3.152 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=152 \[ \frac {2 (332 A+3 C) \sin (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {4 A \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {4 A x}{a^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A+C) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-4*A*x/a^4+2/105*(332*A+3*C)*sin(d*x+c)/a^4/d-1/105*(88*A-3*C)*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2-4*A*sin(d*x+c
)/a^4/d/(1+sec(d*x+c))-1/7*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-2/35*(6*A-C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^
3

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Rubi [A]  time = 0.50, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 (332 A+3 C) \sin (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {4 A \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {4 A x}{a^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A+C) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(-4*A*x)/a^4 + (2*(332*A + 3*C)*Sin[c + d*x])/(105*a^4*d) - ((88*A - 3*C)*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c
+ d*x])^2) - (4*A*Sin[c + d*x])/(a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d*x])/(7*d*(a + a*Sec[c + d*x])^
4) - (2*(6*A - C)*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]

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))^4} \, dx &=-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {\cos (c+d x) (-a (8 A+C)+a (4 A-3 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) \left (-a^2 (52 A+3 C)+6 a^2 (6 A-C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) \left (-2 a^3 (122 A+3 C)+2 a^3 (88 A-3 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 A \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {\int \cos (c+d x) \left (-2 a^4 (332 A+3 C)+420 a^4 A \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 A \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {(4 A) \int 1 \, dx}{a^4}+\frac {(2 (332 A+3 C)) \int \cos (c+d x) \, dx}{105 a^4}\\ &=-\frac {4 A x}{a^4}+\frac {2 (332 A+3 C) \sin (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (6 A-C) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 A \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.31, size = 371, normalized size = 2.44 \[ -\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (46130 A \sin \left (c+\frac {d x}{2}\right )-46116 A \sin \left (c+\frac {3 d x}{2}\right )+18060 A \sin \left (2 c+\frac {3 d x}{2}\right )-19292 A \sin \left (2 c+\frac {5 d x}{2}\right )+2100 A \sin \left (3 c+\frac {5 d x}{2}\right )-3791 A \sin \left (3 c+\frac {7 d x}{2}\right )-735 A \sin \left (4 c+\frac {7 d x}{2}\right )-105 A \sin \left (4 c+\frac {9 d x}{2}\right )-105 A \sin \left (5 c+\frac {9 d x}{2}\right )+29400 A d x \cos \left (c+\frac {d x}{2}\right )+17640 A d x \cos \left (c+\frac {3 d x}{2}\right )+17640 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+5880 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+5880 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+840 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+840 A d x \cos \left (4 c+\frac {7 d x}{2}\right )-60830 A \sin \left (\frac {d x}{2}\right )+29400 A d x \cos \left (\frac {d x}{2}\right )+2520 C \sin \left (c+\frac {d x}{2}\right )-1764 C \sin \left (c+\frac {3 d x}{2}\right )+1260 C \sin \left (2 c+\frac {3 d x}{2}\right )-588 C \sin \left (2 c+\frac {5 d x}{2}\right )+420 C \sin \left (3 c+\frac {5 d x}{2}\right )-144 C \sin \left (3 c+\frac {7 d x}{2}\right )-2520 C \sin \left (\frac {d x}{2}\right )\right )}{26880 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-1/26880*(Sec[c/2]*Sec[(c + d*x)/2]^7*(29400*A*d*x*Cos[(d*x)/2] + 29400*A*d*x*Cos[c + (d*x)/2] + 17640*A*d*x*C
os[c + (3*d*x)/2] + 17640*A*d*x*Cos[2*c + (3*d*x)/2] + 5880*A*d*x*Cos[2*c + (5*d*x)/2] + 5880*A*d*x*Cos[3*c +
(5*d*x)/2] + 840*A*d*x*Cos[3*c + (7*d*x)/2] + 840*A*d*x*Cos[4*c + (7*d*x)/2] - 60830*A*Sin[(d*x)/2] - 2520*C*S
in[(d*x)/2] + 46130*A*Sin[c + (d*x)/2] + 2520*C*Sin[c + (d*x)/2] - 46116*A*Sin[c + (3*d*x)/2] - 1764*C*Sin[c +
(3*d*x)/2] + 18060*A*Sin[2*c + (3*d*x)/2] + 1260*C*Sin[2*c + (3*d*x)/2] - 19292*A*Sin[2*c + (5*d*x)/2] - 588*
C*Sin[2*c + (5*d*x)/2] + 2100*A*Sin[3*c + (5*d*x)/2] + 420*C*Sin[3*c + (5*d*x)/2] - 3791*A*Sin[3*c + (7*d*x)/2
] - 144*C*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]))/(a^4*d)

