\(\int (a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [782]

Optimal result

Integrand size = 40, antiderivative size = 60 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=b (b B+2 a C) x+\frac {a (2 b B+a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 C \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d} \]

[Out]

b*(B*b+2*C*a)*x+a*(2*B*b+C*a)*arctanh(sin(d*x+c))/d+b^2*C*sin(d*x+c)/d+a^2*B*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3108, 3067, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^2 B \tan (c+d x)}{d}+\frac {a (a C+2 b B) \text {arctanh}(\sin (c+d x))}{d}+b x (2 a C+b B)+\frac {b^2 C \sin (c+d x)}{d} \]

[In]

Int[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

b*(b*B + 2*a*C)*x + (a*(2*b*B + a*C)*ArcTanh[Sin[c + d*x]])/d + (b^2*C*Sin[c + d*x])/d + (a^2*B*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 3067

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

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 3108

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 (a+b \cos (c+d x))^2 (B+C \cos (c+d x)) \sec ^2(c+d x) \, dx \\ & = \frac {a^2 B \tan (c+d x)}{d}-\int \left (-a (2 b B+a C)-b (b B+2 a C) \cos (c+d x)-b^2 C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {b^2 C \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d}-\int (-a (2 b B+a C)-b (b B+2 a C) \cos (c+d x)) \sec (c+d x) \, dx \\ & = b (b B+2 a C) x+\frac {b^2 C \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d}+(a (2 b B+a C)) \int \sec (c+d x) \, dx \\ & = b (b B+2 a C) x+\frac {a (2 b B+a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 C \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.82 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {b (b B+2 a C) (c+d x)-a (2 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a (2 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 C \sin (c+d x)+a^2 B \tan (c+d x)}{d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(b*(b*B + 2*a*C)*(c + d*x) - a*(2*b*B + a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + a*(2*b*B + a*C)*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2*C*Sin[c + d*x] + a^2*B*Tan[c + d*x])/d

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32

method	result	size
parts	\(\frac {\left (B \,b^{2}+2 C a b \right ) \left (d x +c \right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {b^{2} C \sin \left (d x +c \right )}{d}\)	\(79\)
derivativedivides	\(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 C a b \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (d x +c \right )}{d}\)	\(86\)
default	\(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 C a b \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (d x +c \right )}{d}\)	\(86\)
parallelrisch	\(\frac {\left (-4 B a b -2 a^{2} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (4 B a b +2 a^{2} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+C \sin \left (2 d x +2 c \right ) b^{2}+2 b d x \left (B b +2 a C \right ) \cos \left (d x +c \right )+2 B \sin \left (d x +c \right ) a^{2}}{2 d \cos \left (d x +c \right )}\)	\(122\)
risch	\(b^{2} B x +2 a b C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{2} C}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{2} C}{2 d}+\frac {2 i B \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{d}\)	\(160\)
norman	\(\frac {\left (B \,b^{2}+2 C a b \right ) x +\left (-4 B \,b^{2}-8 C a b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}-2 C a b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}-2 C a b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B \,b^{2}+2 C a b \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,b^{2}+4 C a b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,b^{2}+4 C a b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 \left (B \,a^{2}-b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (B \,a^{2}-b^{2} C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (B \,a^{2}+b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (B \,a^{2}+b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (3 B \,a^{2}-b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (3 B \,a^{2}+b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a \left (2 B b +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (2 B b +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(412\)

[In]

int((a+cos(d*x+c)*b)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

(B*b^2+2*C*a*b)/d*(d*x+c)+(2*B*a*b+C*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+a^2*B*tan(d*x+c)/d+b^2*C*sin(d*x+c)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.95 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (2 \, C a b + B b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((a+b*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.72 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a b + 2 \, {\left (d x + c\right )} B b^{2} + C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C b^{2} \sin \left (d x + c\right ) + 2 \, B a^{2} \tan \left (d x + c\right )}{2 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/2*(4*(d*x + c)*C*a*b + 2*(d*x + c)*B*b^2 + C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*B*a*b*(
log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*C*b^2*sin(d*x + c) + 2*B*a^2*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (60) = 120\).

Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.53 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (2 \, C a b + B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

((2*C*a*b + B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (C*a^2 + 2*B*a*b)*log(ab
s(tan(1/2*d*x + 1/2*c) - 1)) - 2*(B*a^2*tan(1/2*d*x + 1/2*c)^3 - C*b^2*tan(1/2*d*x + 1/2*c)^3 + B*a^2*tan(1/2*
d*x + 1/2*c) + C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1))/d

Mupad [B] (verification not implemented)

Time = 2.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.82 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )}+\frac {4\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \]

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d*x)^3,x)

[Out]

(B*a^2*tan(c + d*x))/d + (2*B*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (C*a^2*atan((sin(c/2 + (d*x
)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (C*b^2*sin(2*c + 2*d*x))/(2*d*cos(c + d*x)) - (B*a*b*atan((sin(c/2 + (d*x
)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (4*C*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d