3.3.19 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [219]

Optimal. Leaf size=35 \[ b B x+\frac {(A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3047, 3100, 2814, 3855} \begin {gather*} \frac {(a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d}+b B x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x]

[Out]

b*B*x + ((A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (a*A*Tan[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 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 43, normalized size = 1.23 \begin {gather*} b B x+\frac {A b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 57, normalized size = 1.63

method	result	size
derivativedivides	\(\frac {a A \tan \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B b \left (d x +c \right )}{d}\)	\(57\)
default	\(\frac {a A \tan \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B b \left (d x +c \right )}{d}\)	\(57\)
risch	\(b B x +\frac {2 i a A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(105\)
norman	\(\frac {b B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b B x -\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-b B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (35) = 70\).
time = 0.29, size = 73, normalized size = 2.09 \begin {gather*} \frac {2 \, {\left (d x + c\right )} B b + B a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + A b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (35) = 70\).
time = 0.38, size = 85, normalized size = 2.43 \begin {gather*} \frac {2 \, B b d x \cos \left (d x + c\right ) + {\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, A a \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (35) = 70\).
time = 0.44, size = 84, normalized size = 2.40 \begin {gather*} \frac {{\left (d x + c\right )} B b + {\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, A 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+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.48, size = 114, normalized size = 3.26 \begin {gather*} \frac {2\,B\,b\,\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 {A\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {A\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x)))/cos(c + d*x)^2,x)

[Out]

(2*B*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (B*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
2i)/d - (A*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (A*a*sin(c + d*x))/(d*cos(c + d*x))

________________________________________________________________________________________