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

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

[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

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Rubi [A]  time = 0.146066, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, 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 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 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 2637

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

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 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 \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*}

Mathematica [A]  time = 0.883213, size = 315, normalized size = 1.69 \[ -\frac{-2 i b d (c+d x) \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )+2 i b d (c+d x) \text{PolyLog}\left (2,i e^{i (a+b x)}\right )+2 d^2 \text{PolyLog}\left (3,-i e^{i (a+b x)}\right )-2 d^2 \text{PolyLog}\left (3,i e^{i (a+b x)}\right )+b^2 c^2 \sin (a+b x)+2 i b^2 c^2 \tan ^{-1}\left (e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )+2 b^2 c d x \sin (a+b x)-b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+b^2 d^2 x^2 \sin (a+b x)+2 b c d \cos (a+b x)-2 d^2 \sin (a+b x)+2 b d^2 x \cos (a+b x)}{b^3} \]

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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Maple [B]  time = 0.361, size = 512, normalized size = 2.8 \begin{align*}{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-{\frac{2\,i{d}^{2}{a}^{2}\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{cd\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}+{\frac{{a}^{2}{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,idc{\it polylog} \left ( 2,i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{\frac{i}{2}} \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}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+{\frac{2\,idc{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{4\,icda\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+2\,{\frac{cd\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,i{c}^{2}\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-2\,{\frac{cd\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}-{\frac{{a}^{2}{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{\frac{i}{2}} \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}}} \end{align*}

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]

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

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Maxima [B]  time = 1.98873, size = 689, normalized size = 3.7 \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(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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Fricas [C]  time = 0.676838, size = 1666, normalized size = 8.96 \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(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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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(b*x+a)**2,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 (b x + a\right )^{2}\,{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(b*x+a)^2,x, algorithm="giac")

[Out]

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