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

Optimal. Leaf size=131 \[ \frac{i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{(c+d x) \csc (a+b x)}{b}+\frac{(c+d x) \tanh ^{-1}(\sin (a+b x))}{b}-\frac{2 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{d x \tanh ^{-1}(\sin (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.134538, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2621, 321, 207, 4420, 6271, 12, 4181, 2279, 2391, 3770} \[ \frac{i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac{d \tanh ^{-1}(\cos (a+b x))}{b^2}-\frac{(c+d x) \csc (a+b x)}{b}+\frac{(c+d x) \tanh ^{-1}(\sin (a+b x))}{b}-\frac{2 i d x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{d x \tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

Mathematica [C]  time = 2.94677, size = 517, normalized size = 3.95 \[ -\frac{c \csc (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\sin ^2(a+b x)\right )}{b}-\frac{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 )}{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 \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} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Csc[a + b*x]^2*Sec[a + b*x],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, 1, 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 - (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]]*Lo
g[(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])))/(b*(a - I*Log[1 - I*Tan[(a + b*x)/2]] + I*Log[1 + I*Tan[(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)

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Maple [A]  time = 0.299, size = 235, normalized size = 1.8 \begin{align*}{\frac{-2\,i{{\rm e}^{i \left ( bx+a \right ) }} \left ( dx+c \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) }}+{\frac{2\,iad\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{2\,ic\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}-{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{id{\it dilog} \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{id{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 0.607381, size = 1215, normalized size = 9.27 \begin{align*} -\frac{2 \, b d x + i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) -{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) +{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + d \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) \sin \left (b x + a\right ) -{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) +{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) -{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) +{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - d \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) \sin \left (b x + a\right ) -{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) +{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + 2 \, b c}{2 \, b^{2} \sin \left (b x + a\right )} \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),x, algorithm="fricas")

[Out]

-1/2*(2*b*d*x + I*d*dilog(I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) + I*d*dilog(I*cos(b*x + a) - sin(b*x + a
))*sin(b*x + a) - I*d*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - I*d*dilog(-I*cos(b*x + a) - sin(b*x
+ a))*sin(b*x + a) - (b*c - a*d)*log(cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) + (b*c - a*d)*log(cos(b*
x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + d*log(1/2*cos(b*x + a) + 1/2)*sin(b*x + a) - (b*d*x + a*d)*log(I*c
os(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d*x + a*d)*log(I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x +
a) - (b*d*x + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d*x + a*d)*log(-I*cos(b*x + a) -
sin(b*x + a) + 1)*sin(b*x + a) - d*log(-1/2*cos(b*x + a) + 1/2)*sin(b*x + a) - (b*c - a*d)*log(-cos(b*x + a) +
I*sin(b*x + a) + I)*sin(b*x + a) + (b*c - a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + 2*b*c)/
(b^2*sin(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \csc ^{2}{\left (a + b x \right )} \sec{\left (a + b x \right )}\, dx \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),x)

[Out]

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

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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 )^{2} \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)^2*sec(b*x+a),x, algorithm="giac")

[Out]

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