3.287 \(\int x \csc ^3(a+b x) \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=126 \[ \frac {3 i \text {Li}_2\left (-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i \text {Li}_2\left (e^{i (a+b x)}\right )}{2 b^2}-\frac {\csc (a+b x)}{2 b^2}-\frac {\tanh ^{-1}(\sin (a+b x))}{b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {3 x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2622, 288, 321, 207, 4420, 6271, 12, 4183, 2279, 2391, 3770, 2621} \[ \frac {3 i \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {\csc (a+b x)}{2 b^2}-\frac {\tanh ^{-1}(\sin (a+b x))}{b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {3 x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(-3*x*ArcTanh[E^(I*(a + b*x))])/b - ArcTanh[Sin[a + b*x]]/b^2 - Csc[a + b*x]/(2*b^2) + (((3*I)/2)*PolyLog[2, -
E^(I*(a + b*x))])/b^2 - (((3*I)/2)*PolyLog[2, E^(I*(a + b*x))])/b^2 + (3*x*Sec[a + b*x])/(2*b) - (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 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 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 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 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 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 x \csc ^3(a+b x) \sec ^2(a+b x) \, dx &=-\frac {3 x \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}-\int \left (-\frac {3 \tanh ^{-1}(\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 x \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac {\int \csc ^2(a+b x) \sec (a+b x) \, dx}{2 b}+\frac {3 \int \tanh ^{-1}(\cos (a+b x)) \, dx}{2 b}-\frac {3 \int \sec (a+b x) \, dx}{2 b}\\ &=-\frac {3 \tanh ^{-1}(\sin (a+b x))}{2 b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b^2}+\frac {3 \int b x \csc (a+b x) \, dx}{2 b}\\ &=-\frac {3 \tanh ^{-1}(\sin (a+b x))}{2 b^2}-\frac {\csc (a+b x)}{2 b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac {3}{2} \int x \csc (a+b x) \, dx-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b^2}\\ &=-\frac {3 x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {\tanh ^{-1}(\sin (a+b x))}{b^2}-\frac {\csc (a+b x)}{2 b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac {3 \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {3 \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=-\frac {3 x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {\tanh ^{-1}(\sin (a+b x))}{b^2}-\frac {\csc (a+b x)}{2 b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=-\frac {3 x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {\tanh ^{-1}(\sin (a+b x))}{b^2}-\frac {\csc (a+b x)}{2 b^2}+\frac {3 i \text {Li}_2\left (-e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i \text {Li}_2\left (e^{i (a+b x)}\right )}{2 b^2}+\frac {3 x \sec (a+b x)}{2 b}-\frac {x \csc ^2(a+b x) \sec (a+b x)}{2 b}\\ \end {align*}

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Mathematica [B]  time = 2.41, size = 282, normalized size = 2.24 \[ \frac {12 i \left (\text {Li}_2\left (-e^{i (a+b x)}\right )-\text {Li}_2\left (e^{i (a+b x)}\right )\right )+12 (a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )-2 \tan \left (\frac {1}{2} (a+b x)\right )-2 \cot \left (\frac {1}{2} (a+b x)\right )-b x \csc ^2\left (\frac {1}{2} (a+b x)\right )+b x \sec ^2\left (\frac {1}{2} (a+b x)\right )-12 a \log \left (\tan \left (\frac {1}{2} (a+b x)\right )\right )+\frac {8 b x \sin \left (\frac {1}{2} (a+b x)\right )}{\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )}-\frac {8 b x \sin \left (\frac {1}{2} (a+b x)\right )}{\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )}+8 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )-8 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )\right )+8 b x}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(8*b*x - 2*Cot[(a + b*x)/2] - b*x*Csc[(a + b*x)/2]^2 + 12*(a + b*x)*(Log[1 - E^(I*(a + b*x))] - Log[1 + E^(I*(
a + b*x))]) + 8*Log[Cos[(a + b*x)/2] - Sin[(a + b*x)/2]] - 8*Log[Cos[(a + b*x)/2] + Sin[(a + b*x)/2]] - 12*a*L
og[Tan[(a + b*x)/2]] + (12*I)*(PolyLog[2, -E^(I*(a + b*x))] - PolyLog[2, E^(I*(a + b*x))]) + b*x*Sec[(a + b*x)
/2]^2 + (8*b*x*Sin[(a + b*x)/2])/(Cos[(a + b*x)/2] - Sin[(a + b*x)/2]) - (8*b*x*Sin[(a + b*x)/2])/(Cos[(a + b*
x)/2] + Sin[(a + b*x)/2]) - 2*Tan[(a + b*x)/2])/(8*b^2)

