3.216 \(\int (c+d x)^3 \sin (a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=275 \[ -\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \cos (a+b x)}{b^4}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \sin (a+b x)}{b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac {(c+d x)^3 \sin (a+b x)}{b}-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

-2*I*(d*x+c)^3*arctan(exp(I*(b*x+a)))/b+6*d^3*cos(b*x+a)/b^4-3*d*(d*x+c)^2*cos(b*x+a)/b^2+3*I*d*(d*x+c)^2*poly
log(2,-I*exp(I*(b*x+a)))/b^2-3*I*d*(d*x+c)^2*polylog(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*p
olylog(4,I*exp(I*(b*x+a)))/b^4+6*d^2*(d*x+c)*sin(b*x+a)/b^3-(d*x+c)^3*sin(b*x+a)/b

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Rubi [A]  time = 0.21, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4407, 3296, 2638, 4181, 2531, 6609, 2282, 6589} \[ -\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^3 \text {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}+\frac {6 d^2 (c+d x) \sin (a+b x)}{b^3}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}+\frac {6 d^3 \cos (a+b x)}{b^4}-\frac {(c+d x)^3 \sin (a+b x)}{b}-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + (6*d^3*Cos[a + b*x])/b^4 - (3*d*(c + d*x)^2*Cos[a + b*x])/b^2
+ ((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
+ (6*d^2*(c + d*x)*Sin[a + b*x])/b^3 - ((c + d*x)^3*Sin[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 2638

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, 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 4407

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Int[
(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 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]

Rule 6609

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 \sin (a+b x) \tan (a+b x) \, dx &=-\int (c+d x)^3 \cos (a+b x) \, dx+\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 {(c+d x)^3 \sin (a+b x)}{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 {(3 d) \int (c+d x)^2 \sin (a+b x) \, dx}{b}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}+\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 {(c+d x)^3 \sin (a+b x)}{b}-\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 {\left (6 d^2\right ) \int (c+d x) \cos (a+b x) \, dx}{b^2}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}+\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 d^2 (c+d x) \sin (a+b x)}{b^3}-\frac {(c+d x)^3 \sin (a+b x)}{b}+\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 {\left (6 d^3\right ) \int \sin (a+b x) \, dx}{b^3}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {6 d^3 \cos (a+b x)}{b^4}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}+\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 d^2 (c+d x) \sin (a+b x)}{b^3}-\frac {(c+d x)^3 \sin (a+b x)}{b}-\frac {\left (6 i d^3\right ) \operatorname {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 ) \operatorname {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 {6 d^3 \cos (a+b x)}{b^4}-\frac {3 d (c+d x)^2 \cos (a+b x)}{b^2}+\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}+\frac {6 d^2 (c+d x) \sin (a+b x)}{b^3}-\frac {(c+d x)^3 \sin (a+b x)}{b}\\ \end {align*}

