3.325 \(\int (c+d x) \csc ^3(a+b x) \sec ^3(a+b x) \, dx\)
Optimal. Leaf size=110 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{4 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 b x)}{b} \]
[Out]
(-4*(c + d*x)*ArcTanh[E^((2*I)*(a + b*x))])/b - (d*Csc[2*a + 2*b*x])/b^2 - (2*(c + d*x)*Cot[2*a + 2*b*x]*Csc[2
*a + 2*b*x])/b + (I*d*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2
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Rubi [A] time = 0.106173, antiderivative size = 110, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used =
{4419, 4185, 4183, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{4 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 b x)}{b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x]^3,x]
[Out]
(-4*(c + d*x)*ArcTanh[E^((2*I)*(a + b*x))])/b - (d*Csc[2*a + 2*b*x])/b^2 - (2*(c + d*x)*Cot[2*a + 2*b*x]*Csc[2
*a + 2*b*x])/b + (I*d*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2
Rule 4419
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]
Rule 4185
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]
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 ^3(a+b x) \, dx &=8 \int (c+d x) \csc ^3(2 a+2 b x) \, dx\\ &=-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 b x)}{b}+4 \int (c+d x) \csc (2 a+2 b x) \, dx\\ &=-\frac{4 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 b x)}{b}-\frac{(2 d) \int \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac{(2 d) \int \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=-\frac{4 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 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 )}{b^2}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{b^2}\\ &=-\frac{4 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{d \csc (2 a+2 b x)}{b^2}-\frac{2 (c+d x) \cot (2 a+2 b x) \csc (2 a+2 b x)}{b}+\frac{i d \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d \text{Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}\\ \end{align*}
Mathematica [B] time = 2.21126, size = 236, normalized size = 2.15 \[ \frac{d \left (i \left (\text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )-\text{PolyLog}\left (2,e^{2 i (a+b x)}\right )\right )+2 (a+b x) \left (\log \left (1-e^{2 i (a+b x)}\right )-\log \left (1+e^{2 i (a+b x)}\right )\right )\right )}{b^2}-\frac{d \tan (a+b x)}{2 b^2}-\frac{d \cot (a+b x)}{2 b^2}+\frac{d (2 a-2 (a+b x)) \csc ^2(a+b x)}{4 b^2}+\frac{d (2 (a+b x)-2 a) \sec ^2(a+b x)}{4 b^2}-\frac{2 a d \log (\tan (a+b x))}{b^2}-\frac{c \csc ^2(a+b x)}{2 b}+\frac{c \sec ^2(a+b x)}{2 b}+\frac{2 c \log (\sin (a+b x))}{b}-\frac{2 c \log (\cos (a+b x))}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x]^3,x]
[Out]
-(d*Cot[a + b*x])/(2*b^2) - (c*Csc[a + b*x]^2)/(2*b) + (d*(2*a - 2*(a + b*x))*Csc[a + b*x]^2)/(4*b^2) - (2*c*L
og[Cos[a + b*x]])/b + (2*c*Log[Sin[a + b*x]])/b - (2*a*d*Log[Tan[a + b*x]])/b^2 + (d*(2*(a + b*x)*(Log[1 - E^(
(2*I)*(a + b*x))] - Log[1 + E^((2*I)*(a + b*x))]) + I*(PolyLog[2, -E^((2*I)*(a + b*x))] - PolyLog[2, E^((2*I)*
(a + b*x))])))/b^2 + (c*Sec[a + b*x]^2)/(2*b) + (d*(-2*a + 2*(a + b*x))*Sec[a + b*x]^2)/(4*b^2) - (d*Tan[a + b
*x])/(2*b^2)
