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

Optimal. Leaf size=165 \[ -\frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}-\frac{(3 A+4 C) \tan (c+d x)}{a d}+\frac{3 (4 A+5 C) \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{(A+C) \tan (c+d x) \sec ^4(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(4 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 (4 A+5 C) \tan (c+d x) \sec (c+d x)}{8 a d} \]

[Out]

(3*(4*A + 5*C)*ArcTanh[Sin[c + d*x]])/(8*a*d) - ((3*A + 4*C)*Tan[c + d*x])/(a*d) + (3*(4*A + 5*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*a*d) + ((4*A + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d) - ((A + C)*Sec[c + d*x]^4*Tan[c + d
*x])/(d*(a + a*Sec[c + d*x])) - ((3*A + 4*C)*Tan[c + d*x]^3)/(3*a*d)

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Rubi [A]  time = 0.201955, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4085, 3787, 3767, 3768, 3770} \[ -\frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}-\frac{(3 A+4 C) \tan (c+d x)}{a d}+\frac{3 (4 A+5 C) \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{(A+C) \tan (c+d x) \sec ^4(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(4 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 (4 A+5 C) \tan (c+d x) \sec (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*A + 5*C)*ArcTanh[Sin[c + d*x]])/(8*a*d) - ((3*A + 4*C)*Tan[c + d*x])/(a*d) + (3*(4*A + 5*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*a*d) + ((4*A + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d) - ((A + C)*Sec[c + d*x]^4*Tan[c + d
*x])/(d*(a + a*Sec[c + d*x])) - ((3*A + 4*C)*Tan[c + d*x]^3)/(3*a*d)

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)]

