\(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [7]

Optimal result

Integrand size = 29, antiderivative size = 56 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a (A+2 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*a*(A+2*B)*arctanh(sin(d*x+c))/d+a*(A+B)*tan(d*x+c)/d+1/2*a*A*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3100, 2827, 3852, 8, 3855} \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a (A+2 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

(a*(A + 2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(A + B)*Tan[c + d*x])/d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/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*} \text {integral}& = \int \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (2 a (A+B)+a (A+2 B) \cos (c+d x)) \sec ^2(c+d x) \, dx \\ & = \frac {a A \sec (c+d x) \tan (c+d x)}{2 d}+(a (A+B)) \int \sec ^2(c+d x) \, dx+\frac {1}{2} (a (A+2 B)) \int \sec (c+d x) \, dx \\ & = \frac {a (A+2 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {(a (A+B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = \frac {a (A+2 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a B \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d}+\frac {a B \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(a*A*ArcTanh[Sin[c + d*x]])/(2*d) + (a*B*ArcTanh[Sin[c + d*x]])/d + (a*A*Tan[c + d*x])/d + (a*B*Tan[c + d*x])/
d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {a A \tan \left (d x +c \right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )}{d}\)	\(75\)
default	\(\frac {a A \tan \left (d x +c \right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )}{d}\)	\(75\)
parts	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (a A +B a \right ) \tan \left (d x +c \right )}{d}+\frac {B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(76\)
parallelrisch	\(-\frac {\left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 B -2 A \right ) \sin \left (2 d x +2 c \right )-2 A \sin \left (d x +c \right )\right ) a}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(106\)
risch	\(-\frac {i a \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-2 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(155\)
norman	\(\frac {\frac {a \left (A -2 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (5 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (A +2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a \left (A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(167\)

[In]

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

[Out]

1/d*(a*A*tan(d*x+c)+B*a*ln(sec(d*x+c)+tan(d*x+c))+a*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))
)+B*a*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.59 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left (A + 2 \, B\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, B\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*((A + 2*B)*a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A + 2*B)*a*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*
(2*(A + B)*a*cos(d*x + c) + A*a)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)

[Out]

a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(B*cos(c + d*x)*sec(
c + d*x)**3, x) + Integral(B*cos(c + d*x)**2*sec(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.70 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=-\frac {A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, B a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A a \tan \left (d x + c\right ) - 4 \, B a \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

-1/4*(A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 2*B*a*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*A*a*tan(d*x + c) - 4*B*a*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (52) = 104\).

Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.21 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left (A a + 2 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + 2 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A\,a+2\,B\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,a+2\,B\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+2\,B\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*A*a + 2*B*a) - tan(c/2 + (d*x)/2)^3*(A*a + 2*B*a))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2
+ (d*x)/2)^2 + 1)) + (a*atanh(tan(c/2 + (d*x)/2))*(A + 2*B))/d