\(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [298]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

1/2*B*arctanh(sin(d*x+c))/d+1/3*(2*A+3*C)*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/3*A*sec(d*x+c)^2*tan(d*
x+c)/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 78, 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 = {3100, 2827, 3853, 3855, 3852, 8} \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*A + 3*C)*Tan[c + d*x])/(3*d) + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (A*
Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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}& = \frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (3 B+(2 A+3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx \\ & = \frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+B \int \sec ^3(c+d x) \, dx+\frac {1}{3} (2 A+3 C) \int \sec ^2(c+d x) \, dx \\ & = \frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} B \int \sec (c+d x) \, dx-\frac {(2 A+3 C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 B \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 (A+C)+3 B \sec (c+d x)+2 A \tan ^2(c+d x)\right )}{6 d} \]

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(3*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + C) + 3*B*Sec[c + d*x] + 2*A*Tan[c + d*x]^2))/(6*d)

Maple [A] (verified)

Time = 7.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) C}{d}\)	\(68\)
default	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) C}{d}\)	\(68\)
parts	\(-\frac {A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \tan \left (d x +c \right )}{d}\)	\(73\)
risch	\(-\frac {i \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 A -6 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(126\)
parallelrisch	\(\frac {-9 B \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 B \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (4 A +6 C \right ) \sin \left (3 d x +3 c \right )+6 \sin \left (2 d x +2 c \right ) B +12 \left (A +\frac {C}{2}\right ) \sin \left (d x +c \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(135\)
norman	\(\frac {-\frac {4 \left (A -3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (4 A -3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (4 A +3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (2 A -B +2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 A +B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(185\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(-A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+tan(d*x
+c)*C)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, B \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(3*B*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*B*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(2*A + 3*C)*
cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A - 3 \, B {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (70) = 140\).

Time = 0.37 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.08 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(3*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*tan(1/2*d*x + 1/
2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 4*A*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d
*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 3.50 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.58 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,A-B+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^4,x)

[Out]

(B*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)*(2*A + B + 2*C) - tan(c/2 + (d*x)/2)^3*((4*A)/3 + 4*C) +
tan(c/2 + (d*x)/2)^5*(2*A - B + 2*C))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2
)^6 - 1))