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

Optimal. Leaf size=114 \[ \frac{a^3 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+a^3 x (A+3 B)-\frac{5 a^3 A \sin (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]

[Out]

a^3*(A + 3*B)*x + (a^3*(7*A + 6*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*A*Sin[c + d*x])/(2*d) + ((2*A + B)*(a
^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/d + (a*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.337916, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{a^3 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+a^3 x (A+3 B)-\frac{5 a^3 A \sin (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(A + 3*B)*x + (a^3*(7*A + 6*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*A*Sin[c + d*x])/(2*d) + ((2*A + B)*(a
^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/d + (a*A*(a + a*Cos[c + d*x])^2*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 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))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+a \cos (c+d x))^2 (2 a (2 A+B)-a (A-2 B) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac{(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+a \cos (c+d x)) \left (a^2 (7 A+6 B)-5 a^2 A \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (a^3 (7 A+6 B)+\left (-5 a^3 A+a^3 (7 A+6 B)\right ) \cos (c+d x)-5 a^3 A \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{5 a^3 A \sin (c+d x)}{2 d}+\frac{(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (a^3 (7 A+6 B)+2 a^3 (A+3 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^3 (A+3 B) x-\frac{5 a^3 A \sin (c+d x)}{2 d}+\frac{(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^3 (7 A+6 B)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (A+3 B) x+\frac{a^3 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 a^3 A \sin (c+d x)}{2 d}+\frac{(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.83876, size = 208, normalized size = 1.82 \[ \frac{a^3 \left (4 (3 A+B) \tan (c+d x)+\frac{A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-14 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+14 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 A c+4 A d x+4 B \sin (c+d x)-12 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 B c+12 B d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(4*A*c + 12*B*c + 4*A*d*x + 12*B*d*x - 14*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 12*B*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]] + 14*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 12*B*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 4*B*Si
n[c + d*x] + 4*(3*A + B)*Tan[c + d*x]))/(4*d)

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Maple [A]  time = 0.111, size = 144, normalized size = 1.3 \begin{align*} A{a}^{3}x+{\frac{A{a}^{3}c}{d}}+{\frac{{a}^{3}B\sin \left ( dx+c \right ) }{d}}+{\frac{7\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{a}^{3}Bx+3\,{\frac{B{a}^{3}c}{d}}+3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}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)^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)

[Out]

A*a^3*x+1/d*A*a^3*c+a^3*B*sin(d*x+c)/d+7/2/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*B*x+3/d*a^3*B*c+3/d*A*a^3*t
an(d*x+c)+3/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*A*a^3*sec(d*x+c)*tan(d*x+c)+1/d*a^3*B*tan(d*x+c)

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Maxima [A]  time = 1.01014, size = 223, normalized size = 1.96 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a^{3} + 12 \,{\left (d x + c\right )} B a^{3} - A a^{3}{\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 )} + 6 \, A a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right ) + 4 \, B a^{3} \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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.47112, size = 342, normalized size = 3. \begin{align*} \frac{4 \,{\left (A + 3 \, B\right )} a^{3} d x \cos \left (d x + c\right )^{2} +{\left (7 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (7 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}\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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(4*(A + 3*B)*a^3*d*x*cos(d*x + c)^2 + (7*A + 6*B)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (7*A + 6*B)*a
^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B*a^3*cos(d*x + c)^2 + 2*(3*A + B)*a^3*cos(d*x + c) + A*a^3)*s
in(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))**3*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.26239, size = 259, normalized size = 2.27 \begin{align*} \frac{\frac{4 \, B a^{3} \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} + 2 \,{\left (A a^{3} + 3 \, B a^{3}\right )}{\left (d x + c\right )} +{\left (7 \, A a^{3} + 6 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (7 \, A a^{3} + 6 \, B a^{3}\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^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a^{3} \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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/2*(4*B*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(A*a^3 + 3*B*a^3)*(d*x + c) + (7*A*a^3 + 6*
B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (7*A*a^3 + 6*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^
3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^3*tan(1/2*d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d