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

Optimal. Leaf size=147 \[ \frac{2 i d^2 \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac{8 d (c+d x) \sin (a+b x)}{b^2}-\frac{4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac{8 d^2 \cos (a+b x)}{b^3}-\frac{4 (c+d x)^2 \cos (a+b x)}{b}-\frac{(c+d x)^2 \sec (a+b x)}{b} \]

[Out]

((-4*I)*d*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 + (8*d^2*Cos[a + b*x])/b^3 - (4*(c + d*x)^2*Cos[a + b*x])/b +
((2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 - ((2*I)*d^2*PolyLog[2, I*E^(I*(a + b*x))])/b^3 - ((c + d*x)
^2*Sec[a + b*x])/b + (8*d*(c + d*x)*Sin[a + b*x])/b^2

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Rubi [A]  time = 0.212481, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4431, 3296, 2638, 4407, 4409, 4181, 2279, 2391} \[ \frac{2 i d^2 \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac{8 d (c+d x) \sin (a+b x)}{b^2}-\frac{4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac{8 d^2 \cos (a+b x)}{b^3}-\frac{4 (c+d x)^2 \cos (a+b x)}{b}-\frac{(c+d x)^2 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*d*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 + (8*d^2*Cos[a + b*x])/b^3 - (4*(c + d*x)^2*Cos[a + b*x])/b +
((2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 - ((2*I)*d^2*PolyLog[2, I*E^(I*(a + b*x))])/b^3 - ((c + d*x)
^2*Sec[a + b*x])/b + (8*d*(c + d*x)*Sin[a + b*x])/b^2

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 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 2638

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

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 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && 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 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]

Rubi steps

\begin{align*} \int (c+d x)^2 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^2 \sin (a+b x)-(c+d x)^2 \sin (a+b x) \tan ^2(a+b x)\right ) \, dx\\ &=3 \int (c+d x)^2 \sin (a+b x) \, dx-\int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx\\ &=-\frac{3 (c+d x)^2 \cos (a+b x)}{b}+\frac{(6 d) \int (c+d x) \cos (a+b x) \, dx}{b}+\int (c+d x)^2 \sin (a+b x) \, dx-\int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx\\ &=-\frac{4 (c+d x)^2 \cos (a+b x)}{b}-\frac{(c+d x)^2 \sec (a+b x)}{b}+\frac{6 d (c+d x) \sin (a+b x)}{b^2}+\frac{(2 d) \int (c+d x) \cos (a+b x) \, dx}{b}+\frac{(2 d) \int (c+d x) \sec (a+b x) \, dx}{b}-\frac{\left (6 d^2\right ) \int \sin (a+b x) \, dx}{b^2}\\ &=-\frac{4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac{6 d^2 \cos (a+b x)}{b^3}-\frac{4 (c+d x)^2 \cos (a+b x)}{b}-\frac{(c+d x)^2 \sec (a+b x)}{b}+\frac{8 d (c+d x) \sin (a+b x)}{b^2}-\frac{\left (2 d^2\right ) \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (2 d^2\right ) \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac{\left (2 d^2\right ) \int \sin (a+b x) \, dx}{b^2}\\ &=-\frac{4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac{8 d^2 \cos (a+b x)}{b^3}-\frac{4 (c+d x)^2 \cos (a+b x)}{b}-\frac{(c+d x)^2 \sec (a+b x)}{b}+\frac{8 d (c+d x) \sin (a+b x)}{b^2}+\frac{\left (2 i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}-\frac{\left (2 i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac{4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac{8 d^2 \cos (a+b x)}{b^3}-\frac{4 (c+d x)^2 \cos (a+b x)}{b}+\frac{2 i d^2 \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}-\frac{2 i d^2 \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{(c+d x)^2 \sec (a+b x)}{b}+\frac{8 d (c+d x) \sin (a+b x)}{b^2}\\ \end{align*}

