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

Optimal. Leaf size=193 \[ -\frac {d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \tanh ^{-1}(\sin (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \sec (a+b x)}{b^2}-\frac {i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x)^2 \tan (a+b x) \sec (a+b x)}{2 b} \]

[Out]

-I*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b+d^2*arctanh(sin(b*x+a))/b^3+I*d*(d*x+c)*polylog(2,-I*exp(I*(b*x+a)))/b^2
-I*d*(d*x+c)*polylog(2,I*exp(I*(b*x+a)))/b^2-d^2*polylog(3,-I*exp(I*(b*x+a)))/b^3+d^2*polylog(3,I*exp(I*(b*x+a
)))/b^3-d*(d*x+c)*sec(b*x+a)/b^2+1/2*(d*x+c)^2*sec(b*x+a)*tan(b*x+a)/b

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4186, 3770, 4181, 2531, 2282, 6589} \[ \frac {i d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \sec (a+b x)}{b^2}-\frac {d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \tanh ^{-1}(\sin (a+b x))}{b^3}-\frac {i (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x)^2 \tan (a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sec[a + b*x]^3,x]

[Out]

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

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 4186

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.92, size = 184, normalized size = 0.95 \[ \frac {-2 i b^2 (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )+b^2 (c+d x)^2 \tan (a+b x) \sec (a+b x)+2 i b d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )-2 i b d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )-2 b d (c+d x) \sec (a+b x)-2 d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )+2 d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )+2 d^2 \tanh ^{-1}(\sin (a+b x))}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 1.24, size = 795, normalized size = 4.12 \[ -\frac {2 \, d^{2} \cos \left (b x + a\right )^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} \cos \left (b x + a\right )^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, d^{2} \cos \left (b x + a\right )^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} \cos \left (b x + a\right )^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} \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 ) - {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} \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 ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \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 ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \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 ) - {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 2\right )} d^{2}\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 ) + {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 2\right )} d^{2}\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 ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\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 ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\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 ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\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 ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\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 ) - {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 2\right )} d^{2}\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 ) + {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 2\right )} d^{2}\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 ) + 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (b x + a\right )}{4 \, b^{3} \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \sec \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.17, size = 584, normalized size = 3.03 \[ -\frac {i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {i c d \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {i d^{2} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{2 b}+\frac {a^{2} d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{2 b}+\frac {d^{2} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {i \left (d^{2} x^{2} b \,{\mathrm e}^{3 i \left (b x +a \right )}+2 c d x b \,{\mathrm e}^{3 i \left (b x +a \right )}+c^{2} b \,{\mathrm e}^{3 i \left (b x +a \right )}-d^{2} x^{2} b \,{\mathrm e}^{i \left (b x +a \right )}-2 c d x b \,{\mathrm e}^{i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{3 i \left (b x +a \right )}-c^{2} b \,{\mathrm e}^{i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{3 i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {i c d \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {d^{2} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {i d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 2.70, size = 1893, normalized size = 9.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(c^2*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1)) - 2*a*c*d*(2*s
