3.217 \(\int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=186 \[ -\frac {2 d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \sin (a+b x)}{b^3}+\frac {2 i d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {2 d (c+d x) \cos (a+b x)}{b^2}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {2 i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4407, 3296, 2637, 4181, 2531, 2282, 6589} \[ \frac {2 i d (c+d x) \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {2 d (c+d x) \cos (a+b x)}{b^2}+\frac {2 d^2 \sin (a+b x)}{b^3}-\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {2 i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sin[a + b*x]*Tan[a + b*x],x]

[Out]

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4181

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 4407

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Int[
(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

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

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Mathematica [A]  time = 0.83, size = 315, normalized size = 1.69 \[ -\frac {b^2 c^2 \sin (a+b x)+2 i b^2 c^2 \tan ^{-1}\left (e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )+2 b^2 c d x \sin (a+b x)-b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+b^2 d^2 x^2 \sin (a+b x)-2 i b d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )+2 b c d \cos (a+b x)+2 d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )-2 d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )-2 d^2 \sin (a+b x)+2 b d^2 x \cos (a+b x)}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Sin[a + b*x]*Tan[a + b*x],x]

[Out]

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

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fricas [C]  time = 0.52, size = 656, normalized size = 3.53 \[ -\frac {2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) - {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.26, size = 512, normalized size = 2.75 \[ \frac {i \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{3}}+\frac {4 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {2 d^{2} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {i \left (d^{2} x^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 b^{3}}-\frac {2 i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {2 i c d \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {2 d^{2} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 i d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {a^{2} d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i c d \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*(d^2*x^2*b^2+2*b^2*c*d*x+b^2*c^2+2*I*b*d^2*x-2*d^2+2*I*b*c*d)/b^3*exp(I*(b*x+a))+4*I/b^2*c*d*a*arctan(ex
p(I*(b*x+a)))-2/b*c*d*ln(1+I*exp(I*(b*x+a)))*x-2*d^2*polylog(3,-I*exp(I*(b*x+a)))/b^3-1/2*I*(d^2*x^2*b^2+2*b^2
*c*d*x+b^2*c^2-2*I*b*d^2*x-2*d^2-2*I*b*c*d)/b^3*exp(-I*(b*x+a))-2*I/b^2*c*d*polylog(2,I*exp(I*(b*x+a)))-2*I/b*
c^2*arctan(exp(I*(b*x+a)))-1/b*d^2*ln(1+I*exp(I*(b*x+a)))*x^2+2*d^2*polylog(3,I*exp(I*(b*x+a)))/b^3-2*I/b^2*d^
2*polylog(2,I*exp(I*(b*x+a)))*x+2/b^2*c*d*ln(1-I*exp(I*(b*x+a)))*a+2/b*c*d*ln(1-I*exp(I*(b*x+a)))*x+2*I/b^2*d^
2*polylog(2,-I*exp(I*(b*x+a)))*x-2/b^2*c*d*ln(1+I*exp(I*(b*x+a)))*a+1/b^3*a^2*d^2*ln(1+I*exp(I*(b*x+a)))-2*I/b
^3*d^2*a^2*arctan(exp(I*(b*x+a)))+2*I/b^2*c*d*polylog(2,-I*exp(I*(b*x+a)))+1/b*d^2*ln(1-I*exp(I*(b*x+a)))*x^2-
1/b^3*a^2*d^2*ln(1-I*exp(I*(b*x+a)))

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maxima [B]  time = 0.60, size = 510, normalized size = 2.74 \[ \frac {c^{2} {\left (\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (\sin \left (b x + a\right ) - 1\right ) - 2 \, \sin \left (b x + a\right )\right )} - \frac {2 \, a c d {\left (\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (\sin \left (b x + a\right ) - 1\right ) - 2 \, \sin \left (b x + a\right )\right )}}{b} + \frac {a^{2} d^{2} {\left (\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (\sin \left (b x + a\right ) - 1\right ) - 2 \, \sin \left (b x + a\right )\right )}}{b^{2}} + \frac {4 \, d^{2} {\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(-i \, 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 (\cos \left (b x + a\right ), \sin \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 (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) - 4 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} \cos \left (b x + a\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 (i \, 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 (-i \, 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 (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \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 \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{b^{2}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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