3.552 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=229 \[ \frac {4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )-\frac {b (4 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}-\frac {a b (12 A-7 C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^4}{d} \]

[Out]

1/8*(8*a^4*C+24*a^2*b^2*(2*A+C)+b^4*(4*A+3*C))*x+4*a^3*A*b*arctanh(sin(d*x+c))/d-1/6*a*b*(a^2*(12*A-19*C)-8*b^
2*(3*A+2*C))*sin(d*x+c)/d-1/24*b^2*(a^2*(24*A-26*C)-3*b^2*(4*A+3*C))*cos(d*x+c)*sin(d*x+c)/d-1/12*a*b*(12*A-7*
C)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d-1/4*b*(4*A-C)*(a+b*cos(d*x+c))^3*sin(d*x+c)/d+A*(a+b*cos(d*x+c))^4*tan(d*x+
c)/d

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Rubi [A]  time = 0.80, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3048, 3049, 3033, 3023, 2735, 3770} \[ -\frac {a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )+\frac {4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b (4 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}-\frac {a b (12 A-7 C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

((8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*x)/8 + (4*a^3*A*b*ArcTanh[Sin[c + d*x]])/d - (a*b*(a^2*(12
*A - 19*C) - 8*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (b^2*(a^2*(24*A - 26*C) - 3*b^2*(4*A + 3*C))*Cos[c + d*x
]*Sin[c + d*x])/(24*d) - (a*b*(12*A - 7*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) - (b*(4*A - C)*(a + b*C
os[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*(a + b*Cos[c + d*x])^4*Tan[c + d*x])/d

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 3023

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 3033

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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^3 \left (4 A b+a C \cos (c+d x)-b (4 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (16 a A b+\left (4 A b^2+4 a^2 C+3 b^2 C\right ) \cos (c+d x)-a b (12 A-7 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (48 a^2 A b+a \left (36 A b^2+12 a^2 C+23 b^2 C\right ) \cos (c+d x)-b \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {1}{24} \int \left (96 a^3 A b+3 \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \cos (c+d x)-4 a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {1}{24} \int \left (96 a^3 A b+3 \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{8} \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x-\frac {a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\left (4 a^3 A b\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x+\frac {4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac {b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.36, size = 274, normalized size = 1.20 \[ \frac {\frac {96 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {96 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-384 a^3 A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+384 a^3 A b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+96 a b \left (4 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x)+24 b^2 \left (C \left (6 a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))+12 (c+d x) \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )+32 a b^3 C \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

(12*(8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*(c + d*x) - 384*a^3*A*b*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 384*a^3*A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (96*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]) + (96*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 96*a*b*(4*A*b^2 +
4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 24*b^2*(A*b^2 + (6*a^2 + b^2)*C)*Sin[2*(c + d*x)] + 32*a*b^3*C*Sin[3*(c + d
*x)] + 3*b^4*C*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 2.26, size = 203, normalized size = 0.89 \[ \frac {48 \, A a^{3} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, A a^{3} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, C a^{4} + 24 \, {\left (2 \, A + C\right )} a^{2} b^{2} + {\left (4 \, A + 3 \, C\right )} b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, C b^{4} \cos \left (d x + c\right )^{4} + 32 \, C a b^{3} \cos \left (d x + c\right )^{3} + 24 \, A a^{4} + 3 \, {\left (24 \, C a^{2} b^{2} + {\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left (3 \, C a^{3} b + {\left (3 \, A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.53, size = 558, normalized size = 2.44 \[ \frac {96 \, A a^{3} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, A a^{3} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (8 \, C a^{4} + 48 \, A a^{2} b^{2} + 24 \, C a^{2} b^{2} + 4 \, A b^{4} + 3 \, C b^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (96 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 288 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C 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 )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

