3.306 \(\int (c+d x) \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=108 \[ -\frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \tan (a+b x)}{2 b^2}+\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x) \tan ^2(a+b x)}{2 b}+\frac{d x}{2 b}-\frac{i (c+d x)^2}{2 d} \]

[Out]

(d*x)/(2*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*Tan[a + b*x])/(2*b^2) + ((c + d*x)*Tan[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.116911, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3720, 3473, 8, 3719, 2190, 2279, 2391} \[ -\frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \tan (a+b x)}{2 b^2}+\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x) \tan ^2(a+b x)}{2 b}+\frac{d x}{2 b}-\frac{i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Tan[a + b*x]^3,x]

[Out]

(d*x)/(2*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*Tan[a + b*x])/(2*b^2) + ((c + d*x)*Tan[a + b*x]^2)/(2*b)

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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) \tan ^3(a+b x) \, dx &=\frac{(c+d x) \tan ^2(a+b x)}{2 b}-\frac{d \int \tan ^2(a+b x) \, dx}{2 b}-\int (c+d x) \tan (a+b x) \, dx\\ &=-\frac{i (c+d x)^2}{2 d}-\frac{d \tan (a+b x)}{2 b^2}+\frac{(c+d x) \tan ^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 1 \, dx}{2 b}\\ &=\frac{d x}{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 \tan (a+b x)}{2 b^2}+\frac{(c+d x) \tan ^2(a+b x)}{2 b}-\frac{d \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{d x}{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 \tan (a+b x)}{2 b^2}+\frac{(c+d x) \tan ^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}{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{i d \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \tan (a+b x)}{2 b^2}+\frac{(c+d x) \tan ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [B]  time = 6.15395, size = 240, normalized size = 2.22 \[ \frac{d \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{2 b^2 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}-\frac{d \sec (a) \sin (b x) \sec (a+b x)}{2 b^2}+\frac{c \left (\tan ^2(a+b x)+2 \log (\cos (a+b x))\right )}{2 b}+\frac{d x \sec ^2(a+b x)}{2 b}-\frac{1}{2} d x^2 \tan (a) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Tan[a + b*x]^3,x]

[Out]

(d*x*Sec[a + b*x]^2)/(2*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]]))]))
/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (d*Sec[a]*Sec[a + b*x]*Sin[b*x])/(
2*b^2) - (d*x^2*Tan[a])/2 + (c*(2*Log[Cos[a + b*x]] + Tan[a + b*x]^2))/(2*b)

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Maple [A]  time = 0.159, size = 183, normalized size = 1.7 \begin{align*} -{\frac{i}{2}}d{x}^{2}+icx+{\frac{2\,bdx{{\rm e}^{2\,i \left ( bx+a \right ) }}+2\,bc{{\rm e}^{2\,i \left ( bx+a \right ) }}-id{{\rm e}^{2\,i \left ( bx+a \right ) }}-id}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) ^{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)*tan(b*x+a)^3,x)

[Out]

-1/2*I*d*x^2+I*c*x+(2*b*d*x*exp(2*I*(b*x+a))+2*b*c*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))-I*d)/b^2/(exp(2*I*(b*
x+a))+1)^2-2/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 [B]  time = 2.04237, size = 701, normalized size = 6.49 \begin{align*} -\frac{b^{2} d x^{2} + 2 \, b^{2} c x -{\left (2 \, b d x + 2 \, b c + 2 \,{\left (b d x + b c\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \,{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (2 i \, b d x + 2 i \, b c\right )} \sin \left (4 \, b x + 4 \, a\right ) +{\left (4 i \, b d x + 4 i \, b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (b^{2} d x^{2} + 2 \, b^{2} c x\right )} \cos \left (4 \, b x + 4 \, a\right ) +{\left (2 \, b^{2} d x^{2} + 4 i \, b c +{\left (4 \, b^{2} c + 4 i \, b d\right )} x + 2 \, d\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (d \cos \left (4 \, b x + 4 \, a\right ) + 2 \, d \cos \left (2 \, b x + 2 \, a\right ) + i \, d \sin \left (4 \, b x + 4 \, a\right ) + 2 i \, d \sin \left (2 \, b x + 2 \, a\right ) + d\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) -{\left (-i \, b d x - i \, b c +{\left (-i \, b d x - i \, b c\right )} \cos \left (4 \, b x + 4 \, a\right ) +{\left (-2 i \, b d x - 2 i \, b c\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right ) + 2 \,{\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )\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 ) -{\left (-i \, b^{2} d x^{2} - 2 i \, b^{2} c x\right )} \sin \left (4 \, b x + 4 \, a\right ) -{\left (-2 i \, b^{2} d x^{2} + 4 \, b c - 4 \,{\left (i \, b^{2} c - b d\right )} x - 2 i \, d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, d}{-2 i \, b^{2} \cos \left (4 \, b x + 4 \, a\right ) - 4 i \, b^{2} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, b^{2} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, b^{2} \sin \left (2 \, b x + 2 \, a\right ) - 2 i \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.492585, size = 447, normalized size = 4.14 \begin{align*} \frac{2 \, b d x + 2 \,{\left (b d x + b c\right )} \tan \left (b x + a\right )^{2} + i \, d{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, d{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \,{\left (b d x + b c\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \,{\left (b d x + b c\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, d \tan \left (b x + a\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \tan ^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \tan \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)*tan(b*x+a)^3,x, algorithm="giac")

[Out]

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