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Mathematica [B]  time = 1.47, size = 557, normalized size = 2.03 \[ -\frac {b^3 c^3 \sin (a+b x)+2 i b^3 c^3 \tan ^{-1}\left (e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1+i e^{i (a+b x)}\right )+3 b^3 c^2 d x \sin (a+b x)-3 b^3 c d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \sin (a+b x)-b^3 d^3 x^3 \log \left (1-i e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1+i e^{i (a+b x)}\right )+b^3 d^3 x^3 \sin (a+b x)+3 b^2 c^2 d \cos (a+b x)+6 b^2 c d^2 x \cos (a+b x)-3 i b^2 d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )+3 i b^2 d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )+3 b^2 d^3 x^2 \cos (a+b x)+6 b c d^2 \text {Li}_3\left (-i e^{i (a+b x)}\right )-6 b c d^2 \text {Li}_3\left (i e^{i (a+b x)}\right )-6 b c d^2 \sin (a+b x)+6 b d^3 x \text {Li}_3\left (-i e^{i (a+b x)}\right )-6 b d^3 x \text {Li}_3\left (i e^{i (a+b x)}\right )+6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )-6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )-6 b d^3 x \sin (a+b x)-6 d^3 \cos (a+b x)}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*I)*b^3*c^3*ArcTan[E^(I*(a + b*x))] + 3*b^2*c^2*d*Cos[a + b*x] - 6*d^3*Cos[a + b*x] + 6*b^2*c*d^2*x*Cos[a
+ b*x] + 3*b^2*d^3*x^2*Cos[a + b*x] - 3*b^3*c^2*d*x*Log[1 - I*E^(I*(a + b*x))] - 3*b^3*c*d^2*x^2*Log[1 - I*E^
(I*(a + b*x))] - b^3*d^3*x^3*Log[1 - I*E^(I*(a + b*x))] + 3*b^3*c^2*d*x*Log[1 + I*E^(I*(a + b*x))] + 3*b^3*c*d
^2*x^2*Log[1 + I*E^(I*(a + b*x))] + b^3*d^3*x^3*Log[1 + I*E^(I*(a + b*x))] - (3*I)*b^2*d*(c + d*x)^2*PolyLog[2
, (-I)*E^(I*(a + b*x))] + (3*I)*b^2*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))] + 6*b*c*d^2*PolyLog[3, (-I)*E^
(I*(a + b*x))] + 6*b*d^3*x*PolyLog[3, (-I)*E^(I*(a + b*x))] - 6*b*c*d^2*PolyLog[3, I*E^(I*(a + b*x))] - 6*b*d^
3*x*PolyLog[3, I*E^(I*(a + b*x))] + (6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))] - (6*I)*d^3*PolyLog[4, I*E^(I*(
a + b*x))] + b^3*c^3*Sin[a + b*x] - 6*b*c*d^2*Sin[a + b*x] + 3*b^3*c^2*d*x*Sin[a + b*x] - 6*b*d^3*x*Sin[a + b*
x] + 3*b^3*c*d^2*x^2*Sin[a + b*x] + b^3*d^3*x^3*Sin[a + b*x])/b^4)

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fricas [C]  time = 0.56, size = 1071, normalized size = 3.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.36, size = 901, normalized size = 3.28 \[ -\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 {3 i d^{3} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{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 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 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 a^{2} c \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{3} a^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\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 {i \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 i b^{2} d^{3} x^{2}+3 b^{3} c^{2} d x +6 i b^{2} c \,d^{2} x +b^{3} c^{3}+3 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b -6 i d^{3}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{4}}-\frac {i \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}-3 i b^{2} d^{3} x^{2}+3 b^{3} c^{2} d x -6 i b^{2} c \,d^{2} x +b^{3} c^{3}-3 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b +6 i d^{3}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 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 {a^{3} d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\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 {2 i c^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\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 d^{3} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{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^{2} c \polylog \left (3, -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)*sin(b*x+a)^2,x)

[Out]

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

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maxima [B]  time = 0.67, size = 924, normalized size = 3.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c^3*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*sin(b*x + a)) - 3*a*c^2*d*(log(sin(b*x + a) + 1) -
log(sin(b*x + a) - 1) - 2*sin(b*x + a))/b + 3*a^2*c*d^2*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*si
n(b*x + a))/b^2 - a^3*d^3*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*sin(b*x + a))/b^3 + (12*I*d^3*pol
ylog(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 + (-6*I*b*c*d^2 +
6*I*a*d^3)*(b*x + a)^2 + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*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 + (-6*I*b*c*d^2 + 6*I*a*d^3)*(b*x + a)^2 + (-6*I*b^2*c^2*d + 12*I*a*b*c
*d^2 - 6*I*a^2*d^3)*(b*x + a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 6*(b^2*c^2*d - 2*a*b*c*d^2 + (b*x +
a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*cos(b*x + a) + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*
I*(b*x + a)^2*d^3 - 6*I*a^2*d^3 + (-12*I*b*c*d^2 + 12*I*a*d^3)*(b*x + a))*dilog(I*e^(I*b*x + I*a)) + (6*I*b^2*
c^2*d - 12*I*a*b*c*d^2 + 6*I*(b*x + a)^2*d^3 + 6*I*a^2*d^3 + (12*I*b*c*d^2 - 12*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)) - 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*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 - 2)*d^3)*(b*x + a))*sin(b*x + a))/b^3)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(a + b*x)^2*(c + d*x)^3)/cos(a + b*x),x)

[Out]

int((sin(a + b*x)^2*(c + d*x)^3)/cos(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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