3.143 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=120 \[ \frac {2 (36 A+C) \sin (c+d x)}{15 a^3 d}-\frac {3 A \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {3 A x}{a^3}-\frac {(9 A-C) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4085, 4020, 3787, 2637, 8} \[ \frac {2 (36 A+C) \sin (c+d x)}{15 a^3 d}-\frac {3 A \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {3 A x}{a^3}-\frac {(9 A-C) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*x)/a^3 + (2*(36*A + C)*Sin[c + d*x])/(15*a^3*d) - ((A + C)*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) -
((9*A - C)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (3*A*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 2637

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

Rule 3787

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 4020

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]

Rule 4085

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

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) (-a (6 A+C)+a (3 A-2 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-C) \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+2 C)+2 a^2 (9 A-C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {3 A \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+C)+45 a^3 A \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {3 A \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(3 A) \int 1 \, dx}{a^3}+\frac {(2 (36 A+C)) \int \cos (c+d x) \, dx}{15 a^3}\\ &=-\frac {3 A x}{a^3}+\frac {2 (36 A+C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {3 A \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.99, size = 283, normalized size = 2.36 \[ -\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (1125 A \sin \left (c+\frac {d x}{2}\right )-1215 A \sin \left (c+\frac {3 d x}{2}\right )+225 A \sin \left (2 c+\frac {3 d x}{2}\right )-363 A \sin \left (2 c+\frac {5 d x}{2}\right )-75 A \sin \left (3 c+\frac {5 d x}{2}\right )-15 A \sin \left (3 c+\frac {7 d x}{2}\right )-15 A \sin \left (4 c+\frac {7 d x}{2}\right )+900 A d x \cos \left (c+\frac {d x}{2}\right )+450 A d x \cos \left (c+\frac {3 d x}{2}\right )+450 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+90 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+90 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-1755 A \sin \left (\frac {d x}{2}\right )+900 A d x \cos \left (\frac {d x}{2}\right )+120 C \sin \left (c+\frac {d x}{2}\right )-80 C \sin \left (c+\frac {3 d x}{2}\right )+60 C \sin \left (2 c+\frac {3 d x}{2}\right )-28 C \sin \left (2 c+\frac {5 d x}{2}\right )-160 C \sin \left (\frac {d x}{2}\right )\right )}{960 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/960*(Sec[c/2]*Sec[(c + d*x)/2]^5*(900*A*d*x*Cos[(d*x)/2] + 900*A*d*x*Cos[c + (d*x)/2] + 450*A*d*x*Cos[c + (
3*d*x)/2] + 450*A*d*x*Cos[2*c + (3*d*x)/2] + 90*A*d*x*Cos[2*c + (5*d*x)/2] + 90*A*d*x*Cos[3*c + (5*d*x)/2] - 1
755*A*Sin[(d*x)/2] - 160*C*Sin[(d*x)/2] + 1125*A*Sin[c + (d*x)/2] + 120*C*Sin[c + (d*x)/2] - 1215*A*Sin[c + (3
*d*x)/2] - 80*C*Sin[c + (3*d*x)/2] + 225*A*Sin[2*c + (3*d*x)/2] + 60*C*Sin[2*c + (3*d*x)/2] - 363*A*Sin[2*c +
(5*d*x)/2] - 28*C*Sin[2*c + (5*d*x)/2] - 75*A*Sin[3*c + (5*d*x)/2] - 15*A*Sin[3*c + (7*d*x)/2] - 15*A*Sin[4*c
+ (7*d*x)/2]))/(a^3*d)

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fricas [A]  time = 0.50, size = 149, normalized size = 1.24 \[ -\frac {45 \, A d x \cos \left (d x + c\right )^{3} + 135 \, A d x \cos \left (d x + c\right )^{2} + 135 \, A d x \cos \left (d x + c\right ) + 45 \, A d x - {\left (15 \, A \cos \left (d x + c\right )^{3} + {\left (117 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (57 \, A + 2 \, C\right )} \cos \left (d x + c\right ) + 72 \, A + 2 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(45*A*d*x*cos(d*x + c)^3 + 135*A*d*x*cos(d*x + c)^2 + 135*A*d*x*cos(d*x + c) + 45*A*d*x - (15*A*cos(d*x
+ c)^3 + (117*A + 7*C)*cos(d*x + c)^2 + 3*(57*A + 2*C)*cos(d*x + c) + 72*A + 2*C)*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)

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giac [A]  time = 0.28, size = 151, normalized size = 1.26 \[ -\frac {\frac {180 \, {\left (d x + c\right )} A}{a^{3}} - \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(180*(d*x + c)*A/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*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 10*C*a^12*tan(1/2*d*x +
1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1/2*c) + 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 1.19, size = 170, normalized size = 1.42 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*A*tan(1/2*d*x+1
/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A

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maxima [A]  time = 0.49, size = 205, normalized size = 1.71 \[ \frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(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) + C*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*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)/d

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mupad [B]  time = 2.74, size = 153, normalized size = 1.28 \[ \frac {\left (\frac {24\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {7\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {3\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}-\frac {4\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}-\frac {3\,A\,x}{a^3}+\frac {2\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*sin(c/2 + (d*x)/2))/20 + (C*sin(c/2 + (d*x)/2))/20 - cos(c/2 + (d*x)/2)^2*((3*A*sin(c/2 + (d*x)/2))/5 + (4
*C*sin(c/2 + (d*x)/2))/15) + cos(c/2 + (d*x)/2)^4*((24*A*sin(c/2 + (d*x)/2))/5 + (7*C*sin(c/2 + (d*x)/2))/15))
/(a^3*d*cos(c/2 + (d*x)/2)^5) - (3*A*x)/a^3 + (2*A*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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