3.319 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

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

[Out]

(a^2*(3*B + 4*C)*x)/2 + (a^2*C*ArcTanh[Sin[c + d*x]])/d + (a^2*(3*B + 2*C)*Sin[c + d*x])/(2*d) + (B*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 = 0.219644, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4072, 4017, 3996, 3770} \[ \frac{a^2 (3 B+2 C) \sin (c+d x)}{2 d}+\frac{B \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{1}{2} a^2 x (3 B+4 C)+\frac{a^2 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 4072

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

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 ^3(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{B \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 B+2 C)+2 a C \sec (c+d x)) \, dx\\ &=\frac{a^2 (3 B+2 C) \sin (c+d x)}{2 d}+\frac{B \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 B+4 C)-2 a^2 C \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 (3 B+4 C) x+\frac{a^2 (3 B+2 C) \sin (c+d x)}{2 d}+\frac{B \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^2 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^2 (3 B+4 C) x+\frac{a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 (3 B+2 C) \sin (c+d x)}{2 d}+\frac{B \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 = 0.151612, size = 96, normalized size = 1.09 \[ \frac{a^2 \left (4 (2 B+C) \sin (c+d x)+B \sin (2 (c+d x))+6 B d x-4 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 C d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.084, size = 108, normalized size = 1.2 \begin{align*}{\frac{B{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Bx}{2}}+{\frac{3\,B{a}^{2}c}{2\,d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{a}^{2}Cx+2\,{\frac{C{a}^{2}c}{d}}+{\frac{{a}^{2}C\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)^3*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.941587, size = 136, normalized size = 1.55 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 4 \,{\left (d x + c\right )} B a^{2} + 8 \,{\left (d x + c\right )} C a^{2} + 2 \, C 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 \, B a^{2} \sin \left (d x + c\right ) + 4 \, C 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)^3*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.516076, size = 194, normalized size = 2.2 \begin{align*} \frac{{\left (3 \, B + 4 \, C\right )} a^{2} d x + C a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - C a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (B a^{2} \cos \left (d x + c\right ) + 2 \,{\left (2 \, B + C\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)^3*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.17739, size = 196, normalized size = 2.23 \begin{align*} \frac{2 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (3 \, B a^{2} + 4 \, C a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, C 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)^3*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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