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

Optimal. Leaf size=165 \[ -\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(11 A+9 B) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+\frac{(2 A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+a^4 x (A+4 B)+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

a^4*(A + 4*B)*x + (a^4*(12*A + 13*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(2*A + B)*Sin[c + d*x])/(2*d) + ((1
1*A + 9*B)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(3*d) + ((2*A + B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*T
an[c + d*x])/(2*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.514103, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2975, 2968, 3023, 2735, 3770} \[ -\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(11 A+9 B) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+\frac{(2 A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+a^4 x (A+4 B)+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(A + 4*B)*x + (a^4*(12*A + 13*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(2*A + B)*Sin[c + d*x])/(2*d) + ((1
1*A + 9*B)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(3*d) + ((2*A + B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*T
an[c + d*x])/(2*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 ^4(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+a \cos (c+d x))^3 (3 a (2 A+B)-a (A-3 B) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+a \cos (c+d x))^2 \left (2 a^2 (11 A+9 B)-a^2 (8 A-3 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+a \cos (c+d x)) \left (3 a^3 (12 A+13 B)-15 a^3 (2 A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int \left (3 a^4 (12 A+13 B)+\left (-15 a^4 (2 A+B)+3 a^4 (12 A+13 B)\right ) \cos (c+d x)-15 a^4 (2 A+B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int \left (3 a^4 (12 A+13 B)+6 a^4 (A+4 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^4 (A+4 B) x-\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (a^4 (12 A+13 B)\right ) \int \sec (c+d x) \, dx\\ &=a^4 (A+4 B) x+\frac{a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac{(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.20254, size = 380, normalized size = 2.3 \[ a^4 \left (\frac{(A+4 B) (c+d x)}{d}+\frac{-13 A-3 B}{12 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{4 \left (5 A \sin \left (\frac{1}{2} (c+d x)\right )+3 B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 \left (5 A \sin \left (\frac{1}{2} (c+d x)\right )+3 B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{13 A+3 B}{12 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(-12 A-13 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{(12 A+13 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{A \sin \left (\frac{1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{A \sin \left (\frac{1}{2} (c+d x)\right )}{6 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{B \sin (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.124, size = 189, normalized size = 1.2 \begin{align*} A{a}^{4}x+{\frac{A{a}^{4}c}{d}}+{\frac{{a}^{4}B\sin \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{a}^{4}Bx+4\,{\frac{B{a}^{4}c}{d}}+{\frac{20\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{13\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}B\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,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)^4,x)

[Out]

A*a^4*x+1/d*A*a^4*c+1/d*a^4*B*sin(d*x+c)+6/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4*a^4*B*x+4/d*a^4*B*c+20/3/d*A*a^
4*tan(d*x+c)+13/2/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+4/d*a^4*B*tan(d*x+c)+1/3/d
*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^4*B*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 1.01896, size = 317, normalized size = 1.92 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \,{\left (d x + c\right )} A a^{4} + 48 \,{\left (d x + c\right )} B a^{4} - 12 \, 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 )} - 3 \, B 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 )} + 24 \, 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 )} + 36 \, 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 )} + 12 \, B a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right )}{12 \, 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)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 12*(d*x + c)*A*a^4 + 48*(d*x + c)*B*a^4 - 12*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)) - 3*B*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)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x
+ c) - 1)) + 36*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^4*sin(d*x + c) + 72*A*a^4*tan(
d*x + c) + 48*B*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.54502, size = 405, normalized size = 2.45 \begin{align*} \frac{12 \,{\left (A + 4 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (12 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (12 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, B a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 3 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

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

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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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.30296, size = 306, normalized size = 1.85 \begin{align*} \frac{\frac{12 \, B 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} + 6 \,{\left (A a^{4} + 4 \, B a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (12 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (12 \, A a^{4} + 13 \, 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 (30 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 76 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 48 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, 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 )}^{3}}}{6 \, 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)^4,x, algorithm="giac")

[Out]

1/6*(12*B*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(A*a^4 + 4*B*a^4)*(d*x + c) + 3*(12*A*a^4
+ 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A*a^4 + 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(30*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 48*B*
a^4*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2
*c)^2 - 1)^3)/d