\(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [228]

Optimal result

Integrand size = 31, antiderivative size = 80 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=b^2 B x+\frac {\left (a^2 A+2 A b^2+4 a b B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

b^2*B*x+1/2*(A*a^2+2*A*b^2+4*B*a*b)*arctanh(sin(d*x+c))/d+a*(2*A*b+B*a)*tan(d*x+c)/d+1/2*a^2*A*sec(d*x+c)*tan(
d*x+c)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3067, 3100, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {\left (a^2 A+4 a b B+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a (a B+2 A b) \tan (c+d x)}{d}+b^2 B x \]

[In]

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

[Out]

b^2*B*x + ((a^2*A + 2*A*b^2 + 4*a*b*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*A*b + a*B)*Tan[c + d*x])/d + (a^2*
A*Sec[c + d*x]*Tan[c + d*x])/(2*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 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*} \text {integral}& = \frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a (2 A b+a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-2 b^2 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-a^2 A-2 A b^2-4 a b B-2 b^2 B \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = b^2 B x+\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \left (-a^2 A-2 A b^2-4 a b B\right ) \int \sec (c+d x) \, dx \\ & = b^2 B x+\frac {\left (a^2 A+2 A b^2+4 a b B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 b^2 B d x+\left (a^2 A+2 A b^2+4 a b B\right ) \text {arctanh}(\sin (c+d x))+a (4 A b+2 a B+a A \sec (c+d x)) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(2*b^2*B*d*x + (a^2*A + 2*A*b^2 + 4*a*b*B)*ArcTanh[Sin[c + d*x]] + a*(4*A*b + 2*a*B + a*A*Sec[c + d*x])*Tan[c
+ d*x])/(2*d)

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30

method	result	size
parts	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,b^{2}+2 B a b \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,b^{2} \left (d x +c \right )}{d}\)	\(104\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+2 A a b \tan \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )}{d}\)	\(112\)
default	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+2 A a b \tan \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )}{d}\)	\(112\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 B \,b^{2} d x \cos \left (2 d x +2 c \right )+\left (4 A a b +2 B \,a^{2}\right ) \sin \left (2 d x +2 c \right )+2 B \,b^{2} d x +2 A \,a^{2} \sin \left (d x +c \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(162\)
risch	\(b^{2} B x -\frac {i a \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}-4 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}-4 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}\)	\(217\)
norman	\(\frac {b^{2} B x +\frac {a \left (a A -4 A b -2 B a \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (a A +4 A b +2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+b^{2} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b^{2} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b^{2} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {6 A \,a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (a A -2 A b -B a \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (a A +2 A b +B a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-2 b^{2} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b^{2} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(326\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, B b^{2} d x \cos \left (d x + c\right )^{2} + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.75 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B b^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, A a b \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*B*b^2 - A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 4*B*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*A*b^2*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 4*B*a^2*tan(d*x + c) + 8*A*a*b*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (76) = 152\).

Time = 0.31 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.38 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} B b^{2} + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*B*b^2 + (A*a^2 + 4*B*a*b + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a^2 + 4*B*a*b + 2
*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^2*tan(1/2*d*x + 1/2*c)^3
- 4*A*a*b*tan(1/2*d*x + 1/2*c)^3 + A*a^2*tan(1/2*d*x + 1/2*c) + 2*B*a^2*tan(1/2*d*x + 1/2*c) + 4*A*a*b*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 = 1.17 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.20 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2\,\left (\frac {A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

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

[Out]

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