3.33 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\)

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

[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)

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Rubi [A]  time = 0.475574, antiderivative size = 162, normalized size of antiderivative = 1., 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 = {2975, 2976, 2968, 3023, 2735, 3770} \[ -\frac{5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac{a^4 (13 A+8 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac{(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+\frac{1}{2} a^4 x (8 A+13 B)+\frac{a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d} \]

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 2975

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*S
in[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 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 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 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 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 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 ^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*}

Mathematica [B]  time = 4.28311, size = 343, normalized size = 2.12 \[ \frac{1}{64} a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 (A+4 B) \sin (c) \cos (d x)}{d}+\frac{4 (A+4 B) \cos (c) \sin (d x)}{d}+\frac{4 (4 A+B) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (4 A+B) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 (13 A+8 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (13 A+8 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+2 x (8 A+13 B)+\frac{A}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{B \sin (2 c) \cos (2 d x)}{d}+\frac{B \cos (2 c) \sin (2 d x)}{d}\right ) \]

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 = 0.127, size = 182, normalized size = 1.1 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}Bx}{2}}+{\frac{13\,{a}^{4}Bc}{2\,d}}+4\,A{a}^{4}x+4\,{\frac{A{a}^{4}c}{d}}+4\,{\frac{{a}^{4}B\sin \left ( dx+c \right ) }{d}}+{\frac{13\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+4\,{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)

[Out]

1/d*A*a^4*sin(d*x+c)+1/2/d*a^4*B*cos(d*x+c)*sin(d*x+c)+13/2*a^4*B*x+13/2/d*a^4*B*c+4*A*a^4*x+4/d*A*a^4*c+4/d*a
^4*B*sin(d*x+c)+13/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a^4*tan(d*x+c)+4/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c)
)+1/2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+1/d*a^4*B*tan(d*x+c)

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Maxima [A]  time = 1.02769, size = 269, normalized size = 1.66 \begin{align*} \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{align*}

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 = 1.45169, size = 390, normalized size = 2.41 \begin{align*} \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{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 = 1.33526, size = 311, normalized size = 1.92 \begin{align*} \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{align*}

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