1/24*(96*A*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 96*A*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 48*A*a^4
*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(8*C*a^4 + 48*A*a^2*b^2 + 24*C*a^2*b^2 + 4*A*b^4 + 3*C*
b^4)*(d*x + c) + 2*(96*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 72*C*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 + 96*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 12*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^4*tan(1/2*d*x
+ 1/2*c)^7 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*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 + 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 9*C*b^4*tan(1/2*d*x + 1
/2*c)^5 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*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 + 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 9*C*b^4*tan(1/2*d*x + 1/2*c
)^3 + 96*C*a^3*b*tan(1/2*d*x + 1/2*c) + 72*C*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) +
96*C*a*b^3*tan(1/2*d*x + 1/2*c) + 12*A*b^4*tan(1/2*d*x + 1/2*c) + 15*C*b^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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maple [A]  time = 0.33, size = 296, normalized size = 1.29 \[ \frac {A \,a^{4} \tan \left (d x +c \right )}{d}+a^{4} C x +\frac {a^{4} C c}{d}+\frac {4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{3} b C \sin \left (d x +c \right )}{d}+6 A x \,a^{2} b^{2}+\frac {6 A \,a^{2} b^{2} c}{d}+\frac {3 C \,a^{2} b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+3 C \,a^{2} b^{2} x +\frac {3 C \,a^{2} b^{2} c}{d}+\frac {4 a A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {4 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a \,b^{3}}{3 d}+\frac {8 C a \,b^{3} \sin \left (d x +c \right )}{3 d}+\frac {A \,b^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {A x \,b^{4}}{2}+\frac {A \,b^{4} c}{2 d}+\frac {C \,b^{4} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 C \,b^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 b^{4} C x}{8}+\frac {3 C \,b^{4} c}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)

[Out]

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

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maxima [A]  time = 0.35, size = 204, normalized size = 0.89 \[ \frac {96 \, {\left (d x + c\right )} C a^{4} + 576 \, {\left (d x + c\right )} A a^{2} b^{2} + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b^{2} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{3} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} + 192 \, A a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, C a^{3} b \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right ) + 96 \, A a^{4} \tan \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

1/96*(96*(d*x + c)*C*a^4 + 576*(d*x + c)*A*a^2*b^2 + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b^2 - 128*(sin
(d*x + c)^3 - 3*sin(d*x + c))*C*a*b^3 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^4 + 3*(12*d*x + 12*c + sin(4*d
*x + 4*c) + 8*sin(2*d*x + 2*c))*C*b^4 + 192*A*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*C*a^
3*b*sin(d*x + c) + 384*A*a*b^3*sin(d*x + c) + 96*A*a^4*tan(d*x + c))/d

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mupad [B]  time = 2.14, size = 395, normalized size = 1.72 \[ \frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,b^4\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {4\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {A\,b^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {8\,C\,a\,b^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,b^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,C\,a^2\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d}+\frac {4\,C\,a\,b^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}-\frac {A\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{4\,d}-\frac {A\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,8{}\mathrm {i}}{d}-\frac {A\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,12{}\mathrm {i}}{d}-\frac {C\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^2,x)

[Out]

(A*a^4*sin(c + d*x))/(d*cos(c + d*x)) - (C*a^4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (C*b^
4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/(4*d) - (A*a^3*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2
+ (d*x)/2))*8i)/d - (A*b^4*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d + (C*b^4*cos(c + d*x)^3*sin
(c + d*x))/(4*d) - (A*a^2*b^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*12i)/d - (C*a^2*b^2*atanh((sin
(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i)/d + (4*A*a*b^3*sin(c + d*x))/d + (A*b^4*cos(c + d*x)*sin(c + d*x))
/(2*d) + (8*C*a*b^3*sin(c + d*x))/(3*d) + (4*C*a^3*b*sin(c + d*x))/d + (3*C*b^4*cos(c + d*x)*sin(c + d*x))/(8*
d) + (3*C*a^2*b^2*cos(c + d*x)*sin(c + d*x))/d + (4*C*a*b^3*cos(c + d*x)^2*sin(c + d*x))/(3*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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