b^2
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Maple [B] time = 0.173, size = 270, normalized size = 1.9 \begin{align*}{\frac{2\,bdx{{\rm e}^{2\,i \left ( bx+a \right ) }}+2\,bc{{\rm e}^{2\,i \left ( bx+a \right ) }}-id{{\rm e}^{2\,i \left ( bx+a \right ) }}+id}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}}}-{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}+{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{id{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-{\frac{id{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{ad\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x)
[Out]
(2*b*d*x*exp(2*I*(b*x+a))+2*b*c*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))+I*d)/b^2/(exp(2*I*(b*x+a))-1)^2-1/b*c*ln
(exp(2*I*(b*x+a))+1)+1/b*c*ln(exp(I*(b*x+a))-1)+1/b*c*ln(exp(I*(b*x+a))+1)+1/b*d*ln(1-exp(I*(b*x+a)))*x+1/b^2*
d*ln(1-exp(I*(b*x+a)))*a-I*d*polylog(2,exp(I*(b*x+a)))/b^2-1/b*d*ln(exp(2*I*(b*x+a))+1)*x+1/2*I*d*polylog(2,-e
xp(2*I*(b*x+a)))/b^2+1/b*d*ln(exp(I*(b*x+a))+1)*x-I*d*polylog(2,-exp(I*(b*x+a)))/b^2-1/b^2*d*a*ln(exp(I*(b*x+a
))-1)
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Maxima [B] time = 2.21863, size = 1397, normalized size = 9.91 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x, algorithm="maxima")
[Out]
-((2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(4*b*x + 4*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a) + (2*I*b*d*x + 2*I*b*
c)*sin(4*b*x + 4*a) + (-4*I*b*d*x - 4*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1)
- (2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(4*b*x + 4*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a) - (-2*I*b*d*x - 2*I*
b*c)*sin(4*b*x + 4*a) - (4*I*b*d*x + 4*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b
*c*cos(4*b*x + 4*a) - 4*b*c*cos(2*b*x + 2*a) + 2*I*b*c*sin(4*b*x + 4*a) - 4*I*b*c*sin(2*b*x + 2*a) + 2*b*c)*ar
ctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(4*b*x + 4*a) - 4*b*d*x*cos(2*b*x + 2*a) + 2*I*b*d*x*sin(4
*b*x + 4*a) - 4*I*b*d*x*sin(2*b*x + 2*a) + 2*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (4*I*b*d*x + 4*
I*b*c + 2*d)*cos(2*b*x + 2*a) - (d*cos(4*b*x + 4*a) - 2*d*cos(2*b*x + 2*a) + I*d*sin(4*b*x + 4*a) - 2*I*d*sin(
2*b*x + 2*a) + d)*dilog(-e^(2*I*b*x + 2*I*a)) + (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*x + 2*a) + 2*I*d*sin(4*b*x
+ 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(-e^(I*b*x + I*a)) + (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*x + 2*a)
+ 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(e^(I*b*x + I*a)) + (-I*b*d*x - I*b*c + (-I*b*d
*x - I*b*c)*cos(4*b*x + 4*a) + (2*I*b*d*x + 2*I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*c)*sin(4*b*x + 4*a) - 2*(b*
d*x + b*c)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + (I*b*d*x
+ I*b*c + (I*b*d*x + I*b*c)*cos(4*b*x + 4*a) + (-2*I*b*d*x - 2*I*b*c)*cos(2*b*x + 2*a) - (b*d*x + b*c)*sin(4*b
*x + 4*a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (I*b
*d*x + I*b*c + (I*b*d*x + I*b*c)*cos(4*b*x + 4*a) + (-2*I*b*d*x - 2*I*b*c)*cos(2*b*x + 2*a) - (b*d*x + b*c)*si
n(4*b*x + 4*a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) -
2*(2*b*d*x + 2*b*c - I*d)*sin(2*b*x + 2*a) - 2*d)/(-2*I*b^2*cos(4*b*x + 4*a) + 4*I*b^2*cos(2*b*x + 2*a) + 2*b
^2*sin(4*b*x + 4*a) - 4*b^2*sin(2*b*x + 2*a) - 2*I*b^2)
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Fricas [B] time = 0.722204, size = 2457, normalized size = 17.43 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x, algorithm="fricas")
[Out]
1/2*(b*d*x + d*cos(b*x + a)*sin(b*x + a) + b*c + (-I*d*cos(b*x + a)^2 + I*d)*dilog(cos(b*x + a) + I*sin(b*x +
a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(I*co
s(b*x + a) + sin(b*x + a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) + (I*d*cos(b*x +
a)^2 - I*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(-I*cos(b*x + a) - sin(b*
x + a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(-cos(b*x + a) + I*sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog
(-cos(b*x + a) - I*sin(b*x + a)) - (b*d*x - (b*d*x + b*c)*cos(b*x + a)^2 + b*c)*log(cos(b*x + a) + I*sin(b*x +
a) + 1) - ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(cos(b*x + a) + I*sin(b*x + a) + I) - (b*d*x - (b*d*x +
b*c)*cos(b*x + a)^2 + b*c)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*
log(cos(b*x + a) - I*sin(b*x + a) + I) + (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(I*cos(b*x + a) + sin
(b*x + a) + 1) + (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b*d*x
- (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d*x - (b*d*x + a*d)*cos(b*x
+ a)^2 + a*d)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos
(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos(b*x + a) - 1/2*I
*sin(b*x + a) + 1/2) - (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) -
((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b*d*x - (b*d*x + a*d)*cos(
b*x + a)^2 + a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) - ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-cos(
b*x + a) - I*sin(b*x + a) + I))/(b^2*cos(b*x + a)^2 - b^2)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)**3*sec(b*x+a),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x, algorithm="giac")
[Out]
integrate((d*x + c)*csc(b*x + a)^3*sec(b*x + a), x)