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

Optimal. Leaf size=82 \[ \frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac{i d^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{i (c+d x)^2}{b} \]

[Out]

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

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Rubi [A]  time = 0.134178, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4184, 3719, 2190, 2279, 2391} \[ \frac{2 d (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac{i d^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^2 \tan (a+b x)}{b}-\frac{i (c+d x)^2}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sec[a + b*x]^2,x]

[Out]

((-I)*(c + d*x)^2)/b + (2*d*(c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b^2 - (I*d^2*PolyLog[2, -E^((2*I)*(a + b*x
))])/b^3 + ((c + d*x)^2*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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.242623, size = 75, normalized size = 0.91 \[ \frac{b (c+d x) \left (b (c+d x) \tan (a+b x)+2 d \log \left (1+e^{2 i (a+b x)}\right )-i b (c+d x)\right )-i d^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.374, size = 170, normalized size = 2.1 \begin{align*}{\frac{2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-{\frac{2\,i{d}^{2}{x}^{2}}{b}}-{\frac{4\,i{d}^{2}ax}{{b}^{2}}}-{\frac{2\,i{d}^{2}{a}^{2}}{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+4\,{\frac{a{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.51338, size = 437, normalized size = 5.33 \begin{align*} \frac{2 \, b^{2} c^{2} +{\left (2 \, b d^{2} x + 2 \, b c d + 2 \,{\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \cos \left (2 \, b x + 2 \, a\right ) -{\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) + d^{2}\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (-i \, b d^{2} x - i \, b c d +{\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, 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 ) +{\left (-2 i \, b^{2} d^{2} x^{2} - 4 i \, b^{2} c d x\right )} \sin \left (2 \, b x + 2 \, a\right )}{-i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + b^{3} \sin \left (2 \, b x + 2 \, a\right ) - i \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2*c^2 + (2*b*d^2*x + 2*b*c*d + 2*(b*d^2*x + b*c*d)*cos(2*b*x + 2*a) + (2*I*b*d^2*x + 2*I*b*c*d)*sin(2*b*x
+ 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x)*cos(2*b*x + 2*a) - (d
^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) + d^2)*dilog(-e^(2*I*b*x + 2*I*a)) + (-I*b*d^2*x - I*b*c*d + (-I*
b*d^2*x - I*b*c*d)*cos(2*b*x + 2*a) + (b*d^2*x + b*c*d)*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) + (-2*I*b^2*d^2*x^2 - 4*I*b^2*c*d*x)*sin(2*b*x + 2*a))/(-I*b^3*cos(2*b*x + 2
*a) + b^3*sin(2*b*x + 2*a) - I*b^3)

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Fricas [B]  time = 1.73393, size = 1187, normalized size = 14.48 \begin{align*} \frac{i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ){\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \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 (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) +{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) +{\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) +{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) +{\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \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 + b^{2} c^{2}\right )} \sin \left (b x + a\right )}{b^{3} \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a)) - I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x +
a)) - I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) - s
in(b*x + a)) + (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x +
a)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) + sin(b*x + a)
+ 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*
log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) +
1) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*log(-
cos(b*x + a) - I*sin(b*x + a) + I) + (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))

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

[Out]

Integral((c + d*x)**2*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 )}^{2} \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)^2*sec(b*x+a)^2,x, algorithm="giac")

[Out]

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