3.1.29 \(\int (c+d x)^3 \sec (a+b x) \, dx\) [29]

Optimal. Leaf size=205 \[ -\frac {2 i (c+d x)^3 \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 \text {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4266, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {2 i (c+d x)^3 \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}-\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sec[a + b*x],x]

[Out]

((-2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + ((3*I)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 -
((3*I)*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b^2 - (6*d^2*(c + d*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b
^3 + (6*d^2*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b^3 - ((6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^4 +
((6*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))])/b^4

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]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int (c+d x)^3 \sec (a+b x) \, dx &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(3 d) \int (c+d x)^2 \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b}+\frac {(3 d) \int (c+d x)^2 \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {\left (6 d^3\right ) \int \text {Li}_3\left (-i e^{i (a+b x)}\right ) \, dx}{b^3}-\frac {\left (6 d^3\right ) \int \text {Li}_3\left (i e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac {\left (6 i d^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac {\left (6 i d^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 196, normalized size = 0.96 \begin {gather*} -\frac {i \left (2 b^3 (c+d x)^3 \text {ArcTan}\left (e^{i (a+b x)}\right )-3 d \left (b^2 (c+d x)^2 \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )-2 d^2 \text {PolyLog}\left (4,-i e^{i (a+b x)}\right )\right )+3 d \left (b^2 (c+d x)^2 \text {PolyLog}\left (2,i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {PolyLog}\left (3,i e^{i (a+b x)}\right )-2 d^2 \text {PolyLog}\left (4,i e^{i (a+b x)}\right )\right )\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Sec[a + b*x],x]

[Out]

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

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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. 684 vs. \(2 (180 ) = 360\).
time = 0.19, size = 685, normalized size = 3.34