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fricas [B]  time = 0.51, size = 527, normalized size = 4.18 \[ \frac {6 \, b x \cos \left (b x + a\right )^{2} - 4 \, b x + {\left (-3 i \, \cos \left (b x + a\right )^{3} + 3 i \, \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 ) + {\left (3 i \, \cos \left (b x + a\right )^{3} - 3 i \, \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 ) + {\left (-3 i \, \cos \left (b x + a\right )^{3} + 3 i \, \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 ) + {\left (3 i \, \cos \left (b x + a\right )^{3} - 3 i \, \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 (b x \cos \left (b x + a\right )^{3} - b x \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 (b x \cos \left (b x + a\right )^{3} - b x \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 (a \cos \left (b x + a\right )^{3} - a \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 (a \cos \left (b x + a\right )^{3} - a \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 x + a\right )} \cos \left (b x + a\right )^{3} - {\left (b x + a\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 x + a\right )} \cos \left (b x + a\right )^{3} - {\left (b x + a\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 (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, {\left (b^{2} \cos \left (b x + a\right )^{3} - b^{2} \cos \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(6*b*x*cos(b*x + a)^2 - 4*b*x + (-3*I*cos(b*x + a)^3 + 3*I*cos(b*x + a))*dilog(cos(b*x + a) + I*sin(b*x +
a)) + (3*I*cos(b*x + a)^3 - 3*I*cos(b*x + a))*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-3*I*cos(b*x + a)^3 + 3*
I*cos(b*x + a))*dilog(-cos(b*x + a) + I*sin(b*x + a)) + (3*I*cos(b*x + a)^3 - 3*I*cos(b*x + a))*dilog(-cos(b*x
+ a) - I*sin(b*x + a)) - 3*(b*x*cos(b*x + a)^3 - b*x*cos(b*x + a))*log(cos(b*x + a) + I*sin(b*x + a) + 1) - 3
*(b*x*cos(b*x + a)^3 - b*x*cos(b*x + a))*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 3*(a*cos(b*x + a)^3 - a*cos(
b*x + a))*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 3*(a*cos(b*x + a)^3 - a*cos(b*x + a))*log(-1/2*c
os(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) + 3*((b*x + a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(-cos(b*x +
a) + I*sin(b*x + a) + 1) + 3*((b*x + a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(-cos(b*x + a) - I*sin(b*
x + a) + 1) - 2*(cos(b*x + a)^3 - cos(b*x + a))*log(sin(b*x + a) + 1) + 2*(cos(b*x + a)^3 - cos(b*x + a))*log(
-sin(b*x + a) + 1) + 2*cos(b*x + a)*sin(b*x + a))/(b^2*cos(b*x + a)^3 - b^2*cos(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 182, normalized size = 1.44 \[ \frac {3 b x \,{\mathrm e}^{5 i \left (b x +a \right )}-2 b x \,{\mathrm e}^{3 i \left (b x +a \right )}-i {\mathrm e}^{5 i \left (b x +a \right )}+3 b x \,{\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (b x +a \right )}}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}-\frac {3 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b^{2}}+\frac {2 i \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i \dilog \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {3 i \dilog \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*csc(b*x+a)^3*sec(b*x+a)^2,x)

[Out]

