3.3.30 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [230]

Optimal. Leaf size=156 \[ \frac {\left (3 a^2 A+4 A b^2+8 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a A b+2 a^2 B+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (2 A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a^2 A \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3067, 3100, 2827, 3853, 3855, 3852, 8} \begin {gather*} \frac {\left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3067

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

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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

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Mathematica [A]
time = 0.78, size = 120, normalized size = 0.77 \begin {gather*} \frac {3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (24 \left (2 a A b+a^2 B+b^2 B\right )+3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x)+6 a^2 A \sec ^3(c+d x)+8 a (2 A b+a B) \tan ^2(c+d x)\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

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

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Maple [A]
time = 0.30, size = 185, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {a^{2} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-2 A a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \tan \left (d x +c \right )}{d}\)	\(185\)
default	\(\frac {a^{2} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-2 A a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \tan \left (d x +c \right )}{d}\)	\(185\)
norman	\(\frac {\frac {\left (7 a^{2} A -4 A \,b^{2}-8 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (27 a^{2} A -32 A a b +12 A \,b^{2}-16 B \,a^{2}+24 B a b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (27 a^{2} A +32 A a b +12 A \,b^{2}+16 B \,a^{2}+24 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (5 a^{2} A -16 A a b +4 A \,b^{2}-8 B \,a^{2}+8 B a b -8 B \,b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (5 a^{2} A +16 A a b +4 A \,b^{2}+8 B \,a^{2}+8 B a b +8 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (81 a^{2} A -16 A a b -12 A \,b^{2}-8 B \,a^{2}-24 B a b -72 B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (81 a^{2} A +16 A a b -12 A \,b^{2}+8 B \,a^{2}-24 B a b +72 B \,b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 a^{2} A +4 A \,b^{2}+8 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a^{2} A +4 A \,b^{2}+8 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(429\)
risch	\(-\frac {i \left (9 A \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+24 B a b \,{\mathrm e}^{7 i \left (d x +c \right )}-24 B \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+33 A \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+24 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}-96 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-33 A \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{3 i \left (d x +c \right )}-128 A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-72 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{i \left (d x +c \right )}-32 A a b -16 B \,a^{2}-24 B \,b^{2}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}+\frac {3 a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}\)	\(447\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 228, normalized size = 1.46 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A a^{2} {\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 )} - 24 \, B a 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 )} - 12 \, A 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 )} + 48 \, B b^{2} \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(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^2 + 32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b - 3*A*a^2*(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)) - 24*B*a*b*(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*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)) + 48*B*b^2
*tan(d*x + c))/d

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Fricas [A]
time = 0.37, size = 180, normalized size = 1.15 \begin {gather*} \frac {3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, B a^{2} + 4 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, A a^{2} + 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (B a^{2} + 2 \, A a 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

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

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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+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (146) = 292\).
time = 0.48, size = 478, normalized size = 3.06 \begin {gather*} \frac {3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B b^{2} \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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*A*
a*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*b^2*tan(
1/2*d*x + 1/2*c)^7 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a*b*tan(1/2*d*x +
1/2*c)^5 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*b^2*tan(1/2*d*x + 1/2*c)^
5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*B*
a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^2*tan(
1/2*d*x + 1/2*c) + 24*B*a^2*tan(1/2*d*x + 1/2*c) + 48*A*a*b*tan(1/2*d*x + 1/2*c) + 24*B*a*b*tan(1/2*d*x + 1/2*
c) + 12*A*b^2*tan(1/2*d*x + 1/2*c) + 24*B*b^2*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 = 3.87, size = 314, normalized size = 2.01 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{8}+B\,a\,b+\frac {A\,b^2}{2}\right )}{\frac {3\,A\,a^2}{2}+4\,B\,a\,b+2\,A\,b^2}\right )\,\left (\frac {3\,A\,a^2}{4}+2\,B\,a\,b+A\,b^2\right )}{d}+\frac {\left (\frac {5\,A\,a^2}{4}+A\,b^2-2\,B\,a^2-2\,B\,b^2-4\,A\,a\,b+2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a^2}{4}-A\,b^2+\frac {10\,B\,a^2}{3}+6\,B\,b^2+\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a^2}{4}-A\,b^2-\frac {10\,B\,a^2}{3}-6\,B\,b^2-\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a^2}{4}+A\,b^2+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+2\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*A*a^2)/8 + (A*b^2)/2 + B*a*b))/((3*A*a^2)/2 + 2*A*b^2 + 4*B*a*b))*((3*A*a^2)/
4 + A*b^2 + 2*B*a*b))/d + (tan(c/2 + (d*x)/2)^7*((5*A*a^2)/4 + A*b^2 - 2*B*a^2 - 2*B*b^2 - 4*A*a*b + 2*B*a*b)
- tan(c/2 + (d*x)/2)^3*(A*b^2 - (3*A*a^2)/4 + (10*B*a^2)/3 + 6*B*b^2 + (20*A*a*b)/3 + 2*B*a*b) + tan(c/2 + (d*
x)/2)^5*((3*A*a^2)/4 - A*b^2 + (10*B*a^2)/3 + 6*B*b^2 + (20*A*a*b)/3 - 2*B*a*b) + tan(c/2 + (d*x)/2)*((5*A*a^2
)/4 + A*b^2 + 2*B*a^2 + 2*B*b^2 + 4*A*a*b + 2*B*a*b))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*
tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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