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Maple [B] time = 0.209, size = 325, normalized size = 3. \begin{align*} 2\,{\frac{2\,dxb{{\rm e}^{6\,i \left ( bx+a \right ) }}+2\,bc{{\rm e}^{6\,i \left ( bx+a \right ) }}-id{{\rm e}^{6\,i \left ( bx+a \right ) }}+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 ) }}}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) ^{2}}}-2\,{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+2\,{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}+2\,{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}+2\,{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+2\,{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{2\,id{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{id{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-{\frac{2\,id{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-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)^3,x)
[Out]
2/b^2/(exp(2*I*(b*x+a))-1)^2/(exp(2*I*(b*x+a))+1)^2*(2*d*x*b*exp(6*I*(b*x+a))+2*b*c*exp(6*I*(b*x+a))-I*d*exp(6
*I*(b*x+a))+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)))-2/b*c*ln(exp(2*I*(b*x+a))+1)
+2/b*c*ln(exp(I*(b*x+a))-1)+2/b*c*ln(exp(I*(b*x+a))+1)+2/b*d*ln(1-exp(I*(b*x+a)))*x+2/b^2*d*ln(1-exp(I*(b*x+a)
))*a-2*I/b^2*d*polylog(2,exp(I*(b*x+a)))-2/b*d*ln(exp(2*I*(b*x+a))+1)*x+I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2+2
/b*d*ln(exp(I*(b*x+a))+1)*x-2*I*d*polylog(2,-exp(I*(b*x+a)))/b^2-2/b^2*d*a*ln(exp(I*(b*x+a))-1)
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Maxima [B] time = 2.46755, size = 1455, normalized size = 13.23 \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)^3,x, algorithm="maxima")
[Out]
-((2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(8*b*x + 8*a) - 4*(b*d*x + b*c)*cos(4*b*x + 4*a) + (2*I*b*d*x + 2*I*b*
c)*sin(8*b*x + 8*a) + (-4*I*b*d*x - 4*I*b*c)*sin(4*b*x + 4*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(8*b*x + 8*a) - 4*(b*d*x + b*c)*cos(4*b*x + 4*a) - (-2*I*b*d*x - 2*I*
b*c)*sin(8*b*x + 8*a) - (4*I*b*d*x + 4*I*b*c)*sin(4*b*x + 4*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b
*c*cos(8*b*x + 8*a) - 4*b*c*cos(4*b*x + 4*a) + 2*I*b*c*sin(8*b*x + 8*a) - 4*I*b*c*sin(4*b*x + 4*a) + 2*b*c)*ar
ctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(8*b*x + 8*a) - 4*b*d*x*cos(4*b*x + 4*a) + 2*I*b*d*x*sin(8
*b*x + 8*a) - 4*I*b*d*x*sin(4*b*x + 4*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(6*b*x + 6*a) + (4*I*b*d*x + 4*I*b*c - 2*d)*cos(2*b*x + 2*a) - (d*cos(8*b*x + 8*a) - 2*d*cos(4
*b*x + 4*a) + I*d*sin(8*b*x + 8*a) - 2*I*d*sin(4*b*x + 4*a) + d)*dilog(-e^(2*I*b*x + 2*I*a)) + (2*d*cos(8*b*x
+ 8*a) - 4*d*cos(4*b*x + 4*a) + 2*I*d*sin(8*b*x + 8*a) - 4*I*d*sin(4*b*x + 4*a) + 2*d)*dilog(-e^(I*b*x + I*a))
+ (2*d*cos(8*b*x + 8*a) - 4*d*cos(4*b*x + 4*a) + 2*I*d*sin(8*b*x + 8*a) - 4*I*d*sin(4*b*x + 4*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(8*b*x + 8*a) + (2*I*b*d*x + 2*I*b*c)*cos(4*b*x
+ 4*a) + (b*d*x + b*c)*sin(8*b*x + 8*a) - 2*(b*d*x + b*c)*sin(4*b*x + 4*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(8*b*x + 8*a) + (-2*I*b*d*x - 2*
I*b*c)*cos(4*b*x + 4*a) - (b*d*x + b*c)*sin(8*b*x + 8*a) + 2*(b*d*x + b*c)*sin(4*b*x + 4*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(8*b*x + 8*a) + (-2*I*b*d*x
