3.1.33 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [33]

Optimal. Leaf size=162 \[ \frac {1}{2} a^4 (8 A+13 B) x+\frac {a^4 (13 A+8 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*a^4*(8*A+13*B)*x+1/2*a^4*(13*A+8*B)*arctanh(sin(d*x+c))/d-5/2*a^4*(A-B)*sin(d*x+c)/d-1/2*(6*A+B)*(a^4+a^4*
cos(d*x+c))*sin(d*x+c)/d+1/2*(5*A+2*B)*(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/d+1/2*a*A*(a+a*cos(d*x+c))^3*sec(d*x+
c)*tan(d*x+c)/d

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Rubi [A]
time = 0.31, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3054, 3055, 3047, 3102, 2814, 3855} \begin {gather*} -\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a^4 (13 A+8 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {1}{2} a^4 x (8 A+13 B)+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 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]^3,x]

[Out]

(a^4*(8*A + 13*B)*x)/2 + (a^4*(13*A + 8*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A - B)*Sin[c + d*x])/(2*d) -
((6*A + B)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(2*d) + ((5*A + 2*B)*(a^2 + a^2*Cos[c + d*x])^2*Tan[c + d*x
])/(2*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*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 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3102

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

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 ^3(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+a \cos (c+d x))^3 (a (5 A+2 B)-2 a (A-B) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+a \cos (c+d x))^2 \left (a^2 (13 A+8 B)-2 a^2 (6 A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int (a+a \cos (c+d x)) \left (2 a^3 (13 A+8 B)-10 a^3 (A-B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a^4 (13 A+8 B)+\left (-10 a^4 (A-B)+2 a^4 (13 A+8 B)\right ) \cos (c+d x)-10 a^4 (A-B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a^4 (13 A+8 B)+2 a^4 (8 A+13 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (8 A+13 B) x-\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^4 (13 A+8 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (8 A+13 B) x+\frac {a^4 (13 A+8 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(162)=324\).
time = 4.43, size = 343, normalized size = 2.12 \begin {gather*} \frac {1}{64} a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (2 (8 A+13 B) x-\frac {2 (13 A+8 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (13 A+8 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (A+4 B) \cos (d x) \sin (c)}{d}+\frac {B \cos (2 d x) \sin (2 c)}{d}+\frac {4 (A+4 B) \cos (c) \sin (d x)}{d}+\frac {B \cos (2 c) \sin (2 d x)}{d}+\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (4 A+B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (4 A+B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \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]^3,x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(2*(8*A + 13*B)*x - (2*(13*A + 8*B)*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]])/d + (2*(13*A + 8*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(A + 4*B)*Cos[d*x]*Sin[c])/d
+ (B*Cos[2*d*x]*Sin[2*c])/d + (4*(A + 4*B)*Cos[c]*Sin[d*x])/d + (B*Cos[2*c]*Sin[2*d*x])/d + A/(d*(Cos[(c + d*
x)/2] - Sin[(c + d*x)/2])^2) + (4*(4*A + B)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])) - A/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(4*A + B)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/
2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/64

