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

Optimal. Leaf size=223 \[ -\frac{8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac{(21 A-8 B) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac{4 (216 A-83 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(129 A-52 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{x (21 A-8 B)}{2 a^4}-\frac{(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

((21*A - 8*B)*x)/(2*a^4) - (8*(216*A - 83*B)*Sin[c + d*x])/(105*a^4*d) + ((21*A - 8*B)*Cos[c + d*x]*Sin[c + d*
x])/(2*a^4*d) - ((129*A - 52*B)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(216*A - 83*B
)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]*Sin[c + d*x])/(7*d*(a + a*
Sec[c + d*x])^4) - ((2*A - B)*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.648995, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4020, 3787, 2635, 8, 2637} \[ -\frac{8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac{(21 A-8 B) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac{4 (216 A-83 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(129 A-52 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{x (21 A-8 B)}{2 a^4}-\frac{(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A - 8*B)*x)/(2*a^4) - (8*(216*A - 83*B)*Sin[c + d*x])/(105*a^4*d) + ((21*A - 8*B)*Cos[c + d*x]*Sin[c + d*
x])/(2*a^4*d) - ((129*A - 52*B)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(216*A - 83*B
)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]*Sin[c + d*x])/(7*d*(a + a*
Sec[c + d*x])^4) - ((2*A - B)*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{\int \frac{\cos ^2(c+d x) (a (9 A-2 B)-5 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\cos ^2(c+d x) \left (a^2 (73 A-24 B)-28 a^2 (2 A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\cos ^2(c+d x) \left (a^3 (477 A-176 B)-3 a^3 (129 A-52 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (216 A-83 B) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{\int \cos ^2(c+d x) \left (105 a^4 (21 A-8 B)-8 a^4 (216 A-83 B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac{(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (216 A-83 B) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{(8 (216 A-83 B)) \int \cos (c+d x) \, dx}{105 a^4}+\frac{(21 A-8 B) \int \cos ^2(c+d x) \, dx}{a^4}\\ &=-\frac{8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac{(21 A-8 B) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac{(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (216 A-83 B) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{(21 A-8 B) \int 1 \, dx}{2 a^4}\\ &=\frac{(21 A-8 B) x}{2 a^4}-\frac{8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac{(21 A-8 B) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac{(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (216 A-83 B) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.11128, size = 555, normalized size = 2.49 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (14700 d x (21 A-8 B) \cos \left (c+\frac{d x}{2}\right )+14700 d x (21 A-8 B) \cos \left (\frac{d x}{2}\right )+386190 A \sin \left (c+\frac{d x}{2}\right )-422478 A \sin \left (c+\frac{3 d x}{2}\right )+132930 A \sin \left (2 c+\frac{3 d x}{2}\right )-181461 A \sin \left (2 c+\frac{5 d x}{2}\right )+3675 A \sin \left (3 c+\frac{5 d x}{2}\right )-36003 A \sin \left (3 c+\frac{7 d x}{2}\right )-9555 A \sin \left (4 c+\frac{7 d x}{2}\right )-945 A \sin \left (4 c+\frac{9 d x}{2}\right )-945 A \sin \left (5 c+\frac{9 d x}{2}\right )+105 A \sin \left (5 c+\frac{11 d x}{2}\right )+105 A \sin \left (6 c+\frac{11 d x}{2}\right )+185220 A d x \cos \left (c+\frac{3 d x}{2}\right )+185220 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+61740 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+61740 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+8820 A d x \cos \left (3 c+\frac{7 d x}{2}\right )+8820 A d x \cos \left (4 c+\frac{7 d x}{2}\right )-539490 A \sin \left (\frac{d x}{2}\right )-184520 B \sin \left (c+\frac{d x}{2}\right )+184464 B \sin \left (c+\frac{3 d x}{2}\right )-72240 B \sin \left (2 c+\frac{3 d x}{2}\right )+77168 B \sin \left (2 c+\frac{5 d x}{2}\right )-8400 B \sin \left (3 c+\frac{5 d x}{2}\right )+15164 B \sin \left (3 c+\frac{7 d x}{2}\right )+2940 B \sin \left (4 c+\frac{7 d x}{2}\right )+420 B \sin \left (4 c+\frac{9 d x}{2}\right )+420 B \sin \left (5 c+\frac{9 d x}{2}\right )-70560 B d x \cos \left (c+\frac{3 d x}{2}\right )-70560 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-23520 B d x \cos \left (2 c+\frac{5 d x}{2}\right )-23520 B d x \cos \left (3 c+\frac{5 d x}{2}\right )-3360 B d x \cos \left (3 c+\frac{7 d x}{2}\right )-3360 B d x \cos \left (4 c+\frac{7 d x}{2}\right )+243320 B \sin \left (\frac{d x}{2}\right )\right )}{6720 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(14700*(21*A - 8*B)*d*x*Cos[(d*x)/2] + 14700*(21*A - 8*B)*d*x*Cos[c + (d*x)/2] + 18
5220*A*d*x*Cos[c + (3*d*x)/2] - 70560*B*d*x*Cos[c + (3*d*x)/2] + 185220*A*d*x*Cos[2*c + (3*d*x)/2] - 70560*B*d
*x*Cos[2*c + (3*d*x)/2] + 61740*A*d*x*Cos[2*c + (5*d*x)/2] - 23520*B*d*x*Cos[2*c + (5*d*x)/2] + 61740*A*d*x*Co
s[3*c + (5*d*x)/2] - 23520*B*d*x*Cos[3*c + (5*d*x)/2] + 8820*A*d*x*Cos[3*c + (7*d*x)/2] - 3360*B*d*x*Cos[3*c +
(7*d*x)/2] + 8820*A*d*x*Cos[4*c + (7*d*x)/2] - 3360*B*d*x*Cos[4*c + (7*d*x)/2] - 539490*A*Sin[(d*x)/2] + 2433
20*B*Sin[(d*x)/2] + 386190*A*Sin[c + (d*x)/2] - 184520*B*Sin[c + (d*x)/2] - 422478*A*Sin[c + (3*d*x)/2] + 1844
64*B*Sin[c + (3*d*x)/2] + 132930*A*Sin[2*c + (3*d*x)/2] - 72240*B*Sin[2*c + (3*d*x)/2] - 181461*A*Sin[2*c + (5
*d*x)/2] + 77168*B*Sin[2*c + (5*d*x)/2] + 3675*A*Sin[3*c + (5*d*x)/2] - 8400*B*Sin[3*c + (5*d*x)/2] - 36003*A*
Sin[3*c + (7*d*x)/2] + 15164*B*Sin[3*c + (7*d*x)/2] - 9555*A*Sin[4*c + (7*d*x)/2] + 2940*B*Sin[4*c + (7*d*x)/2
] - 945*A*Sin[4*c + (9*d*x)/2] + 420*B*Sin[4*c + (9*d*x)/2] - 945*A*Sin[5*c + (9*d*x)/2] + 420*B*Sin[5*c + (9*
d*x)/2] + 105*A*Sin[5*c + (11*d*x)/2] + 105*A*Sin[6*c + (11*d*x)/2]))/(6720*a^4*d*(1 + Cos[c + d*x])^4)

