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

Optimal. Leaf size=162 \[ \frac{3 i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac{d \sec (a+b x)}{2 b^2}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{3 (c+d x) \csc (a+b x)}{2 b}+\frac{(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}+\frac{3 c \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.195867, antiderivative size = 182, normalized size of antiderivative = 1.12, 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 = {2621, 288, 321, 207, 4420, 6271, 12, 4181, 2279, 2391, 3770, 2622} \[ \frac{3 i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac{d \sec (a+b x)}{2 b^2}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{3 (c+d x) \csc (a+b x)}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}-\frac{3 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 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])

Rubi steps

\begin{align*} \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx &=\frac{3 (c+d x) \tanh ^{-1}(\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}-d \int \left (\frac{3 \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 \csc (a+b x)}{2 b}+\frac{\csc (a+b x) \sec ^2(a+b x)}{2 b}\right ) \, dx\\ &=\frac{3 (c+d x) \tanh ^{-1}(\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}-\frac{d \int \csc (a+b x) \sec ^2(a+b x) \, dx}{2 b}-\frac{(3 d) \int \tanh ^{-1}(\sin (a+b x)) \, dx}{2 b}+\frac{(3 d) \int \csc (a+b x) \, dx}{2 b}\\ &=-\frac{3 d \tanh ^{-1}(\cos (a+b x))}{2 b^2}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\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}-\frac{d \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{2 b^2}+\frac{(3 d) \int b x \sec (a+b x) \, dx}{2 b}\\ &=-\frac{3 d \tanh ^{-1}(\cos (a+b x))}{2 b^2}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 (c+d x) \csc (a+b x)}{2 b}-\frac{d \sec (a+b x)}{2 b^2}+\frac{(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}+\frac{1}{2} (3 d) \int x \sec (a+b x) \, dx-\frac{d \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{2 b^2}\\ &=-\frac{3 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 (c+d x) \csc (a+b x)}{2 b}-\frac{d \sec (a+b x)}{2 b^2}+\frac{(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}-\frac{(3 d) \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac{(3 d) \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=-\frac{3 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 (c+d x) \csc (a+b x)}{2 b}-\frac{d \sec (a+b x)}{2 b^2}+\frac{(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}+\frac{(3 i d) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac{(3 i d) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=-\frac{3 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{3 d x \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{3 (c+d x) \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{3 (c+d x) \csc (a+b x)}{2 b}+\frac{3 i d \text{Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}-\frac{3 i d \text{Li}_2\left (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}\\ \end{align*}

Mathematica [C]  time = 6.57654, size = 669, normalized size = 4.13 \[ -\frac{c \csc (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{2},2,\frac{1}{2},\sin ^2(a+b x)\right )}{b}-\frac{3 d x \left (-i \left (\text{PolyLog}\left (2,\frac{1}{2} \left ((1+i)-(1-i) \tan \left (\frac{1}{2} (a+b x)\right )\right )\right )+\log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )\right )+i \left (\text{PolyLog}\left (2,\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+i\right )\right )+\log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (\frac{1}{2}+\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )\right )-i \left (\text{PolyLog}\left (2,\frac{1}{2} \left ((1-i) \tan \left (\frac{1}{2} (a+b x)\right )+(1+i)\right )\right )+\log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac{1}{2}+\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )-1\right )\right )\right )+i \left (\text{PolyLog}\left (2,\frac{1}{2} \left ((1+i) \tan \left (\frac{1}{2} (a+b x)\right )+(1-i)\right )\right )+\log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )-1\right )\right )\right )+a \log \left (1-\tan \left (\frac{1}{2} (a+b x)\right )\right )-a \log \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )}{2 b \left (-i \log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right )+a\right )}+\frac{d \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}-\frac{d \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}-\frac{d \sin \left (\frac{1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}+\frac{d \sin \left (\frac{1}{2} (a+b x)\right )}{2 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 \cos \left (\frac{1}{2} (a+b x)\right )-(a+b x) \cos \left (\frac{1}{2} (a+b x)\right )\right ) \csc \left (\frac{1}{2} (a+b x)\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 ) \sec \left (\frac{1}{2} (a+b x)\right )}{2 b^2}+\frac{d x}{4 b \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )^2}-\frac{d x}{4 b \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(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]*Hypergeometri
c2F1[-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*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])] + PolyLog[2, ((1 - I) + (1 + I)*Tan
[(a + b*x)/2])/2])))/(2*b*(a - I*Log[1 - I*Tan[(a + b*x)/2]] + I*Log[1 + I*Tan[(a + b*x)/2]])) + (d*x)/(4*b*(C
os[(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)

________________________________________________________________________________________

Maple [B]  time = 0.355, size = 344, normalized size = 2.1 \begin{align*}{\frac{-i \left ( 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 ) }} \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\,ic\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{3\,ida\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}+{\frac{{\frac{3\,i}{2}}d{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{3\,d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{2\,b}}-{\frac{3\,d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{2\,{b}^{2}}}+{\frac{3\,d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{2\,b}}+{\frac{3\,d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{2\,{b}^{2}}}-{\frac{{\frac{3\,i}{2}}d{\it dilog} \left ( 1-i{{\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)^2*sec(b*x+a)^3,x)

[Out]

-I/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*ex
p(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*I/b*c*arctan(exp(I*(b*x+a)))+3*I/b^2*d*a*arctan(exp(I*(b*x+a)))+d/b^2*ln(exp(I*(b*x+a))-1)-d/b^
2*ln(exp(I*(b*x+a))+1)+3/2*I/b^2*d*dilog(1+I*exp(I*(b*x+a)))-3/2/b*d*ln(1+I*exp(I*(b*x+a)))*x-3/2/b^2*d*ln(1+I
*exp(I*(b*x+a)))*a+3/2/b*d*ln(1-I*exp(I*(b*x+a)))*x+3/2/b^2*d*ln(1-I*exp(I*(b*x+a)))*a-3/2*I/b^2*d*dilog(1-I*e
xp(I*(b*x+a)))

________________________________________________________________________________________

Maxima [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)^2*sec(b*x+a)^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 0.696318, size = 1644, normalized size = 10.15 \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)^2*sec(b*x+a)^3,x, algorithm="fricas")

[Out]

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*co
s(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*l
og(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*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) - 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)*s
in(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(-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)**2*sec(b*x+a)**3,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 )^{2} \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)^2*sec(b*x+a)^3,x, algorithm="giac")

[Out]

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