\(\int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) [281]

Optimal result

Integrand size = 22, antiderivative size = 154 \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {3 d x \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {3 c \text {arctanh}(\cos (a+b x))}{2 b}-\frac {d \text {arctanh}(\sin (a+b x))}{b^2}-\frac {d \csc (a+b x)}{2 b^2}+\frac {3 i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}+\frac {3 (c+d x) \sec (a+b x)}{2 b}-\frac {(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b} \]

[Out]

-3*d*x*arctanh(exp(I*(b*x+a)))/b-3/2*c*arctanh(cos(b*x+a))/b-d*arctanh(sin(b*x+a))/b^2-1/2*d*csc(b*x+a)/b^2+3/
2*I*d*polylog(2,-exp(I*(b*x+a)))/b^2-3/2*I*d*polylog(2,exp(I*(b*x+a)))/b^2+3/2*(d*x+c)*sec(b*x+a)/b-1/2*(d*x+c
)*csc(b*x+a)^2*sec(b*x+a)/b

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {2702, 294, 327, 213, 4505, 6406, 12, 4268, 2317, 2438, 3855, 2701} \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {d \text {arctanh}(\sin (a+b x))}{b^2}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 d x \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d x \text {arctanh}(\cos (a+b x))}{2 b}+\frac {3 i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {d \csc (a+b x)}{2 b^2}+\frac {3 (c+d x) \sec (a+b x)}{2 b}-\frac {(c+d x) \csc ^2(a+b x) \sec (a+b x)}{2 b} \]

[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 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 294

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 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 2702

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 3855

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

Rule 4268

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 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 {3 (c+d x) \text {arctanh}(\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 \text {arctanh}(\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) \text {arctanh}(\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 \text {arctanh}(\cos (a+b x)) \, dx}{2 b}-\frac {(3 d) \int \sec (a+b x) \, dx}{2 b} \\ & = \frac {3 d x \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 d \text {arctanh}(\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 \text {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 \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 d \text {arctanh}(\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 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b^2} \\ & = -\frac {3 d x \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d x \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {d \text {arctanh}(\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 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d x \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {d \text {arctanh}(\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) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(3 i d) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2} \\ & = -\frac {3 d x \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {3 d x \text {arctanh}(\cos (a+b x))}{2 b}-\frac {3 (c+d x) \text {arctanh}(\cos (a+b x))}{2 b}-\frac {d \text {arctanh}(\sin (a+b x))}{b^2}-\frac {d \csc (a+b x)}{2 b^2}+\frac {3 i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d \operatorname {PolyLog}\left (2,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] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(520\) vs. \(2(154)=308\).

Time = 6.16 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.38 \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\frac {d x}{b}-\frac {d \cot \left (\frac {1}{2} (a+b x)\right )}{4 b^2}-\frac {c \csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {d x \csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {3 c \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {d \log \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {3 c \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {d \log \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}-\frac {3 a d \log \left (\tan \left (\frac {1}{2} (a+b x)\right )\right )}{2 b^2}+\frac {3 d \left ((a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )+i \left (\operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )-\operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )\right )\right )}{2 b^2}+\frac {c \sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {d x \sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {c \sin \left (\frac {1}{2} (a+b x)\right )}{b \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {c \sin \left (\frac {1}{2} (a+b x)\right )}{b \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}+\frac {d \left (a \sin \left (\frac {1}{2} (a+b x)\right )-(a+b x) \sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}+\frac {d \left (-a \sin \left (\frac {1}{2} (a+b x)\right )+(a+b x) \sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {d \tan \left (\frac {1}{2} (a+b x)\right )}{4 b^2} \]

[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] (verified)

