3.223 \(\int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=184 \[ -\frac {d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {c d x}{2 b}+\frac {d^2 x^2}{4 b}+\frac {i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4407, 4404, 3310, 3719, 2190, 2531, 2282, 6589} \[ \frac {i d (c+d x) \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {c d x}{2 b}+\frac {d^2 x^2}{4 b}+\frac {i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sin[a + b*x]^2*Tan[a + b*x],x]

[Out]

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

Rule 2190

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

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 3310

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

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]

Rubi steps

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

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Mathematica [B]  time = 6.47, size = 518, normalized size = 2.82 \[ -\frac {c d \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac {\cot (a) \left (i \text {Li}_2\left (e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\cot ^2(a)+1}}\right )}{b^2 \sqrt {\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac {\cos (2 b x) \left (2 b^2 c^2 \cos (2 a)+4 b^2 c d x \cos (2 a)+2 b^2 d^2 x^2 \cos (2 a)-2 b c d \sin (2 a)-2 b d^2 x \sin (2 a)-d^2 \cos (2 a)\right )}{8 b^3}-\frac {\sin (2 b x) \left (2 b^2 c^2 \sin (2 a)+4 b^2 c d x \sin (2 a)+2 b^2 d^2 x^2 \sin (2 a)+2 b c d \cos (2 a)+2 b d^2 x \cos (2 a)-d^2 \sin (2 a)\right )}{8 b^3}-\frac {i e^{-i a} d^2 \sec (a) \left (2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 \left (1+e^{2 i a}\right ) b x \text {Li}_2\left (-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \text {Li}_3\left (-e^{-2 i (a+b x)}\right )\right )}{12 b^3}-\frac {c^2 \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac {1}{3} x \tan (a) \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Sin[a + b*x]^2*Tan[a + b*x],x]

[Out]

((-1/12*I)*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + 6*b*(1 + E^((2*I)*
a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(
b^3*E^(I*a)) - (c^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]
^2)) - (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((
-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan
[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])
*Sec[a])/(b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + (Cos[2*b*x]*(2*b^2*c^2*Cos[2*a] - d^2*Cos[2*a] + 4*b^2*c
*d*x*Cos[2*a] + 2*b^2*d^2*x^2*Cos[2*a] - 2*b*c*d*Sin[2*a] - 2*b*d^2*x*Sin[2*a]))/(8*b^3) - ((2*b*c*d*Cos[2*a]
+ 2*b*d^2*x*Cos[2*a] + 2*b^2*c^2*Sin[2*a] - d^2*Sin[2*a] + 4*b^2*c*d*x*Sin[2*a] + 2*b^2*d^2*x^2*Sin[2*a])*Sin[
2*b*x])/(8*b^3) + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Tan[a])/3

