3.328 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=125 \[ \frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{(5 B+3 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac{1}{2} a^3 x (5 B+7 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

(a^3*(5*B + 7*C)*x)/2 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(B + C)*Sin[c + d*x])/(2*d) + (a*B*Cos[c + d*
x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((5*B + 3*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d
*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.339268, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4072, 4017, 3996, 3770} \[ \frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{(5 B+3 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac{1}{2} a^3 x (5 B+7 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(5*B + 7*C)*x)/2 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(B + C)*Sin[c + d*x])/(2*d) + (a*B*Cos[c + d*
x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((5*B + 3*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d
*x])/(6*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 ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 B+3 C)+3 a C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^2 (B+C)+6 a^2 C \sec (c+d x)\right ) \, dx\\ &=\frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac{1}{6} \int \left (-3 a^3 (5 B+7 C)-6 a^3 C \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^3 (5 B+7 C) x+\frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (5 B+7 C) x+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.249272, size = 113, normalized size = 0.9 \[ \frac{a^3 \left (9 (5 B+4 C) \sin (c+d x)+3 (3 B+C) \sin (2 (c+d x))+B \sin (3 (c+d x))+30 B d x-12 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+42 C d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(30*B*d*x + 42*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]] + 9*(5*B + 4*C)*Sin[c + d*x] + 3*(3*B + C)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d)

________________________________________________________________________________________

Maple [A]  time = 0.088, size = 153, normalized size = 1.2 \begin{align*}{\frac{B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{3\,d}}+{\frac{11\,B{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}Cx}{2}}+{\frac{7\,{a}^{3}Cc}{2\,d}}+{\frac{3\,B{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}Bx}{2}}+{\frac{5\,B{a}^{3}c}{2\,d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}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)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/3/d*B*cos(d*x+c)^2*sin(d*x+c)*a^3+11/3*a^3*B*sin(d*x+c)/d+1/2/d*a^3*C*sin(d*x+c)*cos(d*x+c)+7/2*a^3*C*x+7/2/
d*C*a^3*c+3/2/d*B*a^3*sin(d*x+c)*cos(d*x+c)+5/2*a^3*B*x+5/2/d*B*a^3*c+3*a^3*C*sin(d*x+c)/d+1/d*a^3*C*ln(sec(d*
x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 0.948515, size = 200, normalized size = 1.6 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 12 \,{\left (d x + c\right )} B a^{3} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 36 \,{\left (d x + c\right )} C a^{3} - 6 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} \sin \left (d x + c\right ) - 36 \, C a^{3} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 12*(d*x + c)*B*a
^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 36*(d*x + c)*C*a^3 - 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(
d*x + c) - 1)) - 36*B*a^3*sin(d*x + c) - 36*C*a^3*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 0.530216, size = 254, normalized size = 2.03 \begin{align*} \frac{3 \,{\left (5 \, B + 7 \, C\right )} a^{3} d x + 3 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 2 \,{\left (11 \, B + 9 \, C\right )} a^{3}\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)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(5*B + 7*C)*a^3*d*x + 3*C*a^3*log(sin(d*x + c) + 1) - 3*C*a^3*log(-sin(d*x + c) + 1) + (2*B*a^3*cos(d*x
+ c)^2 + 3*(3*B + C)*a^3*cos(d*x + c) + 2*(11*B + 9*C)*a^3)*sin(d*x + c))/d

________________________________________________________________________________________

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)**4*(a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.2378, size = 243, normalized size = 1.94 \begin{align*} \frac{6 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (5 \, B a^{3} + 7 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, C a^{3} \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}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(5*B*a^3 + 7*
C*a^3)*(d*x + c) + 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 36*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*B*a^3*tan(1/2*d*x + 1/2*c) + 21*C*a^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d