3.247 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=216 \[ \frac{a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{a (4 a B+7 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^4 B x \]

[Out]

b^4*B*x + ((3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(16*
a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^2*B)*Tan[c + d*x])/(6*d) + (a^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Sec[c +
d*x]*Tan[c + d*x])/(24*d) + (a*(7*A*b + 4*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(12*d) + (a
*A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.597489, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3047, 3031, 3021, 2735, 3770} \[ \frac{a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{a (4 a B+7 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^4 B x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

b^4*B*x + ((3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(16*
a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^2*B)*Tan[c + d*x])/(6*d) + (a^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Sec[c +
d*x]*Tan[c + d*x])/(24*d) + (a*(7*A*b + 4*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(12*d) + (a
*A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 2989

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

Rule 3047

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

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (a (7 A b+4 a B)+\left (3 a^2 A+4 A b^2+8 a b B\right ) \cos (c+d x)+4 b^2 B \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (a \left (9 a^2 A+26 A b^2+32 a b B\right )+\left (23 a^2 A b+12 A b^3+8 a^3 B+36 a b^2 B\right ) \cos (c+d x)+12 b^3 B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )-3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \cos (c+d x)-24 b^4 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right )-24 b^4 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^4 B x+\frac{a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{8} \left (-3 a^4 A-24 a^2 A b^2-8 A b^4-16 a^3 b B-32 a b^3 B\right ) \int \sec (c+d x) \, dx\\ &=b^4 B x+\frac{\left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.01386, size = 160, normalized size = 0.74 \[ \frac{3 \left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))+3 a \tan (c+d x) \left (a \left (3 a^2 A+16 a b B+24 A b^2\right ) \sec (c+d x)+8 \left (4 a^2 A b+a^3 B+6 a b^2 B+4 A b^3\right )+2 a^3 A \sec ^3(c+d x)\right )+8 a^3 (a B+4 A b) \tan ^3(c+d x)+24 b^4 B d x}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(24*b^4*B*d*x + 3*(3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*ArcTanh[Sin[c + d*x]] + 3*a*(8*
(4*a^2*A*b + 4*A*b^3 + a^3*B + 6*a*b^2*B) + a*(3*a^2*A + 24*A*b^2 + 16*a*b*B)*Sec[c + d*x] + 2*a^3*A*Sec[c + d
*x]^3)*Tan[c + d*x] + 8*a^3*(4*A*b + a*B)*Tan[c + d*x]^3)/(24*d)

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Maple [A]  time = 0.098, size = 338, normalized size = 1.6 \begin{align*}{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{4}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{8\,A{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{B{a}^{3}b\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{Ba{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{4}Bx+{\frac{B{b}^{4}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^5,x)

[Out]

1/4/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*a^4*sec(d*x+c)*tan(d*x+c)+3/8/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+2/
3/d*a^4*B*tan(d*x+c)+1/3/d*a^4*B*tan(d*x+c)*sec(d*x+c)^2+8/3/d*A*a^3*b*tan(d*x+c)+4/3/d*A*a^3*b*tan(d*x+c)*sec
(d*x+c)^2+2/d*B*a^3*b*tan(d*x+c)*sec(d*x+c)+2/d*B*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+3/d*A*a^2*b^2*tan(d*x+c)*sec
(d*x+c)+3/d*A*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+6/d*B*a^2*b^2*tan(d*x+c)+4/d*A*a*b^3*tan(d*x+c)+4/d*B*a*b^3*ln
(sec(d*x+c)+tan(d*x+c))+1/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+b^4*B*x+1/d*B*b^4*c

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Maxima [A]  time = 1.13339, size = 428, normalized size = 1.98 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} b + 48 \,{\left (d x + c\right )} B b^{4} - 3 \, A a^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{3} b{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{2} b^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 288 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 192 \, A a b^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3*b + 48*(d*x + c)
*B*b^4 - 3*A*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*
x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 48*B*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) +
1) + log(sin(d*x + c) - 1)) - 72*A*a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(
sin(d*x + c) - 1)) + 96*B*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*b^4*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 288*B*a^2*b^2*tan(d*x + c) + 192*A*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 1.54221, size = 603, normalized size = 2.79 \begin{align*} \frac{48 \, B b^{4} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, A a^{4} + 16 \,{\left (B a^{4} + 4 \, A a^{3} b + 9 \, B a^{2} b^{2} + 6 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(48*B*b^4*d*x*cos(d*x + c)^4 + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c
)^4*log(sin(d*x + c) + 1) - 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^4*log(
-sin(d*x + c) + 1) + 2*(6*A*a^4 + 16*(B*a^4 + 4*A*a^3*b + 9*B*a^2*b^2 + 6*A*a*b^3)*cos(d*x + c)^3 + 3*(3*A*a^4
+ 16*B*a^3*b + 24*A*a^2*b^2)*cos(d*x + c)^2 + 8*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^4)

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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((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.86199, size = 857, normalized size = 3.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*B*b^4 + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x
+ 1/2*c) + 1)) - 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(15*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 96*A*a^3*b*tan(1/2*d*x + 1/2*c)^
7 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 144*B*a^2*b^2*tan(1/2*d*x + 1/2*
c)^7 - 96*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^4*tan(1/2*d*x + 1/2*c)^5 +
160*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 +
432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 -
40*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*
A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 +
15*A*a^4*tan(1/2*d*x + 1/2*c) + 24*B*a^4*tan(1/2*d*x + 1/2*c) + 96*A*a^3*b*tan(1/2*d*x + 1/2*c) + 48*B*a^3*b*
tan(1/2*d*x + 1/2*c) + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 96*A*a*b^3*tan
(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d