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Maple [A]
time = 0.30, size = 177, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {A \,a^{4} \sin \left (d x +c \right )+a^{4} B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{4} \left (d x +c \right )+4 a^{4} B \sin \left (d x +c \right )+6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} B \left (d x +c \right )+4 A \,a^{4} \tan \left (d x +c \right )+4 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} B \tan \left (d x +c \right )}{d}\)	\(177\)
default	\(\frac {A \,a^{4} \sin \left (d x +c \right )+a^{4} B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{4} \left (d x +c \right )+4 a^{4} B \sin \left (d x +c \right )+6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} B \left (d x +c \right )+4 A \,a^{4} \tan \left (d x +c \right )+4 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} B \tan \left (d x +c \right )}{d}\)	\(177\)
risch	\(4 a^{4} x A +\frac {13 a^{4} B x}{2}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} B}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{4}}{2 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{4} B}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{4}}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} B}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} B}{8 d}-\frac {i a^{4} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-8 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-8 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 A \,a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {13 A \,a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(294\)
norman	\(\frac {\left (4 A \,a^{4}+\frac {13}{2} a^{4} B \right ) x +\left (-20 A \,a^{4}-\frac {65}{2} a^{4} B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-20 A \,a^{4}-\frac {65}{2} a^{4} B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A \,a^{4}+\frac {13}{2} a^{4} B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A \,a^{4}+\frac {13}{2} a^{4} B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A \,a^{4}+\frac {13}{2} a^{4} B \right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{4}+\frac {39}{2} a^{4} B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{4}+\frac {39}{2} a^{4} B \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{4} \left (53 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{4} \left (A -B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{4} \left (3 A -8 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{4} \left (7 A +4 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{4} \left (9 A +5 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{4} \left (11 A -4 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 \left (A +B \right ) a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{4} \left (13 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (13 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(458\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 199, normalized size = 1.23 \begin {gather*} \frac {16 \, {\left (d x + c\right )} A a^{4} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 24 \, {\left (d x + c\right )} B a^{4} - 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 )} + 12 \, 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 )} + 8 \, 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 )} + 4 \, A a^{4} \sin \left (d x + c\right ) + 16 \, B a^{4} \sin \left (d x + c\right ) + 16 \, A a^{4} \tan \left (d x + c\right ) + 4 \, B a^{4} \tan \left (d x + c\right )}{4 \, 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)^3,x, algorithm="maxima")

[Out]

1/4*(16*(d*x + c)*A*a^4 + (2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 24*(d*x + c)*B*a^4 - 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)) + 12*A*a^4*(log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) + 8*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a^4*sin(d*x + c) + 16*B*a^4
*sin(d*x + c) + 16*A*a^4*tan(d*x + c) + 4*B*a^4*tan(d*x + c))/d

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Fricas [A]
time = 0.37, size = 156, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (8 \, A + 13 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{2} + {\left (13 \, A + 8 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (13 \, A + 8 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + A a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \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)^3,x, algorithm="fricas")

[Out]

1/4*(2*(8*A + 13*B)*a^4*d*x*cos(d*x + c)^2 + (13*A + 8*B)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (13*A + 8
*B)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a^4*cos(d*x + c)^3 + 2*(A + 4*B)*a^4*cos(d*x + c)^2 + 2*(
4*A + B)*a^4*cos(d*x + c) + A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**3,x)

[Out]

Timed out

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Giac [A]
time = 0.48, size = 230, normalized size = 1.42 \begin {gather*} \frac {{\left (8 \, A a^{4} + 13 \, B a^{4}\right )} {\left (d x + c\right )} + {\left (13 \, A a^{4} + 8 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (13 \, A a^{4} + 8 \, 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 (5 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11 \, 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 )^{4} - 1\right )}^{2}}}{2 \, 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)^3,x, algorithm="giac")

[Out]

1/2*((8*A*a^4 + 13*B*a^4)*(d*x + c) + (13*A*a^4 + 8*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (13*A*a^4 + 8*
B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 5*B*a^4*tan(1/2*d*x + 1/2*c)^7
+ 7*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 7*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*B*a^4*
tan(1/2*d*x + 1/2*c)^3 - 11*A*a^4*tan(1/2*d*x + 1/2*c) - 11*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
4 - 1)^2)/d

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Mupad [B]
time = 0.40, size = 243, normalized size = 1.50 \begin {gather*} \frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,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 {13\,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 )}{d}+\frac {13\,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 {8\,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 {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \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)^3,x)

[Out]

(A*a^4*sin(c + d*x))/d + (4*B*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(13*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2)))/d + (8*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^4*sin(c + d*x))/(d*cos(c + d*x)) +
(A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (B*a^4*cos(c + d*x)*sin(c
+ d*x))/(2*d)

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