1/b^2/(exp(2*I*(b*x+a))-1)^2/(1+exp(2*I*(b*x+a)))*(3*b*x*exp(5*I*(b*x+a))-2*b*x*exp(3*I*(b*x+a))-I*exp(5*I*(b*
x+a))+3*b*x*exp(I*(b*x+a))+I*exp(I*(b*x+a)))-3/2/b^2*a*ln(exp(I*(b*x+a))-1)+2*I/b^2*arctan(exp(I*(b*x+a)))+3/2
*I/b^2*dilog(exp(I*(b*x+a)))+3/2*I/b^2*dilog(exp(I*(b*x+a))+1)-3/2/b*ln(exp(I*(b*x+a))+1)*x

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maxima [B]  time = 0.69, size = 1184, normalized size = 9.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*I*b*x*cos(3*b*x + 3*a) - 8*b*x*sin(3*b*x + 3*a) - (4*cos(6*b*x + 6*a) - 4*cos(4*b*x + 4*a) - 4*cos(2*b*x +
2*a) + 4*I*sin(6*b*x + 6*a) - 4*I*sin(4*b*x + 4*a) - 4*I*sin(2*b*x + 2*a) + 4)*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)) - (6*b*x*cos(6*b*
x + 6*a) - 6*b*x*cos(4*b*x + 4*a) - 6*b*x*cos(2*b*x + 2*a) + 6*I*b*x*sin(6*b*x + 6*a) - 6*I*b*x*sin(4*b*x + 4*
a) - 6*I*b*x*sin(2*b*x + 2*a) + 6*b*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (6*b*x*cos(6*b*x + 6*a) - 6*b
*x*cos(4*b*x + 4*a) - 6*b*x*cos(2*b*x + 2*a) + 6*I*b*x*sin(6*b*x + 6*a) - 6*I*b*x*sin(4*b*x + 4*a) - 6*I*b*x*s
in(2*b*x + 2*a) + 6*b*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 4*(3*I*b*x + 1)*cos(5*b*x + 5*a) - 4*(3*I*
b*x - 1)*cos(b*x + a) + (6*cos(6*b*x + 6*a) - 6*cos(4*b*x + 4*a) - 6*cos(2*b*x + 2*a) + 6*I*sin(6*b*x + 6*a) -
6*I*sin(4*b*x + 4*a) - 6*I*sin(2*b*x + 2*a) + 6)*dilog(-e^(I*b*x + I*a)) - (6*cos(6*b*x + 6*a) - 6*cos(4*b*x
+ 4*a) - 6*cos(2*b*x + 2*a) + 6*I*sin(6*b*x + 6*a) - 6*I*sin(4*b*x + 4*a) - 6*I*sin(2*b*x + 2*a) + 6)*dilog(e^
(I*b*x + I*a)) + (3*I*b*x*cos(6*b*x + 6*a) - 3*I*b*x*cos(4*b*x + 4*a) - 3*I*b*x*cos(2*b*x + 2*a) - 3*b*x*sin(6
*b*x + 6*a) + 3*b*x*sin(4*b*x + 4*a) + 3*b*x*sin(2*b*x + 2*a) + 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 +
2*cos(b*x + a) + 1) + (-3*I*b*x*cos(6*b*x + 6*a) + 3*I*b*x*cos(4*b*x + 4*a) + 3*I*b*x*cos(2*b*x + 2*a) + 3*b*
x*sin(6*b*x + 6*a) - 3*b*x*sin(4*b*x + 4*a) - 3*b*x*sin(2*b*x + 2*a) - 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x +
a)^2 - 2*cos(b*x + a) + 1) + (-2*I*cos(6*b*x + 6*a) + 2*I*cos(4*b*x + 4*a) + 2*I*cos(2*b*x + 2*a) + 2*sin(6*b
*x + 6*a) - 2*sin(4*b*x + 4*a) - 2*sin(2*b*x + 2*a) - 2*I)*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)) + (12*b*x - 4*I)*sin(5*b*x + 5*a) + (12*b*
x + 4*I)*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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \csc ^{3}{\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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