3.115 \(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=166 \[ \frac {8 (83 A-20 B) \sin (c+d x)}{105 a^4 d}-\frac {(4 A-B) \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {x (4 A-B)}{a^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-(4*A-B)*x/a^4+8/105*(83*A-20*B)*sin(d*x+c)/a^4/d-1/105*(88*A-25*B)*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2-(4*A-B)*
sin(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-1/35*(12*A-5*B)*sin(d*x+c)/a/d/(a+a*
sec(d*x+c))^3

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Rubi [A]  time = 0.57, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4020, 3787, 2637, 8} \[ \frac {8 (83 A-20 B) \sin (c+d x)}{105 a^4 d}-\frac {(4 A-B) \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {x (4 A-B)}{a^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

-(((4*A - B)*x)/a^4) + (8*(83*A - 20*B)*Sin[c + d*x])/(105*a^4*d) - ((88*A - 25*B)*Sin[c + d*x])/(105*a^4*d*(1
+ Sec[c + d*x])^2) - ((4*A - B)*Sin[c + d*x])/(a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Sin[c + d*x])/(7*d*(a + a
*Sec[c + d*x])^4) - ((12*A - 5*B)*Sin[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

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))^4} \, dx &=-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\cos (c+d x) (a (8 A-B)-4 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos (c+d x) \left (2 a^2 (26 A-5 B)-3 a^2 (12 A-5 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos (c+d x) \left (a^3 (244 A-55 B)-2 a^3 (88 A-25 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(4 A-B) \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {\int \cos (c+d x) \left (8 a^4 (83 A-20 B)-105 a^4 (4 A-B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(4 A-B) \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(8 (83 A-20 B)) \int \cos (c+d x) \, dx}{105 a^4}-\frac {(4 A-B) \int 1 \, dx}{a^4}\\ &=-\frac {(4 A-B) x}{a^4}+\frac {8 (83 A-20 B) \sin (c+d x)}{105 a^4 d}-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(4 A-B) \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.11, size = 485, normalized size = 2.92 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-7350 d x (4 A-B) \cos \left (c+\frac {d x}{2}\right )-7350 d x (4 A-B) \cos \left (\frac {d x}{2}\right )-46130 A \sin \left (c+\frac {d x}{2}\right )+46116 A \sin \left (c+\frac {3 d x}{2}\right )-18060 A \sin \left (2 c+\frac {3 d x}{2}\right )+19292 A \sin \left (2 c+\frac {5 d x}{2}\right )-2100 A \sin \left (3 c+\frac {5 d x}{2}\right )+3791 A \sin \left (3 c+\frac {7 d x}{2}\right )+735 A \sin \left (4 c+\frac {7 d x}{2}\right )+105 A \sin \left (4 c+\frac {9 d x}{2}\right )+105 A \sin \left (5 c+\frac {9 d x}{2}\right )-17640 A d x \cos \left (c+\frac {3 d x}{2}\right )-17640 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-5880 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-5880 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-840 A d x \cos \left (3 c+\frac {7 d x}{2}\right )-840 A d x \cos \left (4 c+\frac {7 d x}{2}\right )+60830 A \sin \left (\frac {d x}{2}\right )+16520 B \sin \left (c+\frac {d x}{2}\right )-14280 B \sin \left (c+\frac {3 d x}{2}\right )+7560 B \sin \left (2 c+\frac {3 d x}{2}\right )-5600 B \sin \left (2 c+\frac {5 d x}{2}\right )+1680 B \sin \left (3 c+\frac {5 d x}{2}\right )-1040 B \sin \left (3 c+\frac {7 d x}{2}\right )+4410 B d x \cos \left (c+\frac {3 d x}{2}\right )+4410 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+1470 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+1470 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+210 B d x \cos \left (3 c+\frac {7 d x}{2}\right )+210 B d x \cos \left (4 c+\frac {7 d x}{2}\right )-19880 B \sin \left (\frac {d x}{2}\right )\right )}{1680 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-7350*(4*A - B)*d*x*Cos[(d*x)/2] - 7350*(4*A - B)*d*x*Cos[c + (d*x)/2] - 17640*A*d
*x*Cos[c + (3*d*x)/2] + 4410*B*d*x*Cos[c + (3*d*x)/2] - 17640*A*d*x*Cos[2*c + (3*d*x)/2] + 4410*B*d*x*Cos[2*c
+ (3*d*x)/2] - 5880*A*d*x*Cos[2*c + (5*d*x)/2] + 1470*B*d*x*Cos[2*c + (5*d*x)/2] - 5880*A*d*x*Cos[3*c + (5*d*x
)/2] + 1470*B*d*x*Cos[3*c + (5*d*x)/2] - 840*A*d*x*Cos[3*c + (7*d*x)/2] + 210*B*d*x*Cos[3*c + (7*d*x)/2] - 840
*A*d*x*Cos[4*c + (7*d*x)/2] + 210*B*d*x*Cos[4*c + (7*d*x)/2] + 60830*A*Sin[(d*x)/2] - 19880*B*Sin[(d*x)/2] - 4
6130*A*Sin[c + (d*x)/2] + 16520*B*Sin[c + (d*x)/2] + 46116*A*Sin[c + (3*d*x)/2] - 14280*B*Sin[c + (3*d*x)/2] -
18060*A*Sin[2*c + (3*d*x)/2] + 7560*B*Sin[2*c + (3*d*x)/2] + 19292*A*Sin[2*c + (5*d*x)/2] - 5600*B*Sin[2*c +
(5*d*x)/2] - 2100*A*Sin[3*c + (5*d*x)/2] + 1680*B*Sin[3*c + (5*d*x)/2] + 3791*A*Sin[3*c + (7*d*x)/2] - 1040*B*
Sin[3*c + (7*d*x)/2] + 735*A*Sin[4*c + (7*d*x)/2] + 105*A*Sin[4*c + (9*d*x)/2] + 105*A*Sin[5*c + (9*d*x)/2]))/
(1680*a^4*d*(1 + Cos[c + d*x])^4)

