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

Optimal. Leaf size=114 \[ -\frac{3 i d^2 (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{3 d^3 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{(c+d x)^3 \tan (a+b x)}{b}-\frac{i (c+d x)^3}{b} \]

[Out]

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

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Rubi [A]  time = 0.215341, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4184, 3719, 2190, 2531, 2282, 6589} \[ -\frac{3 i d^2 (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{3 d^3 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{(c+d x)^3 \tan (a+b x)}{b}-\frac{i (c+d x)^3}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sec[a + b*x]^2,x]

[Out]

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

Rule 4184

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

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

Mathematica [A]  time = 0.486614, size = 109, normalized size = 0.96 \[ \frac{2 b^2 (c+d x)^2 \left (b (c+d x) \tan (a+b x)+3 d \log \left (1+e^{2 i (a+b x)}\right )-i b (c+d x)\right )-6 i b d^2 (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )+3 d^3 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.432, size = 316, normalized size = 2.8 \begin{align*}{\frac{-3\,i{d}^{2}c{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-6\,{\frac{{d}^{3}{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+{\frac{2\,i \left ({d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}-{\frac{12\,i{d}^{2}cax}{{b}^{2}}}+6\,{\frac{c{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}+3\,{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{{b}^{2}}}-{\frac{2\,i{d}^{3}{x}^{3}}{b}}+{\frac{3\,{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{4}}}+12\,{\frac{ac{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{6\,i{d}^{3}{a}^{2}x}{{b}^{3}}}-{\frac{3\,i{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{3}}}+{\frac{4\,i{d}^{3}{a}^{3}}{{b}^{4}}}-{\frac{6\,ic{d}^{2}{x}^{2}}{b}}-{\frac{6\,ic{d}^{2}{a}^{2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.733, size = 1426, normalized size = 12.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^3*tan(b*x + a) - 6*a*c^2*d*tan(b*x + a)/b + 6*a^2*c*d^2*tan(b*x + a)/b^2 - 2*a^3*d^3*tan(b*x + a)/b^3
+ 3*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x +
2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 4*(b*x + a)*sin(2*b*x + 2*a))*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*((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(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 4*(b*x + a)*sin(2*b*x + 2*a))*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*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)
^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 4*(b*x +
a)*sin(2*b*x + 2*a))*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*((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(sin(2*b*x
+ 2*a), cos(2*b*x + 2*a) + 1) - 4*((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2)*cos(2*b*x + 2*a) - (6*b*
c*d^2 + 6*(b*x + a)*d^3 - 6*a*d^3 + 6*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*cos(2*b*x + 2*a) - (-6*I*b*c*d^2 - 6*I
*(b*x + a)*d^3 + 6*I*a*d^3)*sin(2*b*x + 2*a))*dilog(-e^(2*I*b*x + 2*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(2*b*x + 2*a)^2 + sin(2*b*x + 2*a
)^2 + 2*cos(2*b*x + 2*a) + 1) + (-3*I*d^3*cos(2*b*x + 2*a) + 3*d^3*sin(2*b*x + 2*a) - 3*I*d^3)*polylog(3, -e^(
2*I*b*x + 2*I*a)) + (-4*I*(b*x + a)^3*d^3 + (-12*I*b*c*d^2 + 12*I*a*d^3)*(b*x + a)^2)*sin(2*b*x + 2*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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Fricas [C]  time = 1.78137, size = 2010, normalized size = 17.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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)*polylo
g(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) + 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 +
3*b^3*c^2*d*x + b^3*c^3)*sin(b*x + a))/(b^4*cos(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \sec ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \sec \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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