3.78 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=173 \[ \frac{5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac{(7 A+4 B) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}+\frac{(35 A+32 B) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (35 A+48 B)+\frac{a^4 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

(a^4*(35*A + 48*B)*x)/8 + (a^4*B*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 8*B)*Sin[c + d*x])/(8*d) + (a*A*Cos[
c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + ((7*A + 4*B)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2
*Sin[c + d*x])/(12*d) + ((35*A + 32*B)*Cos[c + d*x]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

________________________________________________________________________________________

Rubi [A]  time = 0.402583, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4017, 3996, 3770} \[ \frac{5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac{(7 A+4 B) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}+\frac{(35 A+32 B) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (35 A+48 B)+\frac{a^4 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(35*A + 48*B)*x)/8 + (a^4*B*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 8*B)*Sin[c + d*x])/(8*d) + (a*A*Cos[
c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + ((7*A + 4*B)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2
*Sin[c + d*x])/(12*d) + ((35*A + 32*B)*Cos[c + d*x]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*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 ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (a (7 A+4 B)+4 a B \sec (c+d x)) \, dx\\ &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{1}{12} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (a^2 (35 A+32 B)+12 a^2 B \sec (c+d x)\right ) \, dx\\ &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{1}{24} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (7 A+8 B)+24 a^3 B \sec (c+d x)\right ) \, dx\\ &=\frac{5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac{1}{24} \int \left (-3 a^4 (35 A+48 B)-24 a^4 B \sec (c+d x)\right ) \, dx\\ &=\frac{1}{8} a^4 (35 A+48 B) x+\frac{5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^4 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^4 (35 A+48 B) x+\frac{a^4 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end{align*}

Mathematica [A]  time = 0.339811, size = 138, normalized size = 0.8 \[ \frac{a^4 \left (24 (28 A+27 B) \sin (c+d x)+24 (7 A+4 B) \sin (2 (c+d x))+32 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+420 A d x+8 B \sin (3 (c+d x))-96 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+576 B d x\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(420*A*d*x + 576*B*d*x - 96*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 96*B*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + 24*(28*A + 27*B)*Sin[c + d*x] + 24*(7*A + 4*B)*Sin[2*(c + d*x)] + 32*A*Sin[3*(c + d*x)] + 8*B*
Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(96*d)

________________________________________________________________________________________

Maple [A]  time = 0.091, size = 199, normalized size = 1.2 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{27\,A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{35\,{a}^{4}Ax}{8}}+{\frac{35\,A{a}^{4}c}{8\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{20\,B{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{20\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+2\,{\frac{B{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+6\,B{a}^{4}x+6\,{\frac{B{a}^{4}c}{d}}+{\frac{B{a}^{4}\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))^4*(A+B*sec(d*x+c)),x)

[Out]

1/4/d*A*a^4*sin(d*x+c)*cos(d*x+c)^3+27/8/d*A*a^4*sin(d*x+c)*cos(d*x+c)+35/8*a^4*A*x+35/8/d*A*a^4*c+1/3/d*B*sin
(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*B*a^4*sin(d*x+c)+4/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*A*a^4*sin(d*x+c)+2
/d*B*a^4*sin(d*x+c)*cos(d*x+c)+6*B*a^4*x+6/d*B*a^4*c+1/d*B*a^4*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.02961, size = 277, normalized size = 1.6 \begin{align*} -\frac{128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \,{\left (d x + c\right )} A a^{4} + 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 96 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 384 \,{\left (d x + c\right )} B a^{4} - 48 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 576 \, B a^{4} \sin \left (d x + c\right )}{96 \, 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))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 96*(d*x + c)*A*a^4 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*B*a^4 - 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 384*(d*x + c)*B*a^4 - 48*B*a^4*(log(sin(d*x + c) + 1) -
log(sin(d*x + c) - 1)) - 384*A*a^4*sin(d*x + c) - 576*B*a^4*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 0.510658, size = 306, normalized size = 1.77 \begin{align*} \frac{3 \,{\left (35 \, A + 48 \, B\right )} a^{4} d x + 12 \, B a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, B a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \,{\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (d x + c\right ) + 160 \,{\left (A + B\right )} a^{4}\right )} \sin \left (d x + c\right )}{24 \, 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))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(3*(35*A + 48*B)*a^4*d*x + 12*B*a^4*log(sin(d*x + c) + 1) - 12*B*a^4*log(-sin(d*x + c) + 1) + (6*A*a^4*co
s(d*x + c)^3 + 8*(4*A + B)*a^4*cos(d*x + c)^2 + 3*(27*A + 16*B)*a^4*cos(d*x + c) + 160*(A + B)*a^4)*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))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.31806, size = 289, normalized size = 1.67 \begin{align*} \frac{24 \, B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (35 \, A a^{4} + 48 \, B a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 216 \, B a^{4} \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 )}^{4}}}{24 \, 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))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(24*B*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(35*A*a^4
+ 48*B*a^4)*(d*x + c) + 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 385*A*a^4*tan
(1/2*d*x + 1/2*c)^5 + 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 511*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*
d*x + 1/2*c)^3 + 279*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)
^4)/d