3.1.30 \(\int (c+d x)^2 \sec (a+b x) \, dx\) [30]

Optimal. Leaf size=137 \[ -\frac {2 i (c+d x)^2 \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {4266, 2611, 2320, 6724} \begin {gather*} -\frac {2 i (c+d x)^2 \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}-\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 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} \end {gather*}

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 2320

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 2611

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 4266

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 6724

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 130, normalized size = 0.95 \begin {gather*} -\frac {2 i \left (b^2 (c+d x)^2 \text {ArcTan}\left (e^{i (a+b x)}\right )-d \left (b (c+d x) \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )+i d \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )\right )+d \left (b (c+d x) \text {PolyLog}\left (2,i e^{i (a+b x)}\right )+i d \text {PolyLog}\left (3,i e^{i (a+b x)}\right )\right )\right )}{b^3} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (120 ) = 240\).
time = 0.10, size = 392, normalized size = 2.86

method	result	size
risch	\(-\frac {2 i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {a^{2} d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i c d \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {2 d^{2} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {2 d^{2} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}\)	\(392\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (111) = 222\).
time = 0.59, size = 402, normalized size = 2.93 \begin {gather*} \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 )}) - 2 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - 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 ) - 2 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - 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 ) - 4 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) - 4 \, {\left (-i \, b c d - i \, {\left (b x + a\right )} d^{2} + 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 {gather*}

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 + 2*(I*b*c*d - 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 + 2*(I*b*c*d - I*a*d^2)*(b*x + a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 4*(I*b*c*d + I*(b*x + a)*
d^2 - I*a*d^2)*dilog(I*e^(I*b*x + I*a)) - 4*(-I*b*c*d - I*(b*x + a)*d^2 + 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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (111) = 222\).
time = 0.44, size = 598, normalized size = 4.36 \begin {gather*} -\frac {2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{3}} \end {gather*}

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
+ I*b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) - sin(b*x + a)
) + 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + 2*(-I*b*d^2*x - 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{2} \sec {\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^2}{\cos \left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/cos(a + b*x),x)

[Out]

int((c + d*x)^2/cos(a + b*x), x)

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