in(b*x + a)/(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1))/b + a^2*d^2*(2*sin(b*x + a)/
(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1))/b^2 + 4*((2*(b*x + a)^2*d^2 + 4*(b*c*d -
a*d^2)*(b*x + a) + 4*d^2 + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(4*b*x + 4*a) + 4*((b
*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(2*b*x + 2*a) - (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d +
4*I*a*d^2)*(b*x + a) - 4*I*d^2)*sin(4*b*x + 4*a) - (-4*I*(b*x + a)^2*d^2 + (-8*I*b*c*d + 8*I*a*d^2)*(b*x + a)
- 8*I*d^2)*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), sin(b*x + a) + 1) + (2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*
(b*x + a) + 4*d^2 + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(4*b*x + 4*a) + 4*((b*x + a)^
2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(2*b*x + 2*a) - (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d + 4*I*a*d^
2)*(b*x + a) - 4*I*d^2)*sin(4*b*x + 4*a) - (-4*I*(b*x + a)^2*d^2 + (-8*I*b*c*d + 8*I*a*d^2)*(b*x + a) - 8*I*d^
2)*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) + (4*(b*x + a)^2*d^2 - 8*I*b*c*d + 8*I*a*d^2 + (
8*b*c*d - (8*a + 8*I)*d^2)*(b*x + a))*cos(3*b*x + 3*a) - (4*(b*x + a)^2*d^2 + 8*I*b*c*d - 8*I*a*d^2 + (8*b*c*d
- (8*a - 8*I)*d^2)*(b*x + a))*cos(b*x + a) + (4*b*c*d + 4*(b*x + a)*d^2 - 4*a*d^2 + 4*(b*c*d + (b*x + a)*d^2
- a*d^2)*cos(4*b*x + 4*a) + 8*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (-4*I*b*c*d - 4*I*(b*x + a)*d
^2 + 4*I*a*d^2)*sin(4*b*x + 4*a) - (-8*I*b*c*d - 8*I*(b*x + a)*d^2 + 8*I*a*d^2)*sin(2*b*x + 2*a))*dilog(I*e^(I
*b*x + I*a)) - (4*b*c*d + 4*(b*x + a)*d^2 - 4*a*d^2 + 4*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) + 8*(
b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (4*I*b*c*d + 4*I*(b*x + a)*d^2 - 4*I*a*d^2)*sin(4*b*x + 4*a)
+ (8*I*b*c*d + 8*I*(b*x + a)*d^2 - 8*I*a*d^2)*sin(2*b*x + 2*a))*dilog(-I*e^(I*b*x + I*a)) - (-I*(b*x + a)^2*d
^2 + (-2*I*b*c*d + 2*I*a*d^2)*(b*x + a) - 2*I*d^2 + (-I*(b*x + a)^2*d^2 + (-2*I*b*c*d + 2*I*a*d^2)*(b*x + a) -
2*I*d^2)*cos(4*b*x + 4*a) + (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d + 4*I*a*d^2)*(b*x + a) - 4*I*d^2)*cos(2*b*x +
2*a) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*sin(4*b*x + 4*a) + 2*((b*x + a)^2*d^2 + 2*(b*c
*d - a*d^2)*(b*x + a) + 2*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - (
I*(b*x + a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)*(b*x + a) + 2*I*d^2 + (I*(b*x + a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)
*(b*x + a) + 2*I*d^2)*cos(4*b*x + 4*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I*a*d^2)*(b*x + a) + 4*I*d^2)*c
os(2*b*x + 2*a) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*sin(4*b*x + 4*a) - 2*((b*x + a)^2*d^
2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a
) + 1) - (-4*I*d^2*cos(4*b*x + 4*a) - 8*I*d^2*cos(2*b*x + 2*a) + 4*d^2*sin(4*b*x + 4*a) + 8*d^2*sin(2*b*x + 2*
a) - 4*I*d^2)*polylog(3, I*e^(I*b*x + I*a)) - (4*I*d^2*cos(4*b*x + 4*a) + 8*I*d^2*cos(2*b*x + 2*a) - 4*d^2*sin
(4*b*x + 4*a) - 8*d^2*sin(2*b*x + 2*a) + 4*I*d^2)*polylog(3, -I*e^(I*b*x + I*a)) - (-4*I*(b*x + a)^2*d^2 - 8*b
*c*d + 8*a*d^2 + (-8*I*b*c*d - 8*(-I*a + 1)*d^2)*(b*x + a))*sin(3*b*x + 3*a) - (4*I*(b*x + a)^2*d^2 - 8*b*c*d
+ 8*a*d^2 - 8*(-I*b*c*d + (I*a + 1)*d^2)*(b*x + a))*sin(b*x + a))/(-4*I*b^2*cos(4*b*x + 4*a) - 8*I*b^2*cos(2*b
*x + 2*a) + 4*b^2*sin(4*b*x + 4*a) + 8*b^2*sin(2*b*x + 2*a) - 4*I*b^2))/b

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/cos(a + b*x)^3,x)

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \sec ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________