3.30 \(\int (c+d x)^2 \sec (a+b x) \, dx\)

Optimal. Leaf size=137 \[ \frac{2 i d (c+d x) \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac{2 i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

((-2*I)*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b + ((2*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*Po
lyLog[3, I*E^(I*(a + b*x))])/b^3

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Rubi [A]  time = 0.0920892, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4181, 2531, 2282, 6589} \[ \frac{2 i d (c+d x) \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac{2 i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sec[a + b*x],x]

[Out]

((-2*I)*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b + ((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*Po
lyLog[3, I*E^(I*(a + b*x))])/b^3

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

Mathematica [A]  time = 0.108983, size = 130, normalized size = 0.95 \[ -\frac{2 i \left (b^2 (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )-d \left (b (c+d x) \text{Li}_2\left (-i e^{i (a+b x)}\right )+i d \text{Li}_3\left (-i e^{i (a+b x)}\right )\right )+d \left (b (c+d x) \text{Li}_2\left (i e^{i (a+b x)}\right )+i d \text{Li}_3\left (i e^{i (a+b x)}\right )\right )\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Sec[a + b*x],x]

[Out]

((-2*I)*(b^2*(c + d*x)^2*ArcTan[E^(I*(a + b*x))] - d*(b*(c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))] + I*d*PolyL
og[3, (-I)*E^(I*(a + b*x))]) + d*(b*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))] + I*d*PolyLog[3, I*E^(I*(a + b*x))
])))/b^3

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Maple [B]  time = 0.366, size = 392, normalized size = 2.9 \begin{align*}{\frac{4\,icda\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\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{2\,idc{\it polylog} \left ( 2,i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{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{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 ) x}{b}}-2\,{\frac{cd\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\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{{a}^{2}{d}^{2}\ln \left ( 1+i{{\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{2\,i{c}^{2}\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-i{{\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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.85081, size = 535, normalized size = 3.91 \begin{align*} \frac{2 \, c^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right ) - \frac{4 \, a c d \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b} + \frac{2 \, a^{2} d^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{2}} + \frac{4 \, d^{2}{\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2}{\rm Li}_{3}(-i \, e^{\left (i \, b x + i \, a\right )}) +{\left (-2 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (-4 i \, b c d + 4 i \, a d^{2}\right )}{\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), \sin \left (b x + a\right ) + 1\right ) +{\left (-2 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (-4 i \, b c d + 4 i \, a d^{2}\right )}{\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) +{\left (-4 i \, b c d - 4 i \,{\left (b x + a\right )} d^{2} + 4 i \, a d^{2}\right )}{\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) +{\left (4 i \, b c d + 4 i \,{\left (b x + a\right )} d^{2} - 4 i \, a d^{2}\right )}{\rm Li}_2\left (-i \, e^{\left (i \, b x + i \, a\right )}\right ) +{\left ({\left (b x + a\right )}^{2} d^{2} + 2 \,{\left (b c d - a d^{2}\right )}{\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) + 1\right ) -{\left ({\left (b x + a\right )}^{2} d^{2} + 2 \,{\left (b c d - a d^{2}\right )}{\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \sin \left (b x + a\right ) + 1\right )}{b^{2}}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.70654, size = 1534, normalized size = 11.2 \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),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)) - (-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))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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