3.384 \(\int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx\)

Optimal. Leaf size=173 \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{2 d (c+d x) \sin (a+b x) \cos (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}-\frac{2 c d x}{b}-\frac{d^2 x^2}{b}-\frac{i (c+d x)^3}{3 d} \]

[Out]

(-2*c*d*x)/b - (d^2*x^2)/b - ((I/3)*(c + d*x)^3)/d + ((c + d*x)^2*Log[1 + E^((2*I)*(a + b*x))])/b - (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) + (2*d*(c + d*x)*C
os[a + b*x]*Sin[a + b*x])/b^2 - (d^2*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^2*Sin[a + b*x]^2)/b

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Rubi [A]  time = 0.328273, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4431, 4404, 3310, 4407, 3719, 2190, 2531, 2282, 6589} \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{2 d (c+d x) \sin (a+b x) \cos (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}-\frac{2 c d x}{b}-\frac{d^2 x^2}{b}-\frac{i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sec[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

(-2*c*d*x)/b - (d^2*x^2)/b - ((I/3)*(c + d*x)^3)/d + ((c + d*x)^2*Log[1 + E^((2*I)*(a + b*x))])/b - (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) + (2*d*(c + d*x)*C
os[a + b*x]*Sin[a + b*x])/b^2 - (d^2*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^2*Sin[a + b*x]^2)/b

Rule 4431

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

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

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 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 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 \sec (a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^2 \cos (a+b x) \sin (a+b x)-(c+d x)^2 \sin ^2(a+b x) \tan (a+b x)\right ) \, dx\\ &=3 \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx-\int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx\\ &=\frac{3 (c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x) \sin ^2(a+b x) \, dx}{b}+\int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx-\int (c+d x)^2 \tan (a+b x) \, dx\\ &=-\frac{i (c+d x)^3}{3 d}+\frac{3 d (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac{3 d^2 \sin ^2(a+b x)}{4 b^3}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}+2 i \int \frac{e^{2 i (a+b x)} (c+d x)^2}{1+e^{2 i (a+b x)}} \, dx-\frac{d \int (c+d x) \sin ^2(a+b x) \, dx}{b}-\frac{(3 d) \int (c+d x) \, dx}{2 b}\\ &=-\frac{3 c d x}{2 b}-\frac{3 d^2 x^2}{4 b}-\frac{i (c+d x)^3}{3 d}+\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 d (c+d x) \cos (a+b x) \sin (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}-\frac{d \int (c+d x) \, dx}{2 b}-\frac{(2 d) \int (c+d x) \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{2 c d x}{b}-\frac{d^2 x^2}{b}-\frac{i (c+d x)^3}{3 d}+\frac{(c+d x)^2 \log \left (1+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{2 d (c+d x) \cos (a+b x) \sin (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}+\frac{\left (i d^2\right ) \int \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 c d x}{b}-\frac{d^2 x^2}{b}-\frac{i (c+d x)^3}{3 d}+\frac{(c+d x)^2 \log \left (1+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{2 d (c+d x) \cos (a+b x) \sin (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=-\frac{2 c d x}{b}-\frac{d^2 x^2}{b}-\frac{i (c+d x)^3}{3 d}+\frac{(c+d x)^2 \log \left (1+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{d^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{2 d (c+d x) \cos (a+b x) \sin (a+b x)}{b^2}-\frac{d^2 \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin ^2(a+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 6.54109, size = 516, normalized size = 2.98 \[ \frac{c d \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{b^2 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac{i e^{-i a} d^2 \sec (a) \left (6 \left (1+e^{2 i a}\right ) b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )\right )}{12 b^3}-\frac{\cos (2 b x) \left (2 b^2 c^2 \cos (2 a)+4 b^2 c d x \cos (2 a)+2 b^2 d^2 x^2 \cos (2 a)-2 b c d \sin (2 a)-2 b d^2 x \sin (2 a)-d^2 \cos (2 a)\right )}{2 b^3}+\frac{\sin (2 b x) \left (2 b^2 c^2 \sin (2 a)+4 b^2 c d x \sin (2 a)+2 b^2 d^2 x^2 \sin (2 a)+2 b c d \cos (2 a)+2 b d^2 x \cos (2 a)-d^2 \sin (2 a)\right )}{2 b^3}+\frac{c^2 \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b \left (\sin ^2(a)+\cos ^2(a)\right )}-\frac{1}{3} x \tan (a) \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Sec[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