Time = 1.43 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.73

method	result	size
risch	\(\frac {3 d x b \,{\mathrm e}^{5 i \left (x b +a \right )}+3 c b \,{\mathrm e}^{5 i \left (x b +a \right )}-2 d x b \,{\mathrm e}^{3 i \left (x b +a \right )}-2 c b \,{\mathrm e}^{3 i \left (x b +a \right )}-i d \,{\mathrm e}^{5 i \left (x b +a \right )}+3 d x b \,{\mathrm e}^{i \left (x b +a \right )}+3 c b \,{\mathrm e}^{i \left (x b +a \right )}+i d \,{\mathrm e}^{i \left (x b +a \right )}}{b^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}-\frac {3 d \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{2 b}-\frac {3 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{2 b}+\frac {3 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{2 b}-\frac {3 d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{2 b^{2}}+\frac {2 i d \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {3 i d \operatorname {dilog}\left ({\mathrm e}^{i \left (x b +a \right )}\right )}{2 b^{2}}+\frac {3 i d \operatorname {dilog}\left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{2 b^{2}}\)	\(267\)

[In]

int((d*x+c)*csc(b*x+a)^3*sec(b*x+a)^2,x,method=_RETURNVERBOSE)

[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*c*b*exp(5*I*(b*x+a))-2*d*x*b*exp
(3*I*(b*x+a))-2*c*b*exp(3*I*(b*x+a))-I*d*exp(5*I*(b*x+a))+3*d*x*b*exp(I*(b*x+a))+3*c*b*exp(I*(b*x+a))+I*d*exp(
I*(b*x+a)))-3/2/b*d*ln(exp(I*(b*x+a))+1)*x-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)))+3/2*I/b^2*d*dilog(e
xp(I*(b*x+a))+1)

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. 621 vs. \(2 (130) = 260\).

Time = 0.28 (sec) , antiderivative size = 621, normalized size of antiderivative = 4.03 \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=-\frac {4 \, b d x - 6 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, b c + 3 \, {\left (i \, d \cos \left (b x + a\right )^{3} - i \, d \cos \left (b x + a\right )\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 3 \, {\left (-i \, d \cos \left (b x + a\right )^{3} + i \, d \cos \left (b x + a\right )\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 3 \, {\left (i \, d \cos \left (b x + a\right )^{3} - i \, d \cos \left (b x + a\right )\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 3 \, {\left (-i \, d \cos \left (b x + a\right )^{3} + i \, d \cos \left (b x + a\right )\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 3 \, {\left ({\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (b d x + b c\right )} \cos \left (b x + a\right )\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left ({\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (b d x + b c\right )} \cos \left (b x + a\right )\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left ({\left (b c - a d\right )} \cos \left (b x + a\right )^{3} - {\left (b c - a d\right )} \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (b c - a d\right )} \cos \left (b x + a\right )^{3} - {\left (b c - a d\right )} \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (b d x + a d\right )} \cos \left (b x + a\right )^{3} - {\left (b d x + a d\right )} \cos \left (b x + a\right )\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left ({\left (b d x + a d\right )} \cos \left (b x + a\right )^{3} - {\left (b d x + a d\right )} \cos \left (b x + a\right )\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (d \cos \left (b x + a\right )^{3} - d \cos \left (b x + a\right )\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (d \cos \left (b x + a\right )^{3} - d \cos \left (b x + a\right )\right )} \log \left (-\sin \left (b x + a\right ) + 1\right )}{4 \, {\left (b^{2} \cos \left (b x + a\right )^{3} - b^{2} \cos \left (b x + a\right )\right )}} \]

[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
- I*d*cos(b*x + a))*dilog(cos(b*x + a) + I*sin(b*x + a)) + 3*(-I*d*cos(b*x + a)^3 + I*d*cos(b*x + a))*dilog(c
os(b*x + a) - I*sin(b*x + a)) + 3*(I*d*cos(b*x + a)^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 + 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*cos(b*x +
a))

Sympy [F]

\[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\int \left (c + d x\right ) \csc ^{3}{\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1481 vs. \(2 (130) = 260\).

