3.1.35 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [35]

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

[Out]

a^4*B*x+1/8*a^4*(35*A+48*B)*arctanh(sin(d*x+c))/d+5/8*a^4*(7*A+8*B)*tan(d*x+c)/d+1/24*(35*A+32*B)*(a^4+a^4*cos
(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/12*(7*A+4*B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*A*(a+a*
cos(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]
time = 0.31, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3054, 3047, 3100, 2814, 3855} \begin {gather*} \frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {a^4 (35 A+48 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+a^4 B x+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2814

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 3047

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

Rule 3054

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^2)*(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)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3100

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 3855

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

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Mathematica [A]
time = 1.98, size = 326, normalized size = 1.88 \begin {gather*} \frac {a^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 \left (-24 (35 A+48 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) (72 B d x \cos (c)+48 B d x \cos (c+2 d x)+48 B d x \cos (3 c+2 d x)+12 B d x \cos (3 c+4 d x)+12 B d x \cos (5 c+4 d x)-480 A \sin (c)-480 B \sin (c)+105 A \sin (d x)+48 B \sin (d x)+105 A \sin (2 c+d x)+48 B \sin (2 c+d x)+544 A \sin (c+2 d x)+496 B \sin (c+2 d x)-96 A \sin (3 c+2 d x)-144 B \sin (3 c+2 d x)+81 A \sin (2 c+3 d x)+48 B \sin (2 c+3 d x)+81 A \sin (4 c+3 d x)+48 B \sin (4 c+3 d x)+160 A \sin (3 c+4 d x)+160 B \sin (3 c+4 d x))\right )}{3072 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(-24*(35*A + 48*B)*Cos[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(72*B*d*x*Cos[c] + 48*B*d*x*Cos[c + 2*d*x] + 4
8*B*d*x*Cos[3*c + 2*d*x] + 12*B*d*x*Cos[3*c + 4*d*x] + 12*B*d*x*Cos[5*c + 4*d*x] - 480*A*Sin[c] - 480*B*Sin[c]
+ 105*A*Sin[d*x] + 48*B*Sin[d*x] + 105*A*Sin[2*c + d*x] + 48*B*Sin[2*c + d*x] + 544*A*Sin[c + 2*d*x] + 496*B*
Sin[c + 2*d*x] - 96*A*Sin[3*c + 2*d*x] - 144*B*Sin[3*c + 2*d*x] + 81*A*Sin[2*c + 3*d*x] + 48*B*Sin[2*c + 3*d*x
] + 81*A*Sin[4*c + 3*d*x] + 48*B*Sin[4*c + 3*d*x] + 160*A*Sin[3*c + 4*d*x] + 160*B*Sin[3*c + 4*d*x])))/(3072*d
)

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Maple [A]
time = 0.26, size = 250, normalized size = 1.45

method	result	size
derivativedivides	\(\frac {A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} B \left (d x +c \right )+4 A \,a^{4} \tan \left (d x +c \right )+4 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} B \tan \left (d x +c \right )-4 A \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{4} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(250\)
default	\(\frac {A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} B \left (d x +c \right )+4 A \,a^{4} \tan \left (d x +c \right )+4 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} B \tan \left (d x +c \right )-4 A \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{4} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(250\)
risch	\(a^{4} B x -\frac {i a^{4} \left (81 A \,{\mathrm e}^{7 i \left (d x +c \right )}+48 B \,{\mathrm e}^{7 i \left (d x +c \right )}-96 A \,{\mathrm e}^{6 i \left (d x +c \right )}-144 B \,{\mathrm e}^{6 i \left (d x +c \right )}+105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+48 B \,{\mathrm e}^{5 i \left (d x +c \right )}-480 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 B \,{\mathrm e}^{4 i \left (d x +c \right )}-105 A \,{\mathrm e}^{3 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-544 A \,{\mathrm e}^{2 i \left (d x +c \right )}-496 B \,{\mathrm e}^{2 i \left (d x +c \right )}-81 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}-160 A -160 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {35 A \,a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {35 A \,a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 307, normalized size = 1.77 \begin {gather*} \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{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 )} - 72 \, A a^{4} {\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 )} - 48 \, B a^{4} {\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 )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, 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 )} + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 157, normalized size = 0.91 \begin {gather*} \frac {48 \, B a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, {\left (A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [A]
time = 0.54, size = 223, normalized size = 1.29 \begin {gather*} \frac {24 \, {\left (d x + c\right )} B a^{4} + 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} \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 )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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*a^4 + 3*(35*A*a^4 + 48*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(35*A*a^4 + 48*B*a^4
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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

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Mupad [B]
time = 0.38, size = 255, normalized size = 1.47 \begin {gather*} \frac {35\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,A\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(35*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (2*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/d + (12*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d
*x)) + (27*A*a^4*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (4*A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (A*a^4*sin
(c + d*x))/(4*d*cos(c + d*x)^4) + (20*B*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*B*a^4*sin(c + d*x))/(d*cos(c
+ d*x)^2) + (B*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3)

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