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fricas [C]  time = 0.54, size = 688, normalized size = 3.74 \[ -\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} + 4 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 4 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 4 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 4 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - {\left (-4 i \, b d^{2} x - 4 i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - {\left (4 i \, b d^{2} x + 4 i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (4 i \, b d^{2} x + 4 i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - {\left (-4 i \, b d^{2} x - 4 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 (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 \, {\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 \, {\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 ) + 2 \, {\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 ) + 2 \, {\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 ) + 2 \, {\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 ) + 2 \, {\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 \, {\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 )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*d^2*x^2 + 2*b^2*c*d*x - (2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - d^2)*cos(b*x + a)^2 + 4*d^2*polyl
og(3, I*cos(b*x + a) + sin(b*x + a)) + 4*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 4*d^2*polylog(3, -I*c
os(b*x + a) + sin(b*x + a)) + 4*d^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) + 2*(b*d^2*x + b*c*d)*cos(b*x +
a)*sin(b*x + a) - (-4*I*b*d^2*x - 4*I*b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) - (4*I*b*d^2*x + 4*I*b*c*d)
*dilog(I*cos(b*x + a) - sin(b*x + a)) - (4*I*b*d^2*x + 4*I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - (-4*
I*b*d^2*x - 4*I*b*c*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x +
a) + I*sin(b*x + a) + I) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) - I*sin(b*x + a) + I) + 2*(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) + 2*(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) + 2*(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) + 2*(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) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + I)
+ 2*(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.33, size = 379, normalized size = 2.06 \[ \frac {i d^{2} x^{3}}{3}-\frac {2 i d^{2} a^{2} x}{b^{2}}+\frac {2 i c d \,a^{2}}{b^{2}}+\frac {\left (2 d^{2} x^{2} b^{2}+4 b^{2} c d x +2 i b \,d^{2} x +2 b^{2} c^{2}+2 i b c d -d^{2}\right ) {\mathrm e}^{2 i \left (b x +a \right )}}{16 b^{3}}+\frac {\left (2 d^{2} x^{2} b^{2}+4 b^{2} c d x -2 i b \,d^{2} x +2 b^{2} c^{2}-2 i b c d -d^{2}\right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{16 b^{3}}-\frac {c^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 i c d a x}{b}-i c^{2} x +i c d \,x^{2}-\frac {d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {4 i d^{2} a^{3}}{3 b^{3}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {4 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i c d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 c d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))+1/3*I*d^2*x^3-2*I/b^2*a^2*d^2*x+1/16*(2*d^2*x^2*b^2+2*I*b*d^2*x+4*b^2*c
*d*x+2*I*b*c*d+2*b^2*c^2-d^2)/b^3*exp(2*I*(b*x+a))+1/16*(2*d^2*x^2*b^2-2*I*b*d^2*x+4*b^2*c*d*x-2*I*b*c*d+2*b^2
*c^2-d^2)/b^3*exp(-2*I*(b*x+a))-1/b*c^2*ln(1+exp(2*I*(b*x+a)))+2/b*c^2*ln(exp(I*(b*x+a)))+2/b^3*d^2*a^2*ln(exp
(I*(b*x+a)))+I/b^2*d^2*polylog(2,-exp(2*I*(b*x+a)))*x+2*I/b^2*a^2*c*d+4*I/b*a*c*d*x-1/b*d^2*ln(1+exp(2*I*(b*x+
a)))*x^2-I*c^2*x-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3-4/b^2*c*d*a*ln(exp(I*(b*x+a)))+I*c*d*x^2-4/3*I/b^3*a
^3*d^2-2/b*c*d*ln(1+exp(2*I*(b*x+a)))*x

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maxima [B]  time = 0.53, size = 379, normalized size = 2.06 \[ -\frac {12 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} c^{2} - \frac {24 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a c d}{b} + \frac {12 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a^{2} d^{2}}{b^{2}} + \frac {-8 i \, {\left (b x + a\right )}^{3} d^{2} + {\left (-24 i \, b c d + 24 i \, a d^{2}\right )} {\left (b x + a\right )}^{2} + 12 \, d^{2} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + {\left (24 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (48 i \, b c d - 48 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (2 \, {\left (b x + a\right )}^{2} d^{2} + 4 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (-24 i \, b c d - 24 i \, {\left (b x + a\right )} d^{2} + 24 i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 12 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{b^{2}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*c^2 - 24*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*a*c*d
/b + 12*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*a^2*d^2/b^2 + (-8*I*(b*x + a)^3*d^2 + (-24*I*b*c*d + 24*I*a
*d^2)*(b*x + a)^2 + 12*d^2*polylog(3, -e^(2*I*b*x + 2*I*a)) + (24*I*(b*x + a)^2*d^2 + (48*I*b*c*d - 48*I*a*d^2
)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 3*(2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*(b*x +
a) - d^2)*cos(2*b*x + 2*a) + (-24*I*b*c*d - 24*I*(b*x + a)*d^2 + 24*I*a*d^2)*dilog(-e^(2*I*b*x + 2*I*a)) + 12
*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2
*a) + 1) + 6*(b*c*d + (b*x + a)*d^2 - a*d^2)*sin(2*b*x + 2*a))/b^2)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^2}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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