\(\int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx\) [237]

Optimal result

Integrand size = 20, antiderivative size = 131 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {2 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}+\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2} \]

[Out]

-2*I*d*x*arctan(exp(I*(b*x+a)))/b-d*arctanh(cos(b*x+a))/b^2-d*x*arctanh(sin(b*x+a))/b+(d*x+c)*arctanh(sin(b*x+
a))/b-(d*x+c)*csc(b*x+a)/b+I*d*polylog(2,-I*exp(I*(b*x+a)))/b^2-I*d*polylog(2,I*exp(I*(b*x+a)))/b^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2701, 327, 213, 4505, 6406, 12, 4266, 2317, 2438, 3855} \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {2 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x) \csc (a+b x)}{b} \]

[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

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 2317

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 2438

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 2701

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4266

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 4505

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 6406

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}-d \int \left (\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b}\right ) \, dx \\ & = \frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}-\frac {d \int \text {arctanh}(\sin (a+b x)) \, dx}{b}+\frac {d \int \csc (a+b x) \, dx}{b} \\ & = -\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\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 \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\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 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\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 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}+\frac {(i d) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\frac {(i d) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2} \\ & = -\frac {2 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}-\frac {d x \text {arctanh}(\sin (a+b x))}{b}+\frac {(c+d x) \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x) \csc (a+b x)}{b}+\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.83 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.95 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\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 {c \csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\sin ^2(a+b x)\right )}{b}-\frac {d \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {d \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}-\frac {d x \left (a \log \left (1-\tan \left (\frac {1}{2} (a+b x)\right )\right )-a \log \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )-i \left (\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 (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)-(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )+i \left (\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 (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )-i \left (\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 (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)+(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )+i \left (\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 (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1-i)+(1+i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )\right )}{b \left (a-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 )\right )}+\frac {d \sec \left (\frac {1}{2} (a+b x)\right ) \left (a \sin \left (\frac {1}{2} (a+b x)\right )-(a+b x) \sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b^2} \]

[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)

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.79

method	result	size
risch	\(-\frac {2 i \left (d x +c \right ) {\mathrm e}^{i \left (x b +a \right )}}{b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b}-\frac {2 i c \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}+\frac {i d \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {i d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {2 i a d \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{2}}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{2}}\)	\(235\)

[In]

int((d*x+c)*csc(b*x+a)^2*sec(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (113) = 226\).

Time = 0.29 (sec) , antiderivative size = 434, normalized size of antiderivative = 3.31 \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=-\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 )} \]

[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))

Sympy [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\int \left (c + d x\right ) \csc ^{2}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

[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)

Maxima [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right ) \,d x } \]

[In]

integrate((d*x+c)*csc(b*x+a)^2*sec(b*x+a),x, algorithm="maxima")

[Out]

-1/2*(4*(b*d*x + b*c)*cos(b*x + a)*sin(2*b*x + 2*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a)*sin(b*x + a) - 4*(b^2*d
*cos(2*b*x + 2*a)^2 + b^2*d*sin(2*b*x + 2*a)^2 - 2*b^2*d*cos(2*b*x + 2*a) + b^2*d)*integrate((x*cos(2*b*x + 2*
a)*cos(b*x + a) + x*sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x + a))/(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 +
2*cos(2*b*x + 2*a) + 1), x) - (b*c*cos(2*b*x + 2*a)^2 + b*c*sin(2*b*x + 2*a)^2 - 2*b*c*cos(2*b*x + 2*a) + b*c
)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + (b*c*cos(2*b*x + 2*a)^2 + b*c*sin(2*b*x + 2*a)^2
- 2*b*c*cos(2*b*x + 2*a) + b*c)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) + (d*cos(2*b*x + 2*
a)^2 + d*sin(2*b*x + 2*a)^2 - 2*d*cos(2*b*x + 2*a) + d)*log(cos(b*x)^2 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*
x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - (d*cos(2*b*x + 2*a)^2 + d*sin(2*b*x + 2*a)^2 - 2*d*cos(2*b*x + 2*a) + d
)*log(cos(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2) + 4*(b*d*x + b*c)
*sin(b*x + a))/(b^2*cos(2*b*x + 2*a)^2 + b^2*sin(2*b*x + 2*a)^2 - 2*b^2*cos(2*b*x + 2*a) + b^2)

Giac [F]

\[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \csc ^2(a+b x) \sec (a+b x) \, dx=\text {Hanged} \]

[In]

int((c + d*x)/(cos(a + b*x)*sin(a + b*x)^2),x)

[Out]

\text{Hanged}