3.218 \(\int (c+d x) \sin (a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=103 \[ \frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {d \cos (a+b x)}{b^2}-\frac {(c+d x) \sin (a+b x)}{b}-\frac {2 i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4407, 3296, 2638, 4181, 2279, 2391} \[ \frac {i d \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {d \cos (a+b x)}{b^2}-\frac {(c+d x) \sin (a+b x)}{b}-\frac {2 i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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]

Rubi steps

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

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Mathematica [B]  time = 0.40, size = 213, normalized size = 2.07 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\text {Li}_2\left (e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )+\left (-a-b x+\frac {\pi }{2}\right ) \left (\log \left (1-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )-\left (\frac {\pi }{2}-a\right ) \log \left (\tan \left (\frac {1}{2} \left (-a-b x+\frac {\pi }{2}\right )\right )\right )\right )}{b^2}-\frac {d \cos (b x) (b x \sin (a)+\cos (a))}{b^2}-\frac {d \sin (b x) (b x \cos (a)-\sin (a))}{b^2}-\frac {c \sin (a+b x)}{b}+\frac {c \tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.04, size = 331, normalized size = 3.21 \[ -\frac {2 \, d \cos \left (b x + a\right ) + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 209, normalized size = 2.03 \[ -\frac {d \sin \left (b x +a \right ) x}{b}-\frac {d \cos \left (b x +a \right )}{b^{2}}-\frac {c \sin \left (b x +a \right )}{b}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {i d \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {i d \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d a \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b^{2}}+\frac {c \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b*d*sin(b*x+a)*x-d*cos(b*x+a)/b^2-1/b*c*sin(b*x+a)-1/b*d*ln(1+I*exp(I*(b*x+a)))*x+1/b*d*ln(1-I*exp(I*(b*x+a
)))*x-I/b^2*d*dilog(1-I*exp(I*(b*x+a)))-1/b^2*d*ln(1+I*exp(I*(b*x+a)))*a+1/b^2*d*ln(1-I*exp(I*(b*x+a)))*a+I/b^
2*d*dilog(1+I*exp(I*(b*x+a)))-1/b^2*d*a*ln(sec(b*x+a)+tan(b*x+a))+1/b*c*ln(sec(b*x+a)+tan(b*x+a))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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