((I/12)*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + 6*b*(1 + E^((2*I)*a))
*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3
*E^(I*a)) + (c^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)
) + (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*
I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Co
t[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Se
c[a])/(b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (Cos[2*b*x]*(2*b^2*c^2*Cos[2*a] - d^2*Cos[2*a] + 4*b^2*c*d*
x*Cos[2*a] + 2*b^2*d^2*x^2*Cos[2*a] - 2*b*c*d*Sin[2*a] - 2*b*d^2*x*Sin[2*a]))/(2*b^3) + ((2*b*c*d*Cos[2*a] + 2
*b*d^2*x*Cos[2*a] + 2*b^2*c^2*Sin[2*a] - d^2*Sin[2*a] + 4*b^2*c*d*x*Sin[2*a] + 2*b^2*d^2*x^2*Sin[2*a])*Sin[2*b
*x])/(2*b^3) - (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Tan[a])/3

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Maple [B]  time = 0.243, size = 377, normalized size = 2.2 \begin{align*} -icd{x}^{2}-{\frac{4\,iacdx}{b}}+i{c}^{2}x-{\frac{ \left ( 2\,{d}^{2}{x}^{2}{b}^{2}+2\,ib{d}^{2}x+4\,{b}^{2}cdx+2\,ibcd+2\,{b}^{2}{c}^{2}-{d}^{2} \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{4\,{b}^{3}}}-{\frac{ \left ( 2\,{d}^{2}{x}^{2}{b}^{2}-2\,ib{d}^{2}x+4\,{b}^{2}cdx-2\,ibcd+2\,{b}^{2}{c}^{2}-{d}^{2} \right ){{\rm e}^{-2\,i \left ( bx+a \right ) }}}{4\,{b}^{3}}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}-2\,{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-2\,{\frac{{d}^{2}{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}-{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-{\frac{i}{3}}{d}^{2}{x}^{3}+{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}-{\frac{idc{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}+4\,{\frac{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{2\,i{a}^{2}cd}{{b}^{2}}}+{\frac{2\,i{a}^{2}{d}^{2}x}{{b}^{2}}}+{\frac{{\frac{4\,i}{3}}{a}^{3}{d}^{2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.39856, size = 406, normalized size = 2.35 \begin{align*} -\frac{c^{2}{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )\right )}}{2 \, b} + \frac{-2 i \, b^{3} d^{2} x^{3} - 6 i \, b^{3} c d x^{2} + 3 \, d^{2}{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) +{\left (6 i \, b^{2} d^{2} x^{2} + 12 i \, b^{2} c d x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \,{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (-6 i \, b d^{2} x - 6 i \, b c d\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 3 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\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 ) + 6 \,{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*(2*cos(2*b*x + 2*a) - log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*
b*x)*sin(2*a) + sin(2*a)^2))/b + 1/6*(-2*I*b^3*d^2*x^3 - 6*I*b^3*c*d*x^2 + 3*d^2*polylog(3, -e^(2*I*b*x + 2*I*
a)) + (6*I*b^2*d^2*x^2 + 12*I*b^2*c*d*x)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 3*(2*b^2*d^2*x^2 +
4*b^2*c*d*x - d^2)*cos(2*b*x + 2*a) + (-6*I*b*d^2*x - 6*I*b*c*d)*dilog(-e^(2*I*b*x + 2*I*a)) + 3*(b^2*d^2*x^2
+ 2*b^2*c*d*x)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(b*d^2*x + b*c*d)*sin
(2*b*x + 2*a))/b^3

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Fricas [C]  time = 0.718819, size = 1728, normalized size = 9.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x - 2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - d^2)*cos(b*x + a)^2 + 2*d^2*po
lylog(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)*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)) + (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))/b^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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