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

Optimal. Leaf size=96 \[ -\frac{i d^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d} \]

[Out]

((-I)*(c + d*x)^2)/b - (c + d*x)^3/(3*d) + (2*d*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^2 - (I*d^2*PolyLog[2
, -E^((2*I)*(a + b*x))])/b^3 + ((c + d*x)^2*Tan[a + b*x])/b

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Rubi [A]  time = 0.141087, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3720, 3719, 2190, 2279, 2391, 32} \[ -\frac{i d^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tan[a + b*x]^2,x]

[Out]

((-I)*(c + d*x)^2)/b - (c + d*x)^3/(3*d) + (2*d*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^2 - (I*d^2*PolyLog[2
, -E^((2*I)*(a + b*x))])/b^3 + ((c + d*x)^2*Tan[a + b*x])/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 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]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int (c+d x)^2 \tan ^2(a+b x) \, dx &=\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{(2 d) \int (c+d x) \tan (a+b x) \, dx}{b}-\int (c+d x)^2 \, dx\\ &=-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d}+\frac{(c+d x)^2 \tan (a+b x)}{b}+\frac{(4 i d) \int \frac{e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d}+\frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{\left (2 d^2\right ) \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d}+\frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tan (a+b x)}{b}+\frac{\left (i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{b^3}\\ &=-\frac{i (c+d x)^2}{b}-\frac{(c+d x)^3}{3 d}+\frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac{i d^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^2 \tan (a+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 6.38096, size = 276, normalized size = 2.88 \[ \frac{d^2 \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{b^3 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac{2 c d \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b^2 \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac{\sec (a) \sec (a+b x) \left (c^2 \sin (b x)+2 c d x \sin (b x)+d^2 x^2 \sin (b x)\right )}{b}-\frac{1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(x*(3*c^2 + 3*c*d*x + d^2*x^2))/3 + (2*c*d*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]
))/(b^2*(Cos[a]^2 + Sin[a]^2)) + (d^2*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]]))] + P
i*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])/(b^3*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + (Sec[a]*Sec[a + b*x]*(c^2*Sin[
b*x] + 2*c*d*x*Sin[b*x] + d^2*x^2*Sin[b*x]))/b

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Maple [B]  time = 0.162, size = 191, normalized size = 2. \begin{align*} -{\frac{{d}^{2}{x}^{3}}{3}}-cd{x}^{2}-{c}^{2}x+{\frac{2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-{\frac{2\,i{d}^{2}{x}^{2}}{b}}-{\frac{4\,i{d}^{2}ax}{{b}^{2}}}-{\frac{2\,i{d}^{2}{a}^{2}}{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+4\,{\frac{{d}^{2}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.84466, size = 563, normalized size = 5.86 \begin{align*} \frac{i \, b^{3} d^{2} x^{3} + 3 i \, b^{3} c d x^{2} + 3 i \, b^{3} c^{2} x + 6 \, b^{2} c^{2} +{\left (6 \, b d^{2} x + 6 \, b c d + 6 \,{\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (6 i \, b d^{2} x + 6 i \, b c d\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 (i \, b^{3} d^{2} x^{3} +{\left (3 i \, b^{3} c d - 6 \, b^{2} d^{2}\right )} x^{2} - 3 \,{\left (-i \, b^{3} c^{2} + 4 \, b^{2} c d\right )} x\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \,{\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) + d^{2}\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (-3 i \, b d^{2} x - 3 i \, b c d +{\left (-3 i \, b d^{2} x - 3 i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \,{\left (b d^{2} x + b c d\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 (b^{3} d^{2} x^{3} + 3 \,{\left (b^{3} c d + 2 i \, b^{2} d^{2}\right )} x^{2} +{\left (3 \, b^{3} c^{2} + 12 i \, b^{2} c d\right )} x\right )} \sin \left (2 \, b x + 2 \, a\right )}{-3 i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b^{3} \sin \left (2 \, b x + 2 \, a\right ) - 3 i \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(I*b^3*d^2*x^3 + 3*I*b^3*c*d*x^2 + 3*I*b^3*c^2*x + 6*b^2*c^2 + (6*b*d^2*x + 6*b*c*d + 6*(b*d^2*x + b*c*d)*cos(
2*b*x + 2*a) + (6*I*b*d^2*x + 6*I*b*c*d)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (
I*b^3*d^2*x^3 + (3*I*b^3*c*d - 6*b^2*d^2)*x^2 - 3*(-I*b^3*c^2 + 4*b^2*c*d)*x)*cos(2*b*x + 2*a) - 3*(d^2*cos(2*
b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) + d^2)*dilog(-e^(2*I*b*x + 2*I*a)) + (-3*I*b*d^2*x - 3*I*b*c*d + (-3*I*b*d
^2*x - 3*I*b*c*d)*cos(2*b*x + 2*a) + 3*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x
+ 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (b^3*d^2*x^3 + 3*(b^3*c*d + 2*I*b^2*d^2)*x^2 + (3*b^3*c^2 + 12*I*b^2*c*d)
*x)*sin(2*b*x + 2*a))/(-3*I*b^3*cos(2*b*x + 2*a) + 3*b^3*sin(2*b*x + 2*a) - 3*I*b^3)

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Fricas [B]  time = 0.490464, size = 522, normalized size = 5.44 \begin{align*} -\frac{2 \, b^{3} d^{2} x^{3} + 6 \, b^{3} c d x^{2} + 6 \, b^{3} c^{2} x - 3 i \, d^{2}{\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 ) + 3 i \, d^{2}{\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 ) - 6 \,{\left (b d^{2} x + b c d\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \,{\left (b d^{2} x + b c d\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \tan \left (b x + a\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^3*d^2*x^3 + 6*b^3*c*d*x^2 + 6*b^3*c^2*x - 3*I*d^2*dilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)
+ 1) + 3*I*d^2*dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 6*(b*d^2*x + b*c*d)*log(-2*(I*tan(b*x
+ a) - 1)/(tan(b*x + a)^2 + 1)) - 6*(b*d^2*x + b*c*d)*log(-2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 6*
(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*tan(b*x + a))/b^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \tan ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \tan \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*tan(b*x+a)^2,x, algorithm="giac")

[Out]

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