Mathematica [B]  time = 3.7371, size = 364, normalized size = 2.48 \[ \frac{2 d^2 \left (2 \tan ^{-1}(\cot (a)) \tanh ^{-1}\left (\cos (a) \tan \left (\frac{b x}{2}\right )+\sin (a)\right )-\frac{\csc (a) \left (i \text{PolyLog}\left (2,-e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-i \text{PolyLog}\left (2,e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+\left (b x-\tan ^{-1}(\cot (a))\right ) \left (\log \left (1-e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-\log \left (1+e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )\right )\right )}{\sqrt{\csc ^2(a)}}\right )-4 \cos (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )-2 b d \sin (a) (c+d x)\right )+4 \sin (b x) \left (\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )+2 b d \cos (a) (c+d x)\right )-b^2 \sec (a) (c+d x)^2-\frac{b^2 \sin \left (\frac{b x}{2}\right ) (c+d x)^2}{\left (\cos \left (\frac{a}{2}\right )-\sin \left (\frac{a}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}+\frac{b^2 \sin \left (\frac{b x}{2}\right ) (c+d x)^2}{\left (\sin \left (\frac{a}{2}\right )+\cos \left (\frac{a}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )}+4 b c d \tanh ^{-1}\left (\cos (a) \tan \left (\frac{b x}{2}\right )+\sin (a)\right )}{b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] + 2*d^2*(2*ArcTan[Cot[a]]*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]]
- (Csc[a]*((b*x - ArcTan[Cot[a]])*(Log[1 - E^(I*(b*x - ArcTan[Cot[a]]))] - Log[1 + E^(I*(b*x - ArcTan[Cot[a]]
))]) + I*PolyLog[2, -E^(I*(b*x - ArcTan[Cot[a]]))] - I*PolyLog[2, E^(I*(b*x - ArcTan[Cot[a]]))]))/Sqrt[Csc[a]^
2]) - b^2*(c + d*x)^2*Sec[a] - 4*Cos[b*x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] - 2*b*d*(c + d*x)*Sin[a]) + 4*(2*
b*d*(c + d*x)*Cos[a] + (-2*d^2 + b^2*(c + d*x)^2)*Sin[a])*Sin[b*x] - (b^2*(c + d*x)^2*Sin[(b*x)/2])/((Cos[a/2]
- Sin[a/2])*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])) + (b^2*(c + d*x)^2*Sin[(b*x)/2])/((Cos[a/2] + Sin[a/2])*(C
os[(a + b*x)/2] + Sin[(a + b*x)/2])))/b^3

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Maple [B]  time = 0.308, size = 345, normalized size = 2.4 \begin{align*} -2\,{\frac{ \left ({d}^{2}{x}^{2}{b}^{2}+2\,{b}^{2}cdx+{b}^{2}{c}^{2}+2\,ib{d}^{2}x-2\,{d}^{2}+2\,ibcd \right ){{\rm e}^{i \left ( bx+a \right ) }}}{{b}^{3}}}-2\,{\frac{ \left ({d}^{2}{x}^{2}{b}^{2}+2\,{b}^{2}cdx+{b}^{2}{c}^{2}-2\,ib{d}^{2}x-2\,{d}^{2}-2\,ibcd \right ){{\rm e}^{-i \left ( bx+a \right ) }}}{{b}^{3}}}-2\,{\frac{{{\rm e}^{i \left ( bx+a \right ) }} \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}-{\frac{4\,idc\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-2\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+2\,{\frac{{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}+{\frac{2\,i{d}^{2}{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,i{d}^{2}{\it dilog} \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{4\,i{d}^{2}a\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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)^2*sin(3*b*x+3*a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 0.638272, size = 1320, normalized size = 8.98 \begin{align*} -\frac{b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 4 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{2} -{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) +{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) -{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) -{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) -{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) +{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 8 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{b^{3} \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a)) + I*d^2*cos(b*
x + a)*dilog(I*cos(b*x + a) - sin(b*x + a)) - I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - I*d^2
*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + 4*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x
+ a)^2 - (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*lo
g(cos(b*x + a) - I*sin(b*x + a) + I) - (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) + sin(b*x + a) + 1) +
(b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I
*cos(b*x + a) + sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (
b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*
x + a) - I*sin(b*x + a) + I) - 8*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a))/(b^3*cos(b*x + a))

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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)**2*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 )^{2} \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)^2*sin(3*b*x+3*a),x, algorithm="giac")

[Out]

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