\(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [105]

Optimal result

Integrand size = 29, antiderivative size = 136 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(3 A-B) x}{a^3}+\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-(3*A-B)*x/a^3+2/15*(36*A-11*B)*sin(d*x+c)/a^3/d-1/5*(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(9*A-4*B)*sin(
d*x+c)/a/d/(a+a*sec(d*x+c))^2-(3*A-B)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4105, 3872, 2717, 8} \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (3 A-B)}{a^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[In]

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

[Out]

-(((3*A - B)*x)/a^3) + (2*(36*A - 11*B)*Sin[c + d*x])/(15*a^3*d) - ((A - B)*Sin[c + d*x])/(5*d*(a + a*Sec[c +
d*x])^3) - ((9*A - 4*B)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((3*A - B)*Sin[c + d*x])/(d*(a^3 + a^3
*Sec[c + d*x]))

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos (c+d x) (a (6 A-B)-3 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos (c+d x) \left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \cos (c+d x) \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 (36 A-11 B)) \int \cos (c+d x) \, dx}{15 a^3}-\frac {(3 A-B) \int 1 \, dx}{a^3} \\ & = -\frac {(3 A-B) x}{a^3}+\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.46 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\sin (c+d x) \left (27 A-7 B-\frac {3 (A-B)}{(1+\sec (c+d x))^3}+\frac {-9 A+4 B}{(1+\sec (c+d x))^2}+\frac {15 (3 A-B) \left (\arcsin (\cos (c+d x)) (1+\cos (c+d x))+\sqrt {\sin ^2(c+d x)}\right )}{\sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{3/2}}\right )}{15 a^3 d} \]

[In]

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

[Out]

(Sin[c + d*x]*(27*A - 7*B - (3*(A - B))/(1 + Sec[c + d*x])^3 + (-9*A + 4*B)/(1 + Sec[c + d*x])^2 + (15*(3*A -
B)*(ArcSin[Cos[c + d*x]]*(1 + Cos[c + d*x]) + Sqrt[Sin[c + d*x]^2]))/(Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x]
)^(3/2))))/(15*a^3*d)

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {729 \left (\left (\frac {26 A}{81}-\frac {64 B}{729}\right ) \cos \left (2 d x +2 c \right )+\frac {5 A \cos \left (3 d x +3 c \right )}{243}+\left (A -\frac {68 B}{243}\right ) \cos \left (d x +c \right )+\frac {58 A}{81}-\frac {152 B}{729}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-720 \left (A -\frac {B}{3}\right ) d x}{240 a^{3} d}\)	\(89\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {8 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-8 \left (3 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(136\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {8 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-8 \left (3 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(136\)
norman	\(\frac {-\frac {\left (3 A -B \right ) x}{a}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}-\frac {\left (3 A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (25 A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (27 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (45 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2}}\)	\(158\)
risch	\(-\frac {3 A x}{a^{3}}+\frac {x B}{a^{3}}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {2 i \left (90 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 B \,{\mathrm e}^{4 i \left (d x +c \right )}+300 A \,{\mathrm e}^{3 i \left (d x +c \right )}-135 B \,{\mathrm e}^{3 i \left (d x +c \right )}+420 A \,{\mathrm e}^{2 i \left (d x +c \right )}-185 B \,{\mathrm e}^{2 i \left (d x +c \right )}+270 \,{\mathrm e}^{i \left (d x +c \right )} A -115 B \,{\mathrm e}^{i \left (d x +c \right )}+72 A -32 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(178\)

[In]

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

[Out]

1/240*(729*((26/81*A-64/729*B)*cos(2*d*x+2*c)+5/243*A*cos(3*d*x+3*c)+(A-68/243*B)*cos(d*x+c)+58/81*A-152/729*B
)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4-720*(A-1/3*B)*d*x)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.27 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (3 \, A - B\right )} d x - {\left (15 \, A \cos \left (d x + c\right )^{3} + {\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 72 \, A - 22 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

-1/15*(15*(3*A - B)*d*x*cos(d*x + c)^3 + 45*(3*A - B)*d*x*cos(d*x + c)^2 + 45*(3*A - B)*d*x*cos(d*x + c) + 15*
(3*A - B)*d*x - (15*A*cos(d*x + c)^3 + (117*A - 32*B)*cos(d*x + c)^2 + 3*(57*A - 17*B)*cos(d*x + c) + 72*A - 2
2*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F]

\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

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

[Out]

(Integral(A*cos(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*cos(c + d
*x)*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.70 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

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

[Out]

1/60*(3*A*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (d x + c\right )} {\left (3 \, A - B\right )}}{a^{3}} - \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

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

[Out]

-1/60*(60*(d*x + c)*(3*A - B)/a^3 - 120*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (3*A*a^12*
tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*B*a^12*tan(1/
2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 14.01 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2\,a^3}+\frac {3\,\left (A-B\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}-\frac {x\,\left (3\,A-B\right )}{a^3}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((3*A)/(2*a^3) + (3*(A - B))/(4*a^3) + (4*A - 2*B)/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*((A
- B)/(6*a^3) + (4*A - 2*B)/(12*a^3)))/d - (x*(3*A - B))/a^3 + (2*A*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x
)/2)^2 + a^3)) + (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)