method	result	size
risch	\(\frac {d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {a^{3} d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {a^{3} d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {2 i c^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {6 d^{2} c \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 d^{3} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 d^{2} c \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{3} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 i d^{3} \polylog \left (4, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i d^{3} \polylog \left (4, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i c \,d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i c^{2} d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 i d^{2} c \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 i d^{2} c \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 i d^{3} a^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i d^{3} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i c^{2} d \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i c^{2} d \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i d^{3} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {3 a^{2} c \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {3 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {3 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {3 a^{2} c \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(685\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*sec(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-6*I/b^3*c*d^2*a^2*arctan(exp(I*(b*x+a)))+6*I/b^2*c^2*d*a*arctan(exp(I*(b*x+a)))-6*I/b^2*d^2*c*polylog(2,I*exp
(I*(b*x+a)))*x+6*I/b^2*d^2*c*polylog(2,-I*exp(I*(b*x+a)))*x+6/b^3*d^2*c*polylog(3,I*exp(I*(b*x+a)))+1/b*d^3*ln
(1-I*exp(I*(b*x+a)))*x^3-1/b*d^3*ln(1+I*exp(I*(b*x+a)))*x^3-1/b^4*a^3*d^3*ln(1+I*exp(I*(b*x+a)))+1/b^4*a^3*d^3
*ln(1-I*exp(I*(b*x+a)))+6/b^3*d^3*polylog(3,I*exp(I*(b*x+a)))*x-6/b^3*d^2*c*polylog(3,-I*exp(I*(b*x+a)))-6/b^3
*d^3*polylog(3,-I*exp(I*(b*x+a)))*x-2*I/b*c^3*arctan(exp(I*(b*x+a)))-3/b*c^2*d*ln(1+I*exp(I*(b*x+a)))*x-3/b^2*
c^2*d*ln(1+I*exp(I*(b*x+a)))*a-3/b^3*a^2*c*d^2*ln(1-I*exp(I*(b*x+a)))+3/b*c^2*d*ln(1-I*exp(I*(b*x+a)))*x+3/b^2
*c^2*d*ln(1-I*exp(I*(b*x+a)))*a+3/b*d^2*c*ln(1-I*exp(I*(b*x+a)))*x^2-3/b*d^2*c*ln(1+I*exp(I*(b*x+a)))*x^2+3/b^
3*a^2*c*d^2*ln(1+I*exp(I*(b*x+a)))+6*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4-6*I*d^3*polylog(4,-I*exp(I*(b*x+a))
)/b^4+3*I/b^2*c^2*d*polylog(2,-I*exp(I*(b*x+a)))+3*I/b^2*d^3*polylog(2,-I*exp(I*(b*x+a)))*x^2-3*I/b^2*d^3*poly
log(2,I*exp(I*(b*x+a)))*x^2+2*I/b^4*d^3*a^3*arctan(exp(I*(b*x+a)))-3*I/b^2*c^2*d*polylog(2,I*exp(I*(b*x+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. 722 vs. \(2 (167) = 334\).
time = 0.63, size = 722, normalized size = 3.52 \begin {gather*} \frac {2 \, c^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right ) - \frac {6 \, a c^{2} d \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b} + \frac {6 \, a^{2} c d^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{2}} - \frac {2 \, a^{3} d^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{3}} + \frac {12 i \, d^{3} {\rm Li}_{4}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 i \, d^{3} {\rm Li}_{4}(-i \, e^{\left (i \, b x + i \, a\right )}) - 2 \, {\left (i \, {\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, a^{2} d^{3}\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 )}^{3} d^{3} + 3 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) - 6 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, {\left (b x + a\right )}^{2} d^{3} + i \, a^{2} d^{3} + 2 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\left (-i \, b^{2} c^{2} d + 2 i \, a b c d^{2} - i \, {\left (b x + a\right )}^{2} d^{3} - i \, a^{2} d^{3} + 2 \, {\left (-i \, b c d^{2} + i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 ) + 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(-i \, e^{\left (i \, b x + i \, a\right )})}{b^{3}}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^3*log(sec(b*x + a) + tan(b*x + a)) - 6*a*c^2*d*log(sec(b*x + a) + tan(b*x + a))/b + 6*a^2*c*d^2*log(s
ec(b*x + a) + tan(b*x + a))/b^2 - 2*a^3*d^3*log(sec(b*x + a) + tan(b*x + a))/b^3 + (12*I*d^3*polylog(4, I*e^(I
*b*x + I*a)) - 12*I*d^3*polylog(4, -I*e^(I*b*x + I*a)) - 2*(I*(b*x + a)^3*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x +
a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*a^2*d^3)*(b*x + a))*arctan2(cos(b*x + a), sin(b*x + a) + 1) - 2*(I*
(b*x + a)^3*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*a^2*d^3)*(b*x + a))
*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 6*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*(b*x + a)^2*d^3 + I*a^2*d^3 + 2
*(I*b*c*d^2 - I*a*d^3)*(b*x + a))*dilog(I*e^(I*b*x + I*a)) - 6*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*(b*x + a)^2*d
^3 - I*a^2*d^3 + 2*(-I*b*c*d^2 + I*a*d^3)*(b*x + a))*dilog(-I*e^(I*b*x + I*a)) + ((b*x + a)^3*d^3 + 3*(b*c*d^2
- a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 +
2*sin(b*x + a) + 1) - ((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d
^3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) + 12*(b*c*d^2 + (b*x + a)*d^3 - a*d^3
)*polylog(3, I*e^(I*b*x + I*a)) - 12*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*polylog(3, -I*e^(I*b*x + I*a)))/b^3)/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. 970 vs. \(2 (167) = 334\).
time = 0.45, size = 970, normalized size = 4.73 \begin {gather*} \frac {6 i \, d^{3} {\rm polylog}\left (4, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 6 i \, d^{3} {\rm polylog}\left (4, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 6 i \, d^{3} {\rm polylog}\left (4, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 i \, d^{3} {\rm polylog}\left (4, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \, {\left (i \, b^{2} d^{3} x^{2} + 2 i \, b^{2} c d^{2} x + i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 3 \, {\left (i \, b^{2} d^{3} x^{2} + 2 i \, b^{2} c d^{2} x + i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \, {\left (-i \, b^{2} d^{3} x^{2} - 2 i \, b^{2} c d^{2} x - i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 3 \, {\left (-i \, b^{2} d^{3} x^{2} - 2 i \, b^{2} c d^{2} x - i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4, I*cos(b*x + a) - sin(b*x + a)) - 6
*I*d^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 3*(I*
b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2
*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2
*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dilog(-I*cos(b*
x + a) - sin(b*x + a)) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) + I*sin(b*x + a)
+ I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b^3*d^3*
x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) + sin(b*x
+ a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*co
s(b*x + a) - sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^
2 + a^3*d^3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^
2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*
b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^
3)*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) + sin(b*x + a)) +
6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x
+ a) + sin(b*x + a)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a) - sin(b*x + a)))/b^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{3} \sec {\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^3}{\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)^3/cos(a + b*x),x)

[Out]

int((c + d*x)^3/cos(a + b*x), x)

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