3.292 \(\int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=115 \[ \frac{3 i d^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}+\frac{3 i d (c+d x)^2}{2 b^2} \]

[Out]

(((3*I)/2)*d*(c + d*x)^2)/b^2 - (3*d^2*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^3 + (((3*I)/2)*d^3*PolyLog[2,
-E^((2*I)*(a + b*x))])/b^4 + ((c + d*x)^3*Sec[a + b*x]^2)/(2*b) - (3*d*(c + d*x)^2*Tan[a + b*x])/(2*b^2)

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Rubi [A]  time = 0.173672, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4409, 4184, 3719, 2190, 2279, 2391} \[ \frac{3 i d^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}+\frac{3 i d (c+d x)^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

(((3*I)/2)*d*(c + d*x)^2)/b^2 - (3*d^2*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^3 + (((3*I)/2)*d^3*PolyLog[2,
-E^((2*I)*(a + b*x))])/b^4 + ((c + d*x)^3*Sec[a + b*x]^2)/(2*b) - (3*d*(c + d*x)^2*Tan[a + b*x])/(2*b^2)

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && 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]

Rubi steps

\begin{align*} \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx &=\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x)^2 \sec ^2(a+b x) \, dx}{2 b}\\ &=\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{\left (3 d^2\right ) \int (c+d x) \tan (a+b x) \, dx}{b^2}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}-\frac{\left (6 i d^2\right ) \int \frac{e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}} \, dx}{b^2}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{\left (3 d^3\right ) \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}-\frac{\left (3 i d^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{3 i d^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}\\ \end{align*}

Mathematica [B]  time = 6.39472, size = 286, normalized size = 2.49 \[ -\frac{3 d^3 \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{2 b^4 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}-\frac{3 \sec (a) \sec (a+b x) \left (c^2 d \sin (b x)+2 c d^2 x \sin (b x)+d^3 x^2 \sin (b x)\right )}{2 b^2}-\frac{3 c d^2 \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b^3 \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

((c + d*x)^3*Sec[a + b*x]^2)/(2*b) - (3*c*d^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[
a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) - (3*d^3*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*Arc
Tan[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[Co
t[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^4*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (3*Sec[a]*Sec[a + b*x]*(
c^2*d*Sin[b*x] + 2*c*d^2*x*Sin[b*x] + d^3*x^2*Sin[b*x]))/(2*b^2)

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Maple [B]  time = 0.158, size = 301, normalized size = 2.6 \begin{align*}{\frac{2\,b{d}^{3}{x}^{3}{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{d}^{3}{x}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}+6\,bc{d}^{2}{x}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}-6\,ic{d}^{2}x{{\rm e}^{2\,i \left ( bx+a \right ) }}+6\,b{c}^{2}dx{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{c}^{2}d{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{d}^{3}{x}^{2}+2\,b{c}^{3}{{\rm e}^{2\,i \left ( bx+a \right ) }}-6\,ic{d}^{2}x-3\,i{c}^{2}d}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) ^{2}}}+6\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-3\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{3}}}+{\frac{3\,i{d}^{3}{x}^{2}}{{b}^{2}}}+{\frac{6\,i{d}^{3}ax}{{b}^{3}}}+{\frac{3\,i{d}^{3}{a}^{2}}{{b}^{4}}}-3\,{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{3}}}+{\frac{{\frac{3\,i}{2}}{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-6\,{\frac{{d}^{3}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x)

[Out]

(2*b*d^3*x^3*exp(2*I*(b*x+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6*b*c*d^2*x^2*exp(2*I*(b*x+a))-6*I*c*d^2*x*exp(2*I*
(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))-3*I*c^2*d*exp(2*I*(b*x+a))-3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))-6*I*c*d^
2*x-3*I*c^2*d)/b^2/(exp(2*I*(b*x+a))+1)^2+6/b^3*d^2*c*ln(exp(I*(b*x+a)))-3*d^2/b^3*c*ln(exp(2*I*(b*x+a))+1)+3*
I*d^3/b^2*x^2+6*I*d^3/b^3*a*x+3*I*d^3/b^4*a^2-3*d^3/b^3*ln(exp(2*I*(b*x+a))+1)*x+3/2*I*d^3*polylog(2,-exp(2*I*
(b*x+a)))/b^4-6/b^4*d^3*a*ln(exp(I*(b*x+a)))

