3.281 \(\int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx\)
Optimal. Leaf size=154 \[ \frac{3 i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac{d \csc (a+b x)}{2 b^2}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{3 c \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]
[Out]
(-3*d*x*ArcTanh[E^(I*(a + b*x))])/b - (3*c*ArcTanh[Cos[a + b*x]])/(2*b) - (d*ArcTanh[Sin[a + b*x]])/b^2 - (d*C
sc[a + b*x])/(2*b^2) + (((3*I)/2)*d*PolyLog[2, -E^(I*(a + b*x))])/b^2 - (((3*I)/2)*d*PolyLog[2, E^(I*(a + b*x)
)])/b^2 + (3*(c + d*x)*Sec[a + b*x])/(2*b) - ((c + d*x)*Csc[a + b*x]^2*Sec[a + b*x])/(2*b)
________________________________________________________________________________________
Rubi [A] time = 0.191378, antiderivative size = 174, normalized size of antiderivative =
1.13, number of steps used = 13, number of rules used = 12, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.546, Rules used = {2622, 288, 321, 207, 4420, 6271, 12, 4183, 2279, 2391, 3770, 2621}
\[ \frac{3 i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac{d \csc (a+b x)}{2 b^2}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{3 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
[Out]
(-3*d*x*ArcTanh[E^(I*(a + b*x))])/b + (3*d*x*ArcTanh[Cos[a + b*x]])/(2*b) - (3*(c + d*x)*ArcTanh[Cos[a + b*x]]
)/(2*b) - (d*ArcTanh[Sin[a + b*x]])/b^2 - (d*Csc[a + b*x])/(2*b^2) + (((3*I)/2)*d*PolyLog[2, -E^(I*(a + b*x))]
)/b^2 - (((3*I)/2)*d*PolyLog[2, E^(I*(a + b*x))])/b^2 + (3*(c + d*x)*Sec[a + b*x])/(2*b) - ((c + d*x)*Csc[a +
b*x]^2*Sec[a + b*x])/(2*b)
Rule 2622
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[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 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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 6271
Int[ArcTanh[u_], x_Symbol] :> Simp[x*ArcTanh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 - u^2), x], x] /; I
nverseFunctionFreeQ[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]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 2621
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rubi steps
\begin{align*} \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx &=-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}-d \int \left (-\frac{3 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 \sec (a+b x)}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b}\right ) \, dx\\ &=-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{d \int \csc ^2(a+b x) \sec (a+b x) \, dx}{2 b}+\frac{(3 d) \int \tanh ^{-1}(\cos (a+b x)) \, dx}{2 b}-\frac{(3 d) \int \sec (a+b x) \, dx}{2 b}\\ &=\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 d \tanh ^{-1}(\sin (a+b x))}{2 b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{d \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b^2}+\frac{(3 d) \int b x \csc (a+b x) \, dx}{2 b}\\ &=\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 d \tanh ^{-1}(\sin (a+b x))}{2 b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{1}{2} (3 d) \int x \csc (a+b x) \, dx-\frac{d \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b^2}\\ &=-\frac{3 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{(3 d) \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{2 b}+\frac{(3 d) \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=-\frac{3 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{(3 