- 2*I*b*c)*cos(4*b*x + 4*a) - (b*d*x + b*c)*sin(8*b*x + 8*a) + 2*(b*d*x + b*c)*sin(4*b*x + 4*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(6*b*x + 6*a) - 2*(2*b*d*x + 2*b*
c + I*d)*sin(2*b*x + 2*a))/(-I*b^2*cos(8*b*x + 8*a) + 2*I*b^2*cos(4*b*x + 4*a) + b^2*sin(8*b*x + 8*a) - 2*b^2*
sin(4*b*x + 4*a) - I*b^2)
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Fricas [B] time = 0.797302, size = 3087, normalized size = 28.06 \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)^3,x, algorithm="fricas")
[Out]
-1/2*(b*d*x - 2*(b*d*x + b*c)*cos(b*x + a)^2 - d*cos(b*x + a)*sin(b*x + a) + b*c - (-2*I*d*cos(b*x + a)^4 + 2*
I*d*cos(b*x + a)^2)*dilog(cos(b*x + a) + I*sin(b*x + a)) - (2*I*d*cos(b*x + a)^4 - 2*I*d*cos(b*x + a)^2)*dilog
(cos(b*x + a) - I*sin(b*x + a)) - (-2*I*d*cos(b*x + a)^4 + 2*I*d*cos(b*x + a)^2)*dilog(I*cos(b*x + a) + sin(b*
x + a)) - (2*I*d*cos(b*x + a)^4 - 2*I*d*cos(b*x + a)^2)*dilog(I*cos(b*x + a) - sin(b*x + a)) - (2*I*d*cos(b*x
+ a)^4 - 2*I*d*cos(b*x + a)^2)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - (-2*I*d*cos(b*x + a)^4 + 2*I*d*cos(b*x
+ a)^2)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - (2*I*d*cos(b*x + a)^4 - 2*I*d*cos(b*x + a)^2)*dilog(-cos(b*x +
a) + I*sin(b*x + a)) - (-2*I*d*cos(b*x + a)^4 + 2*I*d*cos(b*x + a)^2)*dilog(-cos(b*x + a) - I*sin(b*x + a)) -
2*((b*d*x + b*c)*cos(b*x + a)^4 - (b*d*x + b*c)*cos(b*x + a)^2)*log(cos(b*x + a) + I*sin(b*x + a) + 1) + 2*((
b*c - a*d)*cos(b*x + a)^4 - (b*c - a*d)*cos(b*x + a)^2)*log(cos(b*x + a) + I*sin(b*x + a) + I) - 2*((b*d*x + b
*c)*cos(b*x + a)^4 - (b*d*x + b*c)*cos(b*x + a)^2)*log(cos(b*x + a) - I*sin(b*x + a) + 1) + 2*((b*c - a*d)*cos
(b*x + a)^4 - (b*c - a*d)*cos(b*x + a)^2)*log(cos(b*x + a) - I*sin(b*x + a) + I) + 2*((b*d*x + a*d)*cos(b*x +
a)^4 - (b*d*x + a*d)*cos(b*x + a)^2)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 2*((b*d*x + a*d)*cos(b*x + a)^4
- (b*d*x + a*d)*cos(b*x + a)^2)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + 2*((b*d*x + a*d)*cos(b*x + a)^4 - (b*
d*x + a*d)*cos(b*x + a)^2)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 2*((b*d*x + a*d)*cos(b*x + a)^4 - (b*d*x
+ a*d)*cos(b*x + a)^2)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - 2*((b*c - a*d)*cos(b*x + a)^4 - (b*c - a*d)*c
os(b*x + a)^2)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 2*((b*c - a*d)*cos(b*x + a)^4 - (b*c - a*d)
*cos(b*x + a)^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - 2*((b*d*x + a*d)*cos(b*x + a)^4 - (b*d*x
+ a*d)*cos(b*x + a)^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) + 2*((b*c - a*d)*cos(b*x + a)^4 - (b*c - a*d)*c
os(b*x + a)^2)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - 2*((b*d*x + a*d)*cos(b*x + a)^4 - (b*d*x + a*d)*cos(b
*x + a)^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) + 2*((b*c - a*d)*cos(b*x + a)^4 - (b*c - a*d)*cos(b*x + a)^
2)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/(b^2*cos(b*x + a)^4 - b^2*cos(b*x + a)^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)**3,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 )^{3}\,{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)^3,x, algorithm="giac")
[Out]
integrate((d*x + c)*csc(b*x + a)^3*sec(b*x + a)^3, x)