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

Optimal. Leaf size=159 \[ \frac {6 d^3 \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^4}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \sec (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4409, 4181, 2531, 2282, 6589} \[ -\frac {6 i d^2 (c+d x) \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^3 \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \text {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*I)*d*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b^2 - ((6*I)*d^2*(c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3
+ ((6*I)*d^2*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))])/b^3 + (6*d^3*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^4 - (6
*d^3*PolyLog[3, I*E^(I*(a + b*x))])/b^4 + ((c + d*x)^3*Sec[a + b*x])/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 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 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 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)^3 \sec (a+b x) \tan (a+b x) \, dx &=\frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {(3 d) \int (c+d x)^2 \sec (a+b x) \, dx}{b}\\ &=\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \sec (a+b x)}{b}+\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \sec (a+b x)}{b}+\frac {\left (6 i d^3\right ) \int \text {Li}_2\left (-i e^{i (a+b x)}\right ) \, dx}{b^3}-\frac {\left (6 i d^3\right ) \int \text {Li}_2\left (i e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \sec (a+b x)}{b}+\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}-\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}\\ &=\frac {6 i d (c+d x)^2 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^3 \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^4}+\frac {(c+d x)^3 \sec (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 256, normalized size = 1.61 \[ \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \left (-2 i b^2 c^2 \tan ^{-1}\left (e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right )-2 i b d (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right )-2 d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )+2 d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.53, size = 779, normalized size = 4.90 \[ \frac {2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \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 + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{4} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 463, normalized size = 2.91 \[ \frac {2 \,{\mathrm e}^{i \left (b x +a \right )} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}-\frac {6 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}+\frac {3 d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {6 i d \,c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {6 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {6 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {3 d^{3} a^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {6 i c \,d^{2} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 d^{3} a^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{3} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 i d^{3} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {12 i d^{2} c a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d^{3} x \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i d^{3} x \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{3} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {6 i c \,d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*exp(I*(b*x+a))*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/b/(1+exp(2*I*(b*x+a)))-6/b^3*d^2*c*ln(1-I*exp(I*(b*x+a)))
*a+3/b^2*d^3*ln(1+I*exp(I*(b*x+a)))*x^2+6*I/b^2*d*c^2*arctan(exp(I*(b*x+a)))+6/b^3*d^2*c*ln(1+I*exp(I*(b*x+a))
)*a-6/b^2*d^2*c*ln(1-I*exp(I*(b*x+a)))*x-3/b^4*d^3*a^2*ln(1+I*exp(I*(b*x+a)))-3/b^2*d^3*ln(1-I*exp(I*(b*x+a)))
*x^2-6*I*c*d^2*polylog(2,-I*exp(I*(b*x+a)))/b^3+3/b^4*d^3*a^2*ln(1-I*exp(I*(b*x+a)))+6*d^3*polylog(3,-I*exp(I*
(b*x+a)))/b^4+6*I/b^4*d^3*a^2*arctan(exp(I*(b*x+a)))-12*I/b^3*d^2*c*a*arctan(exp(I*(b*x+a)))-6*I*d^3*x*polylog
(2,-I*exp(I*(b*x+a)))/b^3+6*I*c*d^2*polylog(2,I*exp(I*(b*x+a)))/b^3-6*d^3*polylog(3,I*exp(I*(b*x+a)))/b^4+6/b^
2*d^2*c*ln(1+I*exp(I*(b*x+a)))*x+6*I*d^3*x*polylog(2,I*exp(I*(b*x+a)))/b^3

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maxima [B]  time = 0.62, size = 1774, normalized size = 11.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b*x + a) + 4*(b*x + a)*co
s(b*x + a) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x +
a)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x +
a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*c^2*d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x +
2*a) + 1)*b) - 6*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b*x + a) + 4*(b
*x + a)*cos(b*x + a) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 +
sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log
(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*a*c*d^2/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*
cos(2*b*x + 2*a) + 1)*b^2) + 3*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b
*x + a) + 4*(b*x + a)*cos(b*x + a) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(co
s(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x +
2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*a^2*d^3/((cos(2*b*x + 2*a)^2 + sin(2*b*x
+ 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*b^3) + 2*c^3/cos(b*x + a) - 6*a*c^2*d/(b*cos(b*x + a)) + 6*a^2*c*d^2/(b^2*
cos(b*x + a)) - 2*a^3*d^3/(b^3*cos(b*x + a)) + 2*((6*(b*x + a)^2*d^3 + 12*(b*c*d^2 - a*d^3)*(b*x + a) + 6*((b*
x + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*cos(2*b*x + 2*a) - (-6*I*(b*x + a)^2*d^3 + (-12*I*b*c*d^2 + 12*I
*a*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), sin(b*x + a) + 1) + (6*(b*x + a)^2*d^3 + 12*(b*c*d^
2 - a*d^3)*(b*x + a) + 6*((b*x + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*cos(2*b*x + 2*a) - (-6*I*(b*x + a)^
2*d^3 + (-12*I*b*c*d^2 + 12*I*a*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - (
4*I*(b*x + a)^3*d^3 + (12*I*b*c*d^2 - 12*I*a*d^3)*(b*x + a)^2)*cos(b*x + a) + (12*b*c*d^2 + 12*(b*x + a)*d^3 -
12*a*d^3 + 12*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*cos(2*b*x + 2*a) - (-12*I*b*c*d^2 - 12*I*(b*x + a)*d^3 + 12*I
*a*d^3)*sin(2*b*x + 2*a))*dilog(I*e^(I*b*x + I*a)) - (12*b*c*d^2 + 12*(b*x + a)*d^3 - 12*a*d^3 + 12*(b*c*d^2 +
(b*x + a)*d^3 - a*d^3)*cos(2*b*x + 2*a) + (12*I*b*c*d^2 + 12*I*(b*x + a)*d^3 - 12*I*a*d^3)*sin(2*b*x + 2*a))*
dilog(-I*e^(I*b*x + I*a)) - (-3*I*(b*x + a)^2*d^3 + (-6*I*b*c*d^2 + 6*I*a*d^3)*(b*x + a) + (-3*I*(b*x + a)^2*d
^3 + (-6*I*b*c*d^2 + 6*I*a*d^3)*(b*x + a))*cos(2*b*x + 2*a) + 3*((b*x + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x +
a))*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - (3*I*(b*x + a)^2*d^3 + (6*I*
b*c*d^2 - 6*I*a*d^3)*(b*x + a) + (3*I*(b*x + a)^2*d^3 + (6*I*b*c*d^2 - 6*I*a*d^3)*(b*x + a))*cos(2*b*x + 2*a)
- 3*((b*x + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 -
2*sin(b*x + a) + 1) - (-12*I*d^3*cos(2*b*x + 2*a) + 12*d^3*sin(2*b*x + 2*a) - 12*I*d^3)*polylog(3, I*e^(I*b*x
+ I*a)) - (12*I*d^3*cos(2*b*x + 2*a) - 12*d^3*sin(2*b*x + 2*a) + 12*I*d^3)*polylog(3, -I*e^(I*b*x + I*a)) + 4*
((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2)*sin(b*x + a))/(-2*I*b^3*cos(2*b*x + 2*a) + 2*b^3*sin(2*b*x
+ 2*a) - 2*I*b^3))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(a + b*x)*(c + d*x)^3)/cos(a + b*x),x)

[Out]

int((tan(a + b*x)*(c + d*x)^3)/cos(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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