i d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac{(3 i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=-\frac{3 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 d x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{3 (c+d x) \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{3 i d \text{Li}_2\left (-e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{Li}_2\left (e^{i (a+b x)}\right )}{2 b^2}+\frac{3 (c+d x) \sec (a+b x)}{2 b}-\frac{(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b}\\ \end{align*}
Mathematica [B] time = 5.32919, size = 520, normalized size = 3.38 \[ \frac{3 d \left (i \left (\text{PolyLog}\left (2,-e^{i (a+b x)}\right )-\text{PolyLog}\left (2,e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )\right )}{2 b^2}-\frac{d \tan \left (\frac{1}{2} (a+b x)\right )}{4 b^2}-\frac{d \cot \left (\frac{1}{2} (a+b x)\right )}{4 b^2}-\frac{3 a d \log \left (\tan \left (\frac{1}{2} (a+b x)\right )\right )}{2 b^2}+\frac{d \left (a \sin \left (\frac{1}{2} (a+b x)\right )-(a+b x) \sin \left (\frac{1}{2} (a+b x)\right )\right )}{b^2 \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )}+\frac{d \left ((a+b x) \sin \left (\frac{1}{2} (a+b x)\right )-a \sin \left (\frac{1}{2} (a+b x)\right )\right )}{b^2 \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}+\frac{d \log \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}-\frac{d \log \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}-\frac{c \csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{c \sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{3 c \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}-\frac{3 c \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}+\frac{c \sin \left (\frac{1}{2} (a+b x)\right )}{b \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}-\frac{c \sin \left (\frac{1}{2} (a+b x)\right )}{b \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )}-\frac{d x \csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{d x \sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{d x}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
[Out]
(d*x)/b - (d*Cot[(a + b*x)/2])/(4*b^2) - (c*Csc[(a + b*x)/2]^2)/(8*b) - (d*x*Csc[(a + b*x)/2]^2)/(8*b) - (3*c*
Log[Cos[(a + b*x)/2]])/(2*b) + (d*Log[Cos[(a + b*x)/2] - Sin[(a + b*x)/2]])/b^2 + (3*c*Log[Sin[(a + b*x)/2]])/
(2*b) - (d*Log[Cos[(a + b*x)/2] + Sin[(a + b*x)/2]])/b^2 - (3*a*d*Log[Tan[(a + b*x)/2]])/(2*b^2) + (3*d*((a +
b*x)*(Log[1 - E^(I*(a + b*x))] - Log[1 + E^(I*(a + b*x))]) + I*(PolyLog[2, -E^(I*(a + b*x))] - PolyLog[2, E^(I
*(a + b*x))])))/(2*b^2) + (c*Sec[(a + b*x)/2]^2)/(8*b) + (d*x*Sec[(a + b*x)/2]^2)/(8*b) + (c*Sin[(a + b*x)/2])
/(b*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])) - (c*Sin[(a + b*x)/2])/(b*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) +
(d*(a*Sin[(a + b*x)/2] - (a + b*x)*Sin[(a + b*x)/2]))/(b^2*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) + (d*(-(a*Si
n[(a + b*x)/2]) + (a + b*x)*Sin[(a + b*x)/2]))/(b^2*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])) - (d*Tan[(a + b*x)/
2])/(4*b^2)
