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

Optimal. Leaf size=115 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}+\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d} \]

[Out]

(d*x)/(4*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])/(4*b^2) - ((c + d*x)*Sin[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.127852, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4407, 4404, 2635, 8, 3719, 2190, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}+\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)/(4*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])/(4*b^2) - ((c + d*x)*Sin[a + b*x]^2)/(2*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 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 2635

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] && Integer
Q[2*n]

Rule 8

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

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

Mathematica [A]  time = 0.303667, size = 134, normalized size = 1.17 \[ \frac{d \left (\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )+\frac{1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )\right )}{b^2}-\frac{d \sin (2 (a+b x))}{8 b^2}+\frac{a d \log (\cos (a+b x))}{b^2}-\frac{c \left (\log (\cos (a+b x))-\frac{1}{2} \cos ^2(a+b x)\right )}{b}+\frac{d x \cos (2 (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Cos[2*(a + b*x)])/(4*b) + (a*d*Log[Cos[a + b*x]])/b^2 - (c*(-Cos[a + b*x]^2/2 + Log[Cos[a + b*x]]))/b + (
d*((I/2)*(a + b*x)^2 - (a + b*x)*Log[1 + E^((2*I)*(a + b*x))] + (I/2)*PolyLog[2, -E^((2*I)*(a + b*x))]))/b^2 -
(d*Sin[2*(a + b*x)])/(8*b^2)

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Maple [A]  time = 0.355, size = 179, normalized size = 1.6 \begin{align*}{\frac{i}{2}}d{x}^{2}-icx+{\frac{ \left ( 2\,dxb+id+2\,bc \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{16\,{b}^{2}}}+{\frac{ \left ( 2\,dxb-id+2\,bc \right ){{\rm e}^{-2\,i \left ( bx+a \right ) }}}{16\,{b}^{2}}}+2\,{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{2\,idax}{b}}+{\frac{id{a}^{2}}{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{ad\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*d*x^2-I*c*x+1/16*(2*d*x*b+I*d+2*b*c)/b^2*exp(2*I*(b*x+a))+1/16*(2*d*x*b-I*d+2*b*c)/b^2*exp(-2*I*(b*x+a))
+2/b*c*ln(exp(I*(b*x+a)))-1/b*c*ln(exp(2*I*(b*x+a))+1)+2*I/b*d*a*x+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)))

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Maxima [A]  time = 1.77019, size = 196, normalized size = 1.7 \begin{align*} -\frac{-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x +{\left (8 i \, b d x + 8 i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \,{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 i \, d{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 4 \,{\left (b d x + b c\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 ) + d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-4*I*b^2*d*x^2 - 8*I*b^2*c*x + (8*I*b*d*x + 8*I*b*c)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 2
*(b*d*x + b*c)*cos(2*b*x + 2*a) - 4*I*d*dilog(-e^(2*I*b*x + 2*I*a)) + 4*(b*d*x + b*c)*log(cos(2*b*x + 2*a)^2 +
sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + d*sin(2*b*x + 2*a))/b^2

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Fricas [B]  time = 0.61366, size = 965, normalized size = 8.39 \begin{align*} -\frac{b d x - 2 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 2 i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \,{\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 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 + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \,{\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 c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*sec(b*x+a)*sin(b*x+a)**3,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 )} \sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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