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Maple [A]  time = 0.1, size = 332, normalized size = 1.5 \begin{align*}{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{9\,A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{7\,B}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{13\,A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{23\,B}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{111\,A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{49\,B}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-9\,{\frac{A \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{B \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-7\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+21\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}}-8\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(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*tan(1/2*d*x+1/2*c)^7*B-9/40/d/a^4*tan(1/2*d*x+1/2*c)^5*A+7/40/d/a
^4*tan(1/2*d*x+1/2*c)^5*B+13/8/d/a^4*A*tan(1/2*d*x+1/2*c)^3-23/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*A*t
an(1/2*d*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)-9/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)^3+2/
d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)^3-7/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c
)+2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+21/d/a^4*A*arctan(tan(1/2*d*x+1/2*c))-8/d/a^4*arctan
(tan(1/2*d*x+1/2*c))*B

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Maxima [A]  time = 1.52523, size = 491, normalized size = 2.2 \begin{align*} -\frac{3 \, A{\left (\frac{280 \,{\left (\frac{7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac{2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{63 \, \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}}}{a^{4}} - \frac{5880 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - B{\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 )}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(3*A*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) - 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 5880*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) - B*(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))/d

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Fricas [A]  time = 0.49927, size = 645, normalized size = 2.89 \begin{align*} \frac{105 \,{\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{4} + 420 \,{\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{3} + 630 \,{\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{2} + 420 \,{\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right ) + 105 \,{\left (21 \, A - 8 \, B\right )} d x +{\left (105 \, A \cos \left (d x + c\right )^{5} - 210 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (1509 \, A - 592 \, B\right )} \cos \left (d x + c\right )^{3} - 4 \,{\left (3411 \, A - 1318 \, B\right )} \cos \left (d x + c\right )^{2} -{\left (11619 \, A - 4472 \, B\right )} \cos \left (d x + c\right ) - 3456 \, A + 1328 \, B\right )} \sin \left (d x + c\right )}{210 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(21*A - 8*B)*d*x*cos(d*x + c)^4 + 420*(21*A - 8*B)*d*x*cos(d*x + c)^3 + 630*(21*A - 8*B)*d*x*cos(d*
x + c)^2 + 420*(21*A - 8*B)*d*x*cos(d*x + c) + 105*(21*A - 8*B)*d*x + (105*A*cos(d*x + c)^5 - 210*(2*A - B)*co
s(d*x + c)^4 - 4*(1509*A - 592*B)*cos(d*x + c)^3 - 4*(3411*A - 1318*B)*cos(d*x + c)^2 - (11619*A - 4472*B)*cos
(d*x + c) - 3456*A + 1328*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.47962, size = 315, normalized size = 1.41 \begin{align*} \frac{\frac{420 \,{\left (d x + c\right )}{\left (21 \, A - 8 \, B\right )}}{a^{4}} - \frac{840 \,{\left (9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} 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} - 189 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 147 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 11655 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5145 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(420*(d*x + c)*(21*A - 8*B)/a^4 - 840*(9*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + 7*A*tan
(1/2*d*x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*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 - 189*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 147*B*a^24*tan(1/2*d*x + 1/2
*c)^5 + 1365*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 805*B*a^24*tan(1/2*d*x + 1/2*c)^3 - 11655*A*a^24*tan(1/2*d*x + 1/
2*c) + 5145*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d