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

Optimal. Leaf size=110 \[ \frac{a^2 (A+C) \sin (c+d x)}{d}+\frac{A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+a^2 x (A+2 C)+\frac{a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.267877, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4017, 3996, 3770} \[ \frac{a^2 (A+C) \sin (c+d x)}{d}+\frac{A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+a^2 x (A+2 C)+\frac{a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 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 (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (2 a A+3 a C \sec (c+d x)) \, dx}{3 a}\\ &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (6 a^2 (A+C)+6 a^2 C \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac{a^2 (A+C) \sin (c+d x)}{d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}-\frac{\int \left (-6 a^3 (A+2 C)-6 a^3 C \sec (c+d x)\right ) \, dx}{6 a}\\ &=a^2 (A+2 C) x+\frac{a^2 (A+C) \sin (c+d x)}{d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\left (a^2 C\right ) \int \sec (c+d x) \, dx\\ &=a^2 (A+2 C) x+\frac{a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 (A+C) \sin (c+d x)}{d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.212069, size = 109, normalized size = 0.99 \[ \frac{a^2 \left (3 (7 A+4 C) \sin (c+d x)+6 A \sin (2 (c+d x))+A \sin (3 (c+d x))+12 A d x-12 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 C d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*d*x + 24*C*d*x - 12*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*C*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 3*(7*A + 4*C)*Sin[c + d*x] + 6*A*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.091, size = 128, normalized size = 1.2 \begin{align*}{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{2}}{3\,d}}+{\frac{5\,{a}^{2}A\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+{a}^{2}Ax+{\frac{{a}^{2}Ac}{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*(A+C*sec(d*x+c)^2),x)

[Out]

1/3/d*A*cos(d*x+c)^2*sin(d*x+c)*a^2+5/3/d*a^2*A*sin(d*x+c)+1/d*a^2*C*sin(d*x+c)+1/d*a^2*A*cos(d*x+c)*sin(d*x+c
)+a^2*A*x+1/d*A*a^2*c+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 = 0.940133, size = 154, normalized size = 1.4 \begin{align*} -\frac{2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 12 \,{\left (d x + c\right )} C a^{2} - 3 \, 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 )} - 6 \, A a^{2} \sin \left (d x + c\right ) - 6 \, C a^{2} \sin \left (d x + c\right )}{6 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 12*(d*x + c)*C*a^
2 - 3*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*a^2*sin(d*x + c) - 6*C*a^2*sin(d*x + c))/d

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Fricas [A]  time = 0.527179, size = 236, normalized size = 2.15 \begin{align*} \frac{6 \,{\left (A + 2 \, C\right )} a^{2} d x + 3 \, C a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{2} \cos \left (d x + c\right )^{2} + 3 \, A a^{2} \cos \left (d x + c\right ) +{\left (5 \, A + 3 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(6*(A + 2*C)*a^2*d*x + 3*C*a^2*log(sin(d*x + c) + 1) - 3*C*a^2*log(-sin(d*x + c) + 1) + 2*(A*a^2*cos(d*x +
c)^2 + 3*A*a^2*cos(d*x + c) + (5*A + 3*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*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.22258, size = 242, normalized size = 2.2 \begin{align*} \frac{3 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (A a^{2} + 2 \, C 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 )^{5} + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 8 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, 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 )}^{3}}}{3 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(A*a^2 + 2*C*
a^2)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 8*A*a^2*tan(1/2*d*x + 1/
2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^2*tan(1/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/
2*d*x + 1/2*c)^2 + 1)^3)/d