________________________________________________________________________________________
Maple [A] time = 0.352, size = 267, normalized size = 1.7 \begin{align*}{\frac{3\,dxb{{\rm e}^{5\,i \left ( bx+a \right ) }}+3\,bc{{\rm e}^{5\,i \left ( bx+a \right ) }}-2\,dxb{{\rm e}^{3\,i \left ( bx+a \right ) }}-2\,bc{{\rm e}^{3\,i \left ( bx+a \right ) }}-id{{\rm e}^{5\,i \left ( bx+a \right ) }}+3\,dxb{{\rm e}^{i \left ( bx+a \right ) }}+3\,bc{{\rm e}^{i \left ( bx+a \right ) }}+id{{\rm e}^{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 ) }}-{\frac{3\,c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{2\,b}}+{\frac{3\,c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{2\,b}}-{\frac{3\,da\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{2\,{b}^{2}}}+{\frac{2\,id\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{{\frac{3\,i}{2}}d{\it dilog} \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-{\frac{3\,d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{2\,b}}+{\frac{{\frac{3\,i}{2}}d{\it dilog} \left ({{\rm e}^{i \left ( bx+a \right ) }} \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)^2,x)
[Out]
1/b^2/(exp(2*I*(b*x+a))-1)^2/(exp(2*I*(b*x+a))+1)*(3*d*x*b*exp(5*I*(b*x+a))+3*b*c*exp(5*I*(b*x+a))-2*d*x*b*exp
(3*I*(b*x+a))-2*b*c*exp(3*I*(b*x+a))-I*d*exp(5*I*(b*x+a))+3*d*x*b*exp(I*(b*x+a))+3*b*c*exp(I*(b*x+a))+I*d*exp(
I*(b*x+a)))-3/2/b*c*ln(exp(I*(b*x+a))+1)+3/2/b*c*ln(exp(I*(b*x+a))-1)-3/2/b^2*d*a*ln(exp(I*(b*x+a))-1)+2*I/b^2
*d*arctan(exp(I*(b*x+a)))+3/2*I/b^2*d*dilog(exp(I*(b*x+a))+1)-3/2/b*d*ln(exp(I*(b*x+a))+1)*x+3/2*I/b^2*d*dilog
(exp(I*(b*x+a)))
________________________________________________________________________________________
Maxima [B] time = 2.80594, size = 2029, normalized size = 13.18 \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)^2,x, algorithm="maxima")
[Out]
-((4*d*cos(6*b*x + 6*a) - 4*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*x + 2*a) + 4*I*d*sin(6*b*x + 6*a) - 4*I*d*sin(4*b
*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 4*d)*arctan2(2*(cos(b*x + 2*a)*cos(a) + sin(b*x + 2*a)*sin(a))/(cos(b*x +
2*a)^2 + cos(a)^2 + 2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 - 2*cos(b*x + 2*a)*sin(a) + sin(a)^2), (cos(b*
x + 2*a)^2 - cos(a)^2 + sin(b*x + 2*a)^2 - sin(a)^2)/(cos(b*x + 2*a)^2 + cos(a)^2 + 2*cos(a)*sin(b*x + 2*a) +
sin(b*x + 2*a)^2 - 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)) + (6*b*d*x + 6*b*c + 6*(b*d*x + b*c)*cos(6*b*x + 6*a)
- 6*(b*d*x + b*c)*cos(4*b*x + 4*a) - 6*(b*d*x + b*c)*cos(2*b*x + 2*a) - (-6*I*b*d*x - 6*I*b*c)*sin(6*b*x + 6*a
) - (6*I*b*d*x + 6*I*b*c)*sin(4*b*x + 4*a) - (6*I*b*d*x + 6*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos
(b*x + a) + 1) - (6*b*c*cos(6*b*x + 6*a) - 6*b*c*cos(4*b*x + 4*a) - 6*b*c*cos(2*b*x + 2*a) + 6*I*b*c*sin(6*b*x
+ 6*a) - 6*I*b*c*sin(4*b*x + 4*a) - 6*I*b*c*sin(2*b*x + 2*a) + 6*b*c)*arctan2(sin(b*x + a), cos(b*x + a) - 1)
+ (6*b*d*x*cos(6*b*x + 6*a) - 6*b*d*x*cos(4*b*x + 4*a) - 6*b*d*x*cos(2*b*x + 2*a) + 6*I*b*d*x*sin(6*b*x + 6*a
) - 6*I*b*d*x*sin(4*b*x + 4*a) - 6*I*b*d*x*sin(2*b*x + 2*a) + 6*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1
) - (-12*I*b*d*x - 12*I*b*c - 4*d)*cos(5*b*x + 5*a) - (8*I*b*d*x + 8*I*b*c)*cos(3*b*x + 3*a) - (-12*I*b*d*x -
12*I*b*c + 4*d)*cos(b*x + a) - (6*d*cos(6*b*x + 6*a) - 6*d*cos(4*b*x + 4*a) - 6*d*cos(2*b*x + 2*a) + 6*I*d*sin
(6*b*x + 6*a) - 6*I*d*sin(4*b*x + 4*a) - 6*I*d*sin(2*b*x + 2*a) + 6*d)*dilog(-e^(I*b*x + I*a)) + (6*d*cos(6*b*
x + 6*a) - 6*d*cos(4*b*x + 4*a) - 6*d*cos(2*b*x + 2*a) + 6*I*d*sin(6*b*x + 6*a) - 6*I*d*sin(4*b*x + 4*a) - 6*I
