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

Optimal. Leaf size=113 \[ \frac {i d \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac {c \sec (a+b x)}{b}-\frac {c \tanh ^{-1}(\cos (a+b x))}{b}+\frac {d x \sec (a+b x)}{b}-\frac {2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 122, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2622, 321, 207, 4420, 6271, 12, 4183, 2279, 2391, 3770} \[ \frac {i d \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac {(c+d x) \sec (a+b x)}{b}-\frac {(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac {2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {d x \tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x*ArcTanh[E^(I*(a + b*x))])/b + (d*x*ArcTanh[Cos[a + b*x]])/b - ((c + d*x)*ArcTanh[Cos[a + b*x]])/b - (d
*ArcTanh[Sin[a + b*x]])/b^2 + (I*d*PolyLog[2, -E^(I*(a + b*x))])/b^2 - (I*d*PolyLog[2, E^(I*(a + b*x))])/b^2 +
((c + d*x)*Sec[a + b*x])/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 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 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 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 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 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 212, normalized size = 1.88 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i (a+b x)}\right )-\text {Li}_2\left (e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )\right )}{b^2}-\frac {a d \log \left (\tan \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {d \log \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}-\frac {d \log \left (\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {c \sec (a+b x)}{b}+\frac {c \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}-\frac {c \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b}+\frac {d x \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Log[Cos[(a + b*x)/2]])/b) + (d*Log[Cos[(a + b*x)/2] - Sin[(a + b*x)/2]])/b^2 + (c*Log[Sin[(a + b*x)/2]])/
b - (d*Log[Cos[(a + b*x)/2] + Sin[(a + b*x)/2]])/b^2 - (a*d*Log[Tan[(a + b*x)/2]])/b^2 + (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))]
)))/b^2 + (c*Sec[a + b*x])/b + (d*x*Sec[a + b*x])/b

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fricas [B]  time = 0.47, size = 366, normalized size = 3.24 \[ \frac {2 \, b d x - i \, d \cos \left (b x + a\right ) {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right ) {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right ) {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right ) {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (b d x + b c\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + b c\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right ) \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right ) \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - d \cos \left (b x + a\right ) \log \left (\sin \left (b x + a\right ) + 1\right ) + d \cos \left (b x + a\right ) \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, b c}{2 \, b^{2} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 160, normalized size = 1.42 \[ \frac {2 \,{\mathrm e}^{i \left (b x +a \right )} \left (d x +c \right )}{b \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {2 i d \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \dilog \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \dilog \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.62, size = 806, normalized size = 7.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(d*cos(2*b*x + 2*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 + 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)) + (2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(2*b
*x + 2*a) - (-2*I*b*d*x - 2*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b*c*cos(2*b*
x + 2*a) + 2*I*b*c*sin(2*b*x + 2*a) + 2*b*c)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(2*b*x + 2*
a) + 2*I*b*d*x*sin(2*b*x + 2*a) + 2*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - (-4*I*b*d*x - 4*I*b*c)*c
os(b*x + a) - 2*(d*cos(2*b*x + 2*a) + I*d*sin(2*b*x + 2*a) + d)*dilog(-e^(I*b*x + I*a)) + 2*(d*cos(2*b*x + 2*a
) + I*d*sin(2*b*x + 2*a) + d)*dilog(e^(I*b*x + I*a)) - (I*b*d*x + I*b*c + (I*b*d*x + I*b*c)*cos(2*b*x + 2*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) - (-I*b*d*x - I*b*c
+ (-I*b*d*x - I*b*c)*cos(2*b*x + 2*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) - (-I*d*cos(2*b*x + 2*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*(b*d*x + b*c)*sin(b*x +
a))/(-2*I*b^2*cos(2*b*x + 2*a) + 2*b^2*sin(2*b*x + 2*a) - 2*I*b^2)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \csc {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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