3.31 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=151 \[ \frac {5 a^4 (8 A+7 B) \sin (c+d x)}{8 d}+\frac {(32 A+35 B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac {1}{8} a^4 x (48 A+35 B)+\frac {a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(4 A+7 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}+\frac {a B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2976, 2968, 3023, 2735, 3770} \[ \frac {5 a^4 (8 A+7 B) \sin (c+d x)}{8 d}+\frac {(4 A+7 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}+\frac {(32 A+35 B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac {1}{8} a^4 x (48 A+35 B)+\frac {a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(48*A + 35*B)*x)/8 + (a^4*A*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(8*A + 7*B)*Sin[c + d*x])/(8*d) + (a*B*(a +
a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((4*A + 7*B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + ((32*A
+ 35*B)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(24*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 2968

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 2976

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

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Mathematica [A]  time = 0.40, size = 138, normalized size = 0.91 \[ \frac {a^4 \left (24 (27 A+28 B) \sin (c+d x)+24 (4 A+7 B) \sin (2 (c+d x))+8 A \sin (3 (c+d x))-96 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+576 A d x+32 B \sin (3 (c+d x))+3 B \sin (4 (c+d x))+420 B d x\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(576*A*d*x + 420*B*d*x - 96*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 96*A*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + 24*(27*A + 28*B)*Sin[c + d*x] + 24*(4*A + 7*B)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + 32*B*
Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.65, size = 118, normalized size = 0.78 \[ \frac {3 \, {\left (48 \, A + 35 \, B\right )} a^{4} d x + 12 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, B a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (16 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right ) + 160 \, {\left (A + B\right )} a^{4}\right )} \sin \left (d x + c\right )}{24 \, d} \]

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),x, algorithm="fricas")

[Out]

1/24*(3*(48*A + 35*B)*a^4*d*x + 12*A*a^4*log(sin(d*x + c) + 1) - 12*A*a^4*log(-sin(d*x + c) + 1) + (6*B*a^4*co
s(d*x + c)^3 + 8*(A + 4*B)*a^4*cos(d*x + c)^2 + 3*(16*A + 27*B)*a^4*cos(d*x + c) + 160*(A + B)*a^4)*sin(d*x +
c))/d

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giac [A]  time = 1.07, size = 214, normalized size = 1.42 \[ \frac {24 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (48 \, A a^{4} + 35 \, B a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 424 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 520 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 216 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 279 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

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),x, algorithm="giac")

[Out]

1/24*(24*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(48*A*a^4
+ 35*B*a^4)*(d*x + c) + 2*(120*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 105*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 424*A*a^4*tan
(1/2*d*x + 1/2*c)^5 + 385*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 520*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 511*B*a^4*tan(1/2*
d*x + 1/2*c)^3 + 216*A*a^4*tan(1/2*d*x + 1/2*c) + 279*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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maple [A]  time = 0.14, size = 199, normalized size = 1.32 \[ \frac {A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 A \,a^{4} \sin \left (d x +c \right )}{3 d}+\frac {a^{4} B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {27 a^{4} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {35 a^{4} B x}{8}+\frac {35 a^{4} B c}{8 d}+\frac {2 A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+6 A \,a^{4} x +\frac {6 A \,a^{4} c}{d}+\frac {4 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 a^{4} B \sin \left (d x +c \right )}{3 d}+\frac {A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

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),x)

[Out]

1/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*A*a^4*sin(d*x+c)+1/4/d*a^4*B*sin(d*x+c)*cos(d*x+c)^3+27/8/d*a^4*B*c
os(d*x+c)*sin(d*x+c)+35/8*a^4*B*x+35/8/d*a^4*B*c+2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+6*A*a^4*x+6/d*A*a^4*c+4/3/d*B
*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*a^4*B*sin(d*x+c)+1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.47, size = 198, normalized size = 1.31 \[ -\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 384 \, {\left (d x + c\right )} A a^{4} + 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 96 \, {\left (d x + c\right )} B a^{4} - 96 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 576 \, A a^{4} \sin \left (d x + c\right ) - 384 \, B a^{4} \sin \left (d x + c\right )}{96 \, d} \]

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),x, algorithm="maxima")

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 384*(d*x + c)*
A*a^4 + 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c)
)*B*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 96*(d*x + c)*B*a^4 - 96*A*a^4*log(sec(d*x + c) + tan(d*
x + c)) - 576*A*a^4*sin(d*x + c) - 384*B*a^4*sin(d*x + c))/d

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mupad [B]  time = 0.67, size = 188, normalized size = 1.25 \[ \frac {144\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+24\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+105\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+12\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+A\,a^4\,\sin \left (3\,c+3\,d\,x\right )+21\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )+4\,B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+81\,A\,a^4\,\sin \left (c+d\,x\right )+84\,B\,a^4\,\sin \left (c+d\,x\right )}{12\,d} \]

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),x)

[Out]

(144*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 24*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
+ 105*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 12*A*a^4*sin(2*c + 2*d*x) + A*a^4*sin(3*c + 3*d*x)
+ 21*B*a^4*sin(2*c + 2*d*x) + 4*B*a^4*sin(3*c + 3*d*x) + (3*B*a^4*sin(4*c + 4*d*x))/8 + 81*A*a^4*sin(c + d*x)
+ 84*B*a^4*sin(c + d*x))/(12*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

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),x)

[Out]

a**4*(Integral(A*sec(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x), x) + Integral(6*A*cos(c + d*x)**2*
sec(c + d*x), x) + Integral(4*A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(A*cos(c + d*x)**4*sec(c + d*x), x)
+ Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(6*B*cos
(c + d*x)**3*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)**4*sec(c + d*x), x) + Integral(B*cos(c + d*x)**5*sec
(c + d*x), x))

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