*d*sin(2*b*x + 2*a) + 6*d)*dilog(e^(I*b*x + I*a)) - (3*I*b*d*x + 3*I*b*c + (3*I*b*d*x + 3*I*b*c)*cos(6*b*x + 6
*a) + (-3*I*b*d*x - 3*I*b*c)*cos(4*b*x + 4*a) + (-3*I*b*d*x - 3*I*b*c)*cos(2*b*x + 2*a) - 3*(b*d*x + b*c)*sin(
6*b*x + 6*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))*log(cos(b*x + a)^2 + sin(b
*x + a)^2 + 2*cos(b*x + a) + 1) - (-3*I*b*d*x - 3*I*b*c + (-3*I*b*d*x - 3*I*b*c)*cos(6*b*x + 6*a) + (3*I*b*d*x
+ 3*I*b*c)*cos(4*b*x + 4*a) + (3*I*b*d*x + 3*I*b*c)*cos(2*b*x + 2*a) + 3*(b*d*x + b*c)*sin(6*b*x + 6*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))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(
b*x + a) + 1) - (-2*I*d*cos(6*b*x + 6*a) + 2*I*d*cos(4*b*x + 4*a) + 2*I*d*cos(2*b*x + 2*a) + 2*d*sin(6*b*x + 6
*a) - 2*d*sin(4*b*x + 4*a) - 2*d*sin(2*b*x + 2*a) - 2*I*d)*log((cos(b*x + 2*a)^2 + cos(a)^2 - 2*cos(a)*sin(b*x
+ 2*a) + sin(b*x + 2*a)^2 + 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)/(cos(b*x + 2*a)^2 + cos(a)^2 + 2*cos(a)*sin(b
*x + 2*a) + sin(b*x + 2*a)^2 - 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)) - 4*(3*b*d*x + 3*b*c - I*d)*sin(5*b*x + 5*
a) + 8*(b*d*x + b*c)*sin(3*b*x + 3*a) - 4*(3*b*d*x + 3*b*c + I*d)*sin(b*x + a))/(-4*I*b^2*cos(6*b*x + 6*a) + 4
*I*b^2*cos(4*b*x + 4*a) + 4*I*b^2*cos(2*b*x + 2*a) + 4*b^2*sin(6*b*x + 6*a) - 4*b^2*sin(4*b*x + 4*a) - 4*b^2*s
in(2*b*x + 2*a) - 4*I*b^2)
________________________________________________________________________________________
Fricas [B] time = 0.633759, size = 1652, normalized size = 10.73 \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)^2,x, algorithm="fricas")
[Out]
-1/4*(4*b*d*x - 6*(b*d*x + b*c)*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x + a) + 4*b*c - (-3*I*d*cos(b*x + a)^
3 + 3*I*d*cos(b*x + a))*dilog(cos(b*x + a) + I*sin(b*x + a)) - (3*I*d*cos(b*x + a)^3 - 3*I*d*cos(b*x + a))*dil
og(cos(b*x + a) - I*sin(b*x + a)) - (-3*I*d*cos(b*x + a)^3 + 3*I*d*cos(b*x + a))*dilog(-cos(b*x + a) + I*sin(b
*x + a)) - (3*I*d*cos(b*x + a)^3 - 3*I*d*cos(b*x + a))*dilog(-cos(b*x + a) - I*sin(b*x + a)) + 3*((b*d*x + b*c
)*cos(b*x + a)^3 - (b*d*x + b*c)*cos(b*x + a))*log(cos(b*x + a) + I*sin(b*x + a) + 1) + 3*((b*d*x + b*c)*cos(b
*x + a)^3 - (b*d*x + b*c)*cos(b*x + a))*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 3*((b*c - a*d)*cos(b*x + a)^3
- (b*c - a*d)*cos(b*x + a))*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 3*((b*c - a*d)*cos(b*x + a)^3
- (b*c - a*d)*cos(b*x + a))*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - 3*((b*d*x + a*d)*cos(b*x + a)
^3 - (b*d*x + a*d)*cos(b*x + a))*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - 3*((b*d*x + a*d)*cos(b*x + a)^3 - (
b*d*x + a*d)*cos(b*x + a))*log(-cos(b*x + a) - I*sin(b*x + a) + 1) + 2*(d*cos(b*x + a)^3 - d*cos(b*x + a))*log
(sin(b*x + a) + 1) - 2*(d*cos(b*x + a)^3 - d*cos(b*x + a))*log(-sin(b*x + a) + 1))/(b^2*cos(b*x + a)^3 - b^2*c
os(b*x + a))
________________________________________________________________________________________
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)**2,x)
[Out]
Timed out
________________________________________________________________________________________
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 )^{2}\,{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)^2,x, algorithm="giac")
[Out]
integrate((d*x + c)*csc(b*x + a)^3*sec(b*x + a)^2, x)