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

Optimal. Leaf size=136 \[ \frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (3 A-B)}{a^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[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))

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Rubi [A]  time = 0.37, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4020, 3787, 2637, 8} \[ \frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (3 A-B)}{a^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=-\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*}

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Mathematica [B]  time = 1.08, size = 365, normalized size = 2.68 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-300 d x (3 A-B) \cos \left (c+\frac {d x}{2}\right )-300 d x (3 A-B) \cos \left (\frac {d x}{2}\right )-1125 A \sin \left (c+\frac {d x}{2}\right )+1215 A \sin \left (c+\frac {3 d x}{2}\right )-225 A \sin \left (2 c+\frac {3 d x}{2}\right )+363 A \sin \left (2 c+\frac {5 d x}{2}\right )+75 A \sin \left (3 c+\frac {5 d x}{2}\right )+15 A \sin \left (3 c+\frac {7 d x}{2}\right )+15 A \sin \left (4 c+\frac {7 d x}{2}\right )-450 A d x \cos \left (c+\frac {3 d x}{2}\right )-450 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-90 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-90 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+1755 A \sin \left (\frac {d x}{2}\right )+540 B \sin \left (c+\frac {d x}{2}\right )-460 B \sin \left (c+\frac {3 d x}{2}\right )+180 B \sin \left (2 c+\frac {3 d x}{2}\right )-128 B \sin \left (2 c+\frac {5 d x}{2}\right )+150 B d x \cos \left (c+\frac {3 d x}{2}\right )+150 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+30 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+30 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-740 B \sin \left (\frac {d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-300*(3*A - B)*d*x*Cos[(d*x)/2] - 300*(3*A - B)*d*x*Cos[c + (d*x)/2] - 450*A*d*x*C
os[c + (3*d*x)/2] + 150*B*d*x*Cos[c + (3*d*x)/2] - 450*A*d*x*Cos[2*c + (3*d*x)/2] + 150*B*d*x*Cos[2*c + (3*d*x
)/2] - 90*A*d*x*Cos[2*c + (5*d*x)/2] + 30*B*d*x*Cos[2*c + (5*d*x)/2] - 90*A*d*x*Cos[3*c + (5*d*x)/2] + 30*B*d*
x*Cos[3*c + (5*d*x)/2] + 1755*A*Sin[(d*x)/2] - 740*B*Sin[(d*x)/2] - 1125*A*Sin[c + (d*x)/2] + 540*B*Sin[c + (d
*x)/2] + 1215*A*Sin[c + (3*d*x)/2] - 460*B*Sin[c + (3*d*x)/2] - 225*A*Sin[2*c + (3*d*x)/2] + 180*B*Sin[2*c + (
3*d*x)/2] + 363*A*Sin[2*c + (5*d*x)/2] - 128*B*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]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.42, size = 173, normalized size = 1.27 \[ -\frac {15 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (3 \, A - B\right )} d x - {\left (15 \, A \cos \left (d x + c\right )^{3} + {\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 72 \, A - 22 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.52, size = 157, normalized size = 1.15 \[ -\frac {\frac {60 \, {\left (d x + c\right )} {\left (3 \, A - B\right )}}{a^{3}} - \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 1.04, size = 189, normalized size = 1.39 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+B*sec(d*x+c))/(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*B*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/3/d/a^3
*B*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)-7/4/d/a^3*B*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+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B

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maxima [A]  time = 0.44, size = 231, normalized size = 1.70 \[ \frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.98, size = 155, normalized size = 1.14 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2\,a^3}+\frac {3\,\left (A-B\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}-\frac {x\,\left (3\,A-B\right )}{a^3}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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