Time = 0.58 (sec) , antiderivative size = 1481, normalized size of antiderivative = 9.62 \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \]

[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) - d*cos(4*b*x + 4*a) - d*cos(2*b*x + 2*a) + I*d*sin(6*b*x + 6*a) - I*d*sin(4*b*x + 4*a
) - I*d*sin(2*b*x + 2*a) + d)*arctan2(2*(cos(b*x + 2*a)*cos(a) + sin(b*x + 2*a)*sin(a))/(cos(b*x + 2*a)^2 + co
s(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 + b*c + (b*d*x + b*c)*cos(6*b*x + 6*a) - (b*d*x + b*c)*
cos(4*b*x + 4*a) - (b*d*x + b*c)*cos(2*b*x + 2*a) + (I*b*d*x + I*b*c)*sin(6*b*x + 6*a) + (-I*b*d*x - I*b*c)*si
n(4*b*x + 4*a) + (-I*b*d*x - 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) - b*c*cos(4*b*x + 4*a) - b*c*cos(2*b*x + 2*a) + I*b*c*sin(6*b*x + 6*a) - I*b*c*sin(4*b*x + 4*a) - I*
b*c*sin(2*b*x + 2*a) + b*c)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + 6*(b*d*x*cos(6*b*x + 6*a) - b*d*x*cos(4*
b*x + 4*a) - b*d*x*cos(2*b*x + 2*a) + I*b*d*x*sin(6*b*x + 6*a) - I*b*d*x*sin(4*b*x + 4*a) - I*b*d*x*sin(2*b*x
+ 2*a) + b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 4*(3*I*b*d*x + 3*I*b*c + d)*cos(5*b*x + 5*a) + 8*(-
I*b*d*x - I*b*c)*cos(3*b*x + 3*a) + 4*(3*I*b*d*x + 3*I*b*c - d)*cos(b*x + a) - 6*(d*cos(6*b*x + 6*a) - d*cos(4
*b*x + 4*a) - d*cos(2*b*x + 2*a) + I*d*sin(6*b*x + 6*a) - I*d*sin(4*b*x + 4*a) - I*d*sin(2*b*x + 2*a) + d)*dil
og(-e^(I*b*x + I*a)) + 6*(d*cos(6*b*x + 6*a) - d*cos(4*b*x + 4*a) - d*cos(2*b*x + 2*a) + I*d*sin(6*b*x + 6*a)
- I*d*sin(4*b*x + 4*a) - I*d*sin(2*b*x + 2*a) + d)*dilog(e^(I*b*x + I*a)) + 3*(-I*b*d*x - I*b*c + (-I*b*d*x -
I*b*c)*cos(6*b*x + 6*a) + (I*b*d*x + I*b*c)*cos(4*b*x + 4*a) + (I*b*d*x + I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b
*c)*sin(6*b*x + 6*a) - (b*d*x + b*c)*sin(4*b*x + 4*a) - (b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + s
in(b*x + a)^2 + 2*cos(b*x + a) + 1) + 3*(I*b*d*x + I*b*c + (I*b*d*x + I*b*c)*cos(6*b*x + 6*a) + (-I*b*d*x - I*
b*c)*cos(4*b*x + 4*a) + (-I*b*d*x - I*b*c)*cos(2*b*x + 2*a) - (b*d*x + b*c)*sin(6*b*x + 6*a) + (b*d*x + b*c)*s
in(4*b*x + 4*a) + (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) - I*d*cos(4*b*x + 4*a) - I*d*cos(2*b*x + 2*a) - d*sin(6*b*x + 6*a) + d*sin(4*b*x + 4*a
) + d*sin(2*b*x + 2*a) + 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*sin(2*b*x + 2*a) - 4*I*b^2)

Giac [F]

\[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \csc ^3(a+b x) \sec ^2(a+b x) \, dx=\text {Hanged} \]

[In]

int((c + d*x)/(cos(a + b*x)^2*sin(a + b*x)^3),x)

[Out]

\text{Hanged}