3.3.45 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [245]

Optimal. Leaf size=209 \[ \frac {1}{2} b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) x+\frac {a^2 \left (a^2 A+12 A b^2+8 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*b^2*(8*A*a*b+12*B*a^2+B*b^2)*x+1/2*a^2*(A*a^2+12*A*b^2+8*B*a*b)*arctanh(sin(d*x+c))/d-1/2*b*(13*A*a^2*b-2*
A*b^3+4*B*a^3-8*B*a*b^2)*sin(d*x+c)/d-1/2*b^2*(6*A*a*b+2*B*a^2-B*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/2*a*(5*A*b+2*B
*a)*(a+b*cos(d*x+c))^2*tan(d*x+c)/d+1/2*a*A*(a+b*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]
time = 0.40, antiderivative size = 209, normalized size of antiderivative = 1.00, 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 = {3068, 3126, 3112, 3102, 2814, 3855} \begin {gather*} \frac {a^2 \left (a^2 A+8 a b B+12 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} b^2 x \left (12 a^2 B+8 a A b+b^2 B\right )-\frac {b \left (4 a^3 B+13 a^2 A b-8 a b^2 B-2 A b^3\right ) \sin (c+d x)}{2 d}+\frac {a (2 a B+5 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {a A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]

[Out]

(b^2*(8*a*A*b + 12*a^2*B + b^2*B)*x)/2 + (a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (b*(
13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Sin[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b^2*B)*Cos[c + d*x
]*Sin[c + d*x])/(2*d) + (a*(5*A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(2*d) + (a*A*(a + b*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 3068

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*Si
n[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 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 3112

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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3126

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 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))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^2 \left (a (5 A b+2 a B)+\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-2 b (a A-b B) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x)) \left (a \left (a^2 A+12 A b^2+8 a b B\right )-b \left (a^2 A-2 A b^2-6 a b B\right ) \cos (c+d x)-2 b \left (6 a A b+2 a^2 B-b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a^2 \left (a^2 A+12 A b^2+8 a b B\right )+2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \cos (c+d x)-2 b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a^2 \left (a^2 A+12 A b^2+8 a b B\right )+2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) x-\frac {b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^2 \left (a^2 A+12 A b^2+8 a b B\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) x+\frac {a^2 \left (a^2 A+12 A b^2+8 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 1.60, size = 310, normalized size = 1.48 \begin {gather*} \frac {2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) (c+d x)-2 a^2 \left (a^2 A+12 A b^2+8 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \left (a^2 A+12 A b^2+8 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^3 (4 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^3 (4 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^3 (A b+4 a B) \sin (c+d x)+b^4 B \sin (2 (c+d x))}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]

[Out]

(2*b^2*(8*a*A*b + 12*a^2*B + b^2*B)*(c + d*x) - 2*a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] + 2*a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^4*A)/(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^2 + (4*a^3*(4*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
- (a^4*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a^3*(4*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]) + 4*b^3*(A*b + 4*a*B)*Sin[c + d*x] + b^4*B*Sin[2*(c + d*x)])/(4*d)

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Maple [A]
time = 0.29, size = 187, normalized size = 0.89 Too large to display

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)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 209, normalized size = 1.00 \begin {gather*} \frac {24 \, {\left (d x + c\right )} B a^{2} b^{2} + 16 \, {\left (d x + c\right )} A a b^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{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 )} + 8 \, B a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, B a b^{3} \sin \left (d x + c\right ) + 4 \, A b^{4} \sin \left (d x + c\right ) + 4 \, B a^{4} \tan \left (d x + c\right ) + 16 \, A a^{3} b \tan \left (d x + c\right )}{4 \, d} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

1/4*(24*(d*x + c)*B*a^2*b^2 + 16*(d*x + c)*A*a*b^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^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)) + 8*B*a^3*b*(log(sin(d*x + c) + 1)
- log(sin(d*x + c) - 1)) + 12*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 16*B*a*b^3*sin(d*x
+ c) + 4*A*b^4*sin(d*x + c) + 4*B*a^4*tan(d*x + c) + 16*A*a^3*b*tan(d*x + c))/d

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Fricas [A]
time = 0.38, size = 202, normalized size = 0.97 \begin {gather*} \frac {2 \, {\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} d x \cos \left (d x + c\right )^{2} + {\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B b^{4} \cos \left (d x + c\right )^{3} + A a^{4} + 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\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+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(2*(12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*d*x*cos(d*x + c)^2 + (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*cos(d*x + c)
^2*log(sin(d*x + c) + 1) - (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*b^4
*cos(d*x + c)^3 + A*a^4 + 2*(4*B*a*b^3 + A*b^4)*cos(d*x + c)^2 + 2*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*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+b*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 [B] Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (197) = 394\).
time = 0.54, size = 526, normalized size = 2.52 \begin {gather*} \frac {{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} {\left (d x + c\right )} + {\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{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+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/2*((12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*(d*x + c) + (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1
/2*c) + 1)) - (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a^4*tan(1/2*d*x + 1
/2*c)^7 - 2*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^7
+ 2*A*b^4*tan(1/2*d*x + 1/2*c)^7 - B*b^4*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 2*B*a^4*ta
n(1/2*d*x + 1/2*c)^5 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 2*A*b^4*tan(1/2*d
*x + 1/2*c)^5 + 3*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c)
^3 + 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*
B*b^4*tan(1/2*d*x + 1/2*c)^3 + A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b*tan(1/2*d
*x + 1/2*c) + 8*B*a*b^3*tan(1/2*d*x + 1/2*c) + 2*A*b^4*tan(1/2*d*x + 1/2*c) + B*b^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 = 2.31, size = 330, normalized size = 1.58 \begin {gather*} \frac {2\,\left (\frac {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 )}{2}+\frac {B\,b^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 )}{2}+4\,A\,a\,b^3\,\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 )+4\,B\,a^3\,b\,\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 )+6\,A\,a^2\,b^2\,\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 )+6\,B\,a^2\,b^2\,\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 )\right )}{d}+\frac {\frac {B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {B\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2}+\frac {A\,b^4\,\sin \left (c+d\,x\right )}{4}+B\,a\,b^3\,\sin \left (c+d\,x\right )+2\,A\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )+B\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\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))^4)/cos(c + d*x)^3,x)

[Out]

(2*((A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (B*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/2 + 4*A*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 4*B*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d
*x)/2)) + 6*A*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*B*a^2*b^2*atan(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2))))/d + ((B*a^4*sin(2*c + 2*d*x))/2 + (A*b^4*sin(3*c + 3*d*x))/4 + (B*b^4*sin(2*c + 2*d*x))/8 + (
B*b^4*sin(4*c + 4*d*x))/16 + (A*a^4*sin(c + d*x))/2 + (A*b^4*sin(c + d*x))/4 + B*a*b^3*sin(c + d*x) + 2*A*a^3*
b*sin(2*c + 2*d*x) + B*a*b^3*sin(3*c + 3*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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