3.58 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=88 \[ \frac{a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac{1}{2} a^2 x (3 A+4 B)+\frac{A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{a^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(a^2*(3*A + 4*B)*x)/2 + (a^2*B*ArcTanh[Sin[c + d*x]])/d + (a^2*(3*A + 2*B)*Sin[c + d*x])/(2*d) + (A*Cos[c + d*
x]*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.144874, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4017, 3996, 3770} \[ \frac{a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac{1}{2} a^2 x (3 A+4 B)+\frac{A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{a^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*(3*A + 4*B)*x)/2 + (a^2*B*ArcTanh[Sin[c + d*x]])/d + (a^2*(3*A + 2*B)*Sin[c + d*x])/(2*d) + (A*Cos[c + d*
x]*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(2*d)

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3770

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x)) (a (3 A+2 B)+2 a B \sec (c+d x)) \, dx\\ &=\frac{a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac{1}{2} \int \left (-a^2 (3 A+4 B)-2 a^2 B \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 (3 A+4 B) x+\frac{a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^2 (3 A+4 B) x+\frac{a^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.159389, size = 96, normalized size = 1.09 \[ \frac{a^2 \left (4 (2 A+B) \sin (c+d x)+A \sin (2 (c+d x))+6 A d x-4 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 B d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*(6*A*d*x + 8*B*d*x - 4*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d
*x)/2]] + 4*(2*A + B)*Sin[c + d*x] + A*Sin[2*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.076, size = 108, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Ax}{2}}+{\frac{3\,{a}^{2}Ac}{2\,d}}+{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+2\,B{a}^{2}x+2\,{\frac{B{a}^{2}c}{d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)

[Out]

1/2*a^2*A*cos(d*x+c)*sin(d*x+c)/d+3/2*a^2*A*x+3/2/d*A*a^2*c+1/d*B*a^2*sin(d*x+c)+2*a^2*A*sin(d*x+c)/d+2*B*a^2*
x+2/d*B*a^2*c+1/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.994061, size = 136, normalized size = 1.55 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \,{\left (d x + c\right )} A a^{2} + 8 \,{\left (d x + c\right )} B a^{2} + 2 \, B a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 8*(d*x + c)*B*a^2 + 2*B*a^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 8*A*a^2*sin(d*x + c) + 4*B*a^2*sin(d*x + c))/d

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Fricas [A]  time = 0.496096, size = 194, normalized size = 2.2 \begin{align*} \frac{{\left (3 \, A + 4 \, B\right )} a^{2} d x + B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (A a^{2} \cos \left (d x + c\right ) + 2 \,{\left (2 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*A + 4*B)*a^2*d*x + B*a^2*log(sin(d*x + c) + 1) - B*a^2*log(-sin(d*x + c) + 1) + (A*a^2*cos(d*x + c) +
2*(2*A + B)*a^2)*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.3544, size = 196, normalized size = 2.23 \begin{align*} \frac{2 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (3 \, A a^{2} + 4 \, B a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*A*a^2 + 4*B*
a^2)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^2*tan(1/2*d*x + 1/
2*c) + 2*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d