\(\int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx\) [385]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {d x}{b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {d \cos (a+b x) \sin (a+b x)}{b^2}+\frac {2 (c+d x) \sin ^2(a+b x)}{b} \]

[Out]

-d*x/b-1/2*I*(d*x+c)^2/d+(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b-1/2*I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2+d*cos(b*x+a
)*sin(b*x+a)/b^2+2*(d*x+c)*sin(b*x+a)^2/b

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4516, 4489, 2715, 8, 4492, 3800, 2221, 2317, 2438} \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {d \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 (c+d x) \sin ^2(a+b x)}{b}-\frac {d x}{b}-\frac {i (c+d x)^2}{2 d} \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 2715

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

Rule 3800

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 4489

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 4492

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 4516

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]

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(107)=214\).

Time = 3.79 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.40 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {c \log (\cos (a+b x))}{b}+\frac {d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {d \cos (2 b x) (2 b x \cos (2 a)-\sin (2 a))}{2 b^2}+\frac {d (\cos (2 a)+2 b x \sin (2 a)) \sin (2 b x)}{2 b^2}+\frac {2 c \sin ^2(a+b x)}{b}-\frac {1}{2} d x^2 \tan (a) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.65

method	result	size
risch	\(-\frac {i d \,x^{2}}{2}+i c x -\frac {\left (2 d x b +2 c b +i d \right ) {\mathrm e}^{2 i \left (x b +a \right )}}{4 b^{2}}-\frac {\left (2 d x b +2 c b -i d \right ) {\mathrm e}^{-2 i \left (x b +a \right )}}{4 b^{2}}+\frac {c \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}-\frac {2 i d x a}{b}-\frac {i d \,a^{2}}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}-\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{2}}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}\)	\(177\)

[In]

int((d*x+c)*sec(b*x+a)*sin(3*b*x+3*a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (94) = 188\).

Time = 0.28 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.18 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {2 \, b d x - 4 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \left (c + d x\right ) \sin {\left (3 a + 3 b x \right )} \sec {\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \]

[In]

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

[Out]

-1/2*c*(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 - 1/2*(2*b*x*cos(2*b*x + 2*a) + 4*b^2*integrate(x*sin(2*b*x + 2*a)/(cos(2*b*x + 2
*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) - sin(2*b*x + 2*a))*d/b^2

Giac [F]

\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )\,\left (c+d\,x\right )}{\cos \left (a+b\,x\right )} \,d x \]

[In]

int((sin(3*a + 3*b*x)*(c + d*x))/cos(a + b*x),x)

[Out]

int((sin(3*a + 3*b*x)*(c + d*x))/cos(a + b*x), x)