3.1.5 \(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [5]

Optimal. Leaf size=32 \[ a (A+B) x+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \sin (c+d x)}{d} \]

[Out]

a*(A+B)*x+a*A*arctanh(sin(d*x+c))/d+a*B*sin(d*x+c)/d

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Rubi [A]
time = 0.07, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3047, 3102, 2814, 3855} \begin {gather*} a x (A+B)+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \sin (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*(A + B)*x + (a*A*ArcTanh[Sin[c + d*x]])/d + (a*B*Sin[c + d*x])/d

Rule 2814

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 3047

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 3102

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 3855

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)) (A+B \cos (c+d x)) \sec (c+d x) \, dx &=\int \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {a B \sin (c+d x)}{d}+\int (a A+a (A+B) \cos (c+d x)) \sec (c+d x) \, dx\\ &=a (A+B) x+\frac {a B \sin (c+d x)}{d}+(a A) \int \sec (c+d x) \, dx\\ &=a (A+B) x+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 46, normalized size = 1.44 \begin {gather*} a A x+a B x+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \cos (d x) \sin (c)}{d}+\frac {a B \cos (c) \sin (d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + a*B*x + (a*A*ArcTanh[Sin[c + d*x]])/d + (a*B*Cos[d*x]*Sin[c])/d + (a*B*Cos[c]*Sin[d*x])/d

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Maple [A]
time = 0.15, size = 48, normalized size = 1.50

method	result	size
derivativedivides	\(\frac {a A \left (d x +c \right )+a B \sin \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a B \left (d x +c \right )}{d}\)	\(48\)
default	\(\frac {a A \left (d x +c \right )+a B \sin \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a B \left (d x +c \right )}{d}\)	\(48\)
risch	\(a x A +a B x -\frac {i a B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(83\)
norman	\(\frac {\left (a A +a B \right ) x +\left (a A +a B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a A +2 a B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 a B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*A*(d*x+c)+a*B*sin(d*x+c)+a*A*ln(sec(d*x+c)+tan(d*x+c))+a*B*(d*x+c))

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Maxima [A]
time = 0.26, size = 47, normalized size = 1.47 \begin {gather*} \frac {{\left (d x + c\right )} A a + {\left (d x + c\right )} B a + A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + B a \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*A*a + (d*x + c)*B*a + A*a*log(sec(d*x + c) + tan(d*x + c)) + B*a*sin(d*x + c))/d

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Fricas [A]
time = 0.37, size = 51, normalized size = 1.59 \begin {gather*} \frac {2 \, {\left (A + B\right )} a d x + A a \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(A + B)*a*d*x + A*a*log(sin(d*x + c) + 1) - A*a*log(-sin(d*x + c) + 1) + 2*B*a*sin(d*x + c))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \cos {\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 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c),x)

[Out]

a*(Integral(A*sec(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*
x), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x), x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (32) = 64\).
time = 0.45, size = 79, normalized size = 2.47 \begin {gather*} \frac {A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (A a + B a\right )} {\left (d x + c\right )} + \frac {2 \, B a \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}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - A*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (A*a + B*a)*(d*x + c) + 2*B
*a*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1))/d

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Mupad [B]
time = 0.28, size = 100, normalized size = 3.12 \begin {gather*} \frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\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 )}{d}+\frac {2\,A\,a\,\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 )}{d}+\frac {2\,B\,a\,\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 )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x)))/cos(c + d*x),x)

[Out]

(B*a*sin(c + d*x))/d + (2*A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*A*a*atanh(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2)))/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d

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