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

Optimal. Leaf size=127 \[ -\frac {d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

[Out]

-2*(d*x+c)^2*arctanh(exp(2*I*(b*x+a)))/b+I*d*(d*x+c)*polylog(2,-exp(2*I*(b*x+a)))/b^2-I*d*(d*x+c)*polylog(2,ex
p(2*I*(b*x+a)))/b^2-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/2*d^2*polylog(3,exp(2*I*(b*x+a)))/b^3

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Rubi [A]  time = 0.12, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4419, 4183, 2531, 2282, 6589} \[ \frac {i d (c+d x) \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/b + (I*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*
(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 - (d^2*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^3) + (d^2*PolyLog
[3, E^((2*I)*(a + b*x))])/(2*b^3)

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx &=2 \int (c+d x)^2 \csc (2 a+2 b x) \, dx\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {(2 d) \int (c+d x) \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac {(2 d) \int (c+d x) \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {\left (i d^2\right ) \int \text {Li}_2\left (-e^{i (2 a+2 b x)}\right ) \, dx}{b^2}+\frac {\left (i d^2\right ) \int \text {Li}_2\left (e^{i (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 213, normalized size = 1.68 \[ \frac {-4 b^2 c^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )+4 b^2 c d x \log \left (1-e^{2 i (a+b x)}\right )-4 b^2 c d x \log \left (1+e^{2 i (a+b x)}\right )+2 b^2 d^2 x^2 \log \left (1-e^{2 i (a+b x)}\right )-2 b^2 d^2 x^2 \log \left (1+e^{2 i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )-2 i b d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )-d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )+d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^2*c^2*ArcTanh[E^((2*I)*(a + b*x))] + 4*b^2*c*d*x*Log[1 - E^((2*I)*(a + b*x))] + 2*b^2*d^2*x^2*Log[1 - E^
((2*I)*(a + b*x))] - 4*b^2*c*d*x*Log[1 + E^((2*I)*(a + b*x))] - 2*b^2*d^2*x^2*Log[1 + E^((2*I)*(a + b*x))] + (
2*I)*b*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))] - (2*I)*b*d*(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))] - d^
2*PolyLog[3, -E^((2*I)*(a + b*x))] + d^2*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3)

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fricas [C]  time = 0.56, size = 1090, normalized size = 8.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 2*d^2*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 2*d^2
*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 2*d^2*polylog(3
, -I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) + 2*d^2*polylog(3, -cos(b
*x + a) + I*sin(b*x + a)) + 2*d^2*polylog(3, -cos(b*x + a) - I*sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*b*c*d)*dilo
g(cos(b*x + a) + I*sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-2*I*b*d^
2*x - 2*I*b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(I*cos(b*x + a) - sin(b
*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*b*c*d)*dilog(
-I*cos(b*x + a) - sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(-cos(b*x + a) + I*sin(b*x + a)) + (-2*I*b*d^
2*x - 2*I*b*c*d)*dilog(-cos(b*x + a) - I*sin(b*x + a)) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(cos(b*x + a
) + I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b^2*d^2*x^
2 + 2*b^2*c*d*x + b^2*c^2)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*
x + a) - I*sin(b*x + a) + I) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos(b*x + a) + sin(b*x
+ a) + 1) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^2*d^
2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^2*d^2*x^2 + 2*b^2*c*d*
x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-1/2*co
s(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x
+ a) + 1/2) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - (b^
2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d
- a^2*d^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) - I*sin
(b*x + a) + I))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.09, size = 469, normalized size = 3.69 \[ -\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {2 d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {c^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {i c d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/b^3*d^2*a^2*ln(exp(I*(b*x+a))-1)+1/b*d^2*ln(1-exp(I*(b*x+a)))*x^2-
1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+1/b*d^2*ln(exp(I*(b*x+a))+1)*x^2+2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2
*polylog(3,exp(I*(b*x+a)))/b^3-1/b*c^2*ln(1+exp(2*I*(b*x+a)))+1/b*c^2*ln(exp(I*(b*x+a))-1)+1/b*c^2*ln(exp(I*(b
*x+a))+1)-1/b*d^2*ln(1+exp(2*I*(b*x+a)))*x^2-2*I/b^2*d^2*polylog(2,-exp(I*(b*x+a)))*x-2*I/b^2*d^2*polylog(2,ex
p(I*(b*x+a)))*x-2/b*c*d*ln(1+exp(2*I*(b*x+a)))*x+2/b*c*d*ln(1-exp(I*(b*x+a)))*x+2/b^2*c*d*ln(1-exp(I*(b*x+a)))
*a+2/b*c*d*ln(exp(I*(b*x+a))+1)*x-2/b^2*c*d*a*ln(exp(I*(b*x+a))-1)-2*I/b^2*c*d*polylog(2,exp(I*(b*x+a)))+I/b^2
*d^2*polylog(2,-exp(2*I*(b*x+a)))*x+I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))-2*I/b^2*c*d*polylog(2,-exp(I*(b*x+a
)))

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maxima [B]  time = 0.60, size = 590, normalized size = 4.65 \[ -\frac {c^{2} {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )} - \frac {2 \, a c d {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )}}{b} + \frac {a^{2} d^{2} {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )}}{b^{2}} + \frac {d^{2} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) + {\left (2 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (4 i \, b c d - 4 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\left (-2 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (-4 i \, b c d + 4 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (2 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (4 i \, b c d - 4 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + {\left (-2 i \, b c d - 2 i \, {\left (b x + a\right )} d^{2} + 2 i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 i \, b c d + 4 i \, {\left (b x + a\right )} d^{2} - 4 i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (4 i \, b c d + 4 i \, {\left (b x + a\right )} d^{2} - 4 i \, a d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right )}{b^{2}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 2*a*c*d*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a
)^2))/b + a^2*d^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 + (d^2*polylog(3, -e^(2*I*b*x + 2*I*a))
- 4*d^2*polylog(3, -e^(I*b*x + I*a)) - 4*d^2*polylog(3, e^(I*b*x + I*a)) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d -
4*I*a*d^2)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d +
4*I*a*d^2)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I*a*d^2
)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (-2*I*b*c*d - 2*I*(b*x + a)*d^2 + 2*I*a*d^2)*dilog(-e^
(2*I*b*x + 2*I*a)) + (4*I*b*c*d + 4*I*(b*x + a)*d^2 - 4*I*a*d^2)*dilog(-e^(I*b*x + I*a)) + (4*I*b*c*d + 4*I*(b
*x + a)*d^2 - 4*I*a*d^2)*dilog(e^(I*b*x + I*a)) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(2*b*
x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*lo
g(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(
cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1))/b^2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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