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Maxima [B]  time = 2.24113, size = 900, normalized size = 7.83 \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)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="maxima")

[Out]

-(6*b^2*c^2*d + (6*b*d^3*x + 6*b*c*d^2 + 6*(b*d^3*x + b*c*d^2)*cos(4*b*x + 4*a) + 12*(b*d^3*x + b*c*d^2)*cos(2
*b*x + 2*a) + (6*I*b*d^3*x + 6*I*b*c*d^2)*sin(4*b*x + 4*a) + (12*I*b*d^3*x + 12*I*b*c*d^2)*sin(2*b*x + 2*a))*a
rctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(b^2*d^3*x^2 + 2*b^2*c*d^2*x)*cos(4*b*x + 4*a) + (4*I*b^3*d
^3*x^3 + 4*I*b^3*c^3 + 6*b^2*c^2*d + (12*I*b^3*c*d^2 - 6*b^2*d^3)*x^2 - 12*(-I*b^3*c^2*d + b^2*c*d^2)*x)*cos(2
*b*x + 2*a) - (3*d^3*cos(4*b*x + 4*a) + 6*d^3*cos(2*b*x + 2*a) + 3*I*d^3*sin(4*b*x + 4*a) + 6*I*d^3*sin(2*b*x
+ 2*a) + 3*d^3)*dilog(-e^(2*I*b*x + 2*I*a)) + (-3*I*b*d^3*x - 3*I*b*c*d^2 + (-3*I*b*d^3*x - 3*I*b*c*d^2)*cos(4
*b*x + 4*a) + (-6*I*b*d^3*x - 6*I*b*c*d^2)*cos(2*b*x + 2*a) + 3*(b*d^3*x + b*c*d^2)*sin(4*b*x + 4*a) + 6*(b*d^
3*x + b*c*d^2)*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) + (-6*I
*b^2*d^3*x^2 - 12*I*b^2*c*d^2*x)*sin(4*b*x + 4*a) - (4*b^3*d^3*x^3 + 4*b^3*c^3 - 6*I*b^2*c^2*d + 6*(2*b^3*c*d^
2 + I*b^2*d^3)*x^2 + (12*b^3*c^2*d + 12*I*b^2*c*d^2)*x)*sin(2*b*x + 2*a))/(-2*I*b^4*cos(4*b*x + 4*a) - 4*I*b^4
*cos(2*b*x + 2*a) + 2*b^4*sin(4*b*x + 4*a) + 4*b^4*sin(2*b*x + 2*a) - 2*I*b^4)

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Fricas [B]  time = 0.690775, size = 1374, normalized size = 11.95 \begin{align*} \frac{b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3} - 3 i \, d^{3} \cos \left (b x + a\right )^{2}{\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 3 i \, d^{3} \cos \left (b x + a\right )^{2}{\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 3 i \, d^{3} \cos \left (b x + a\right )^{2}{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 3 i \, d^{3} \cos \left (b x + a\right )^{2}{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \,{\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - 3 \,{\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 3 \,{\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - 3 \,{\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \,{\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - 3 \,{\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \,{\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - 3 \,{\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 3 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b^{4} \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="fricas")

[Out]

1/2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3 - 3*I*d^3*cos(b*x + a)^2*dilog(I*cos(b*x + a) + s
in(b*x + a)) + 3*I*d^3*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a)) + 3*I*d^3*cos(b*x + a)^2*dilog(-I*c
os(b*x + a) + sin(b*x + a)) - 3*I*d^3*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 3*(b*c*d^2 - a*d^
3)*cos(b*x + a)^2*log(cos(b*x + a) + I*sin(b*x + a) + I) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(cos(b*x + a)
- I*sin(b*x + a) + I) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(I*cos(b*x + a) + sin(b*x + a) + 1) - 3*(b*d^3*
x + a*d^3)*cos(b*x + a)^2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(-I*c
os(b*x + a) + sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(-I*cos(b*x + a) - sin(b*x + a) + 1) -
3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(-cos(b*x + a) + I*sin(b*x + a) + I) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)
^2*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos(b*x + a)*sin(b*x
+ a))/(b^4*cos(b*x + a)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \tan{\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*sec(b*x+a)**2*tan(b*x+a),x)

[Out]

Integral((c + d*x)**3*tan(a + b*x)*sec(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 )}^{3} \sec \left (b x + a\right )^{2} \tan \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*sec(b*x + a)^2*tan(b*x + a), x)