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fricas [A]  time = 0.45, size = 223, normalized size = 1.34 \[ -\frac {105 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (4 \, A - B\right )} d x - {\left (105 \, A \cos \left (d x + c\right )^{4} + 4 \, {\left (296 \, A - 65 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (659 \, A - 155 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (2236 \, A - 535 \, B\right )} \cos \left (d x + c\right ) + 664 \, A - 160 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

-1/105*(105*(4*A - B)*d*x*cos(d*x + c)^4 + 420*(4*A - B)*d*x*cos(d*x + c)^3 + 630*(4*A - B)*d*x*cos(d*x + c)^2
+ 420*(4*A - B)*d*x*cos(d*x + c) + 105*(4*A - B)*d*x - (105*A*cos(d*x + c)^4 + 4*(296*A - 65*B)*cos(d*x + c)^
3 + 4*(659*A - 155*B)*cos(d*x + c)^2 + (2236*A - 535*B)*cos(d*x + c) + 664*A - 160*B)*sin(d*x + c))/(a^4*d*cos
(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [A]  time = 0.70, size = 190, normalized size = 1.14 \[ -\frac {\frac {840 \, {\left (d x + c\right )} {\left (4 \, A - B\right )}}{a^{4}} - \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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))^4,x, algorithm="giac")

[Out]

-1/840*(840*(d*x + c)*(4*A - B)/a^4 - 1680*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^4) + (15*A*a
^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 - 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*B*a^24
*tan(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*tan(1/2*d*x + 1/2*c)^3 - 5145*A*a^24*
tan(1/2*d*x + 1/2*c) + 1575*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 1.23, size = 229, normalized size = 1.38 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{4}} \]

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))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A+1/56/d/a^4*B*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a
^4*B*tan(1/2*d*x+1/2*c)^5-23/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A+11/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*A*t
an(1/2*d*x+1/2*c)-15/8/d/a^4*B*tan(1/2*d*x+1/2*c)+2/d/a^4*A*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-8/d/a^
4*A*arctan(tan(1/2*d*x+1/2*c))+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B

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maxima [A]  time = 0.71, size = 271, normalized size = 1.63 \[ \frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

1/840*(A*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d
*x + c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
- 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) - 5*B*((315
*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4))/d

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mupad [B]  time = 2.06, size = 202, normalized size = 1.22 \[ \frac {\left (\frac {764\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-\frac {52\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {16\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}-\frac {143\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {8\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {5\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{28}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}-\frac {4\,A\,d\,x-B\,d\,x}{a^4\,d}+\frac {2\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \]

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))^4,x)

[Out]

((B*sin(c/2 + (d*x)/2))/56 - (A*sin(c/2 + (d*x)/2))/56 + cos(c/2 + (d*x)/2)^2*((8*A*sin(c/2 + (d*x)/2))/35 - (
5*B*sin(c/2 + (d*x)/2))/28) - cos(c/2 + (d*x)/2)^4*((143*A*sin(c/2 + (d*x)/2))/105 - (16*B*sin(c/2 + (d*x)/2))
/21) + cos(c/2 + (d*x)/2)^6*((764*A*sin(c/2 + (d*x)/2))/105 - (52*B*sin(c/2 + (d*x)/2))/21))/(a^4*d*cos(c/2 +
(d*x)/2)^7) - (4*A*d*x - B*d*x)/(a^4*d) + (2*A*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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))**4,x)

[Out]

(Integral(A*cos(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) +
Integral(B*cos(c + d*x)*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x)
+ 1), x))/a**4

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