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fricas [A]  time = 0.45, size = 193, normalized size = 1.27 \[ -\frac {420 \, A d x \cos \left (d x + c\right )^{4} + 1680 \, A d x \cos \left (d x + c\right )^{3} + 2520 \, A d x \cos \left (d x + c\right )^{2} + 1680 \, A d x \cos \left (d x + c\right ) + 420 \, A d x - {\left (105 \, A \cos \left (d x + c\right )^{4} + 4 \, {\left (296 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2636 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (559 \, A + 6 \, C\right )} \cos \left (d x + c\right ) + 664 \, A + 6 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/105*(420*A*d*x*cos(d*x + c)^4 + 1680*A*d*x*cos(d*x + c)^3 + 2520*A*d*x*cos(d*x + c)^2 + 1680*A*d*x*cos(d*x
+ c) + 420*A*d*x - (105*A*cos(d*x + c)^4 + 4*(296*A + 9*C)*cos(d*x + c)^3 + (2636*A + 39*C)*cos(d*x + c)^2 + 4
*(559*A + 6*C)*cos(d*x + c) + 664*A + 6*C)*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 = 0.50, size = 184, normalized size = 1.21 \[ -\frac {\frac {3360 \, {\left (d x + c\right )} A}{a^{4}} - \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 63 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(3360*(d*x + c)*A/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*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 63*C*a^24*tan(1/2
*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 5145*A*a^24*tan(1/2*
d*x + 1/2*c) - 105*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 1.36, size = 210, normalized size = 1.38 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {3 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}} \]

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))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5+3/40/d/
a^4*C*tan(1/2*d*x+1/2*c)^5-23/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A-1/8/d/a^4*C*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*A*ta
n(1/2*d*x+1/2*c)+1/8/d/a^4*C*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))

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maxima [A]  time = 0.46, size = 246, normalized size = 1.62 \[ \frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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))^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) + 3*C*(35*s
in(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)
^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

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mupad [B]  time = 2.75, size = 192, normalized size = 1.26 \[ \frac {2\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d}-\frac {\left (-\frac {764\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-\frac {12\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {143\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {23\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{70}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {8\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {9\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{70}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}-\frac {4\,A\,x}{a^4} \]

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))^4,x)

[Out]

(2*A*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^4*d) - ((A*sin(c/2 + (d*x)/2))/56 + (C*sin(c/2 + (d*x)/2))/56 -
cos(c/2 + (d*x)/2)^2*((8*A*sin(c/2 + (d*x)/2))/35 + (9*C*sin(c/2 + (d*x)/2))/70) + cos(c/2 + (d*x)/2)^4*((143
*A*sin(c/2 + (d*x)/2))/105 + (23*C*sin(c/2 + (d*x)/2))/70) - cos(c/2 + (d*x)/2)^6*((764*A*sin(c/2 + (d*x)/2))/
105 + (12*C*sin(c/2 + (d*x)/2))/35))/(a^4*d*cos(c/2 + (d*x)/2)^7) - (4*A*x)/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)**2)/(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(C*cos(c + d*x)*sec(c + d*x)**2/(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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