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{\int \sec ^4(c+d x) (a (3 A+4 C)-a (4 A+5 C) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{(3 A+4 C) \int \sec ^4(c+d x) \, dx}{a}+\frac{(4 A+5 C) \int \sec ^5(c+d x) \, dx}{a}\\ &=\frac{(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac{(3 (4 A+5 C)) \int \sec ^3(c+d x) \, dx}{4 a}+\frac{(3 A+4 C) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac{(3 A+4 C) \tan (c+d x)}{a d}+\frac{3 (4 A+5 C) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}+\frac{(3 (4 A+5 C)) \int \sec (c+d x) \, dx}{8 a}\\ &=\frac{3 (4 A+5 C) \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{(3 A+4 C) \tan (c+d x)}{a d}+\frac{3 (4 A+5 C) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [B]  time = 6.3334, size = 792, normalized size = 4.8 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \left (204 A \sin \left (c-\frac{d x}{2}\right )-60 A \sin \left (c+\frac{d x}{2}\right )+84 A \sin \left (2 c+\frac{d x}{2}\right )+36 A \sin \left (c+\frac{3 d x}{2}\right )+36 A \sin \left (2 c+\frac{3 d x}{2}\right )+132 A \sin \left (3 c+\frac{3 d x}{2}\right )-156 A \sin \left (c+\frac{5 d x}{2}\right )-60 A \sin \left (2 c+\frac{5 d x}{2}\right )-60 A \sin \left (3 c+\frac{5 d x}{2}\right )+36 A \sin \left (4 c+\frac{5 d x}{2}\right )-12 A \sin \left (2 c+\frac{7 d x}{2}\right )+12 A \sin \left (3 c+\frac{7 d x}{2}\right )+12 A \sin \left (4 c+\frac{7 d x}{2}\right )+36 A \sin \left (5 c+\frac{7 d x}{2}\right )-48 A \sin \left (3 c+\frac{9 d x}{2}\right )-24 A \sin \left (4 c+\frac{9 d x}{2}\right )-24 A \sin \left (5 c+\frac{9 d x}{2}\right )-60 A \sin \left (\frac{d x}{2}\right )-60 A \sin \left (\frac{3 d x}{2}\right )+219 C \sin \left (c-\frac{d x}{2}\right )+21 C \sin \left (c+\frac{d x}{2}\right )+165 C \sin \left (2 c+\frac{d x}{2}\right )+5 C \sin \left (c+\frac{3 d x}{2}\right )+69 C \sin \left (2 c+\frac{3 d x}{2}\right )+165 C \sin \left (3 c+\frac{3 d x}{2}\right )-211 C \sin \left (c+\frac{5 d x}{2}\right )-115 C \sin \left (2 c+\frac{5 d x}{2}\right )-51 C \sin \left (3 c+\frac{5 d x}{2}\right )+45 C \sin \left (4 c+\frac{5 d x}{2}\right )-19 C \sin \left (2 c+\frac{7 d x}{2}\right )+5 C \sin \left (3 c+\frac{7 d x}{2}\right )+21 C \sin \left (4 c+\frac{7 d x}{2}\right )+45 C \sin \left (5 c+\frac{7 d x}{2}\right )-64 C \sin \left (3 c+\frac{9 d x}{2}\right )-40 C \sin \left (4 c+\frac{9 d x}{2}\right )-24 C \sin \left (5 c+\frac{9 d x}{2}\right )-75 C \sin \left (\frac{d x}{2}\right )-91 C \sin \left (\frac{3 d x}{2}\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{192 d (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}-\frac{3 (4 A+5 C) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{2 d (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}+\frac{3 (4 A+5 C) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{2 d (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*A + 5*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + C*Sec[c +
d*x]^2))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (3*(4*A + 5*C)*Cos[c/2 + (d*x)/2]^2*Cos[c
+ d*x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x
])*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2)*(-60*A*Si
n[(d*x)/2] - 75*C*Sin[(d*x)/2] - 60*A*Sin[(3*d*x)/2] - 91*C*Sin[(3*d*x)/2] + 204*A*Sin[c - (d*x)/2] + 219*C*Si
n[c - (d*x)/2] - 60*A*Sin[c + (d*x)/2] + 21*C*Sin[c + (d*x)/2] + 84*A*Sin[2*c + (d*x)/2] + 165*C*Sin[2*c + (d*
x)/2] + 36*A*Sin[c + (3*d*x)/2] + 5*C*Sin[c + (3*d*x)/2] + 36*A*Sin[2*c + (3*d*x)/2] + 69*C*Sin[2*c + (3*d*x)/
2] + 132*A*Sin[3*c + (3*d*x)/2] + 165*C*Sin[3*c + (3*d*x)/2] - 156*A*Sin[c + (5*d*x)/2] - 211*C*Sin[c + (5*d*x
)/2] - 60*A*Sin[2*c + (5*d*x)/2] - 115*C*Sin[2*c + (5*d*x)/2] - 60*A*Sin[3*c + (5*d*x)/2] - 51*C*Sin[3*c + (5*
d*x)/2] + 36*A*Sin[4*c + (5*d*x)/2] + 45*C*Sin[4*c + (5*d*x)/2] - 12*A*Sin[2*c + (7*d*x)/2] - 19*C*Sin[2*c + (
7*d*x)/2] + 12*A*Sin[3*c + (7*d*x)/2] + 5*C*Sin[3*c + (7*d*x)/2] + 12*A*Sin[4*c + (7*d*x)/2] + 21*C*Sin[4*c +
(7*d*x)/2] + 36*A*Sin[5*c + (7*d*x)/2] + 45*C*Sin[5*c + (7*d*x)/2] - 48*A*Sin[3*c + (9*d*x)/2] - 64*C*Sin[3*c
+ (9*d*x)/2] - 24*A*Sin[4*c + (9*d*x)/2] - 40*C*Sin[4*c + (9*d*x)/2] - 24*A*Sin[5*c + (9*d*x)/2] - 24*C*Sin[5*
c + (9*d*x)/2]))/(192*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x]))

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Maple [B]  time = 0.068, size = 386, normalized size = 2.3 \begin{align*} -{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{4\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{5\,C}{6\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{15\,C}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{A}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{15\,C}{8\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,A}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{25\,C}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,A}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{C}{4\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}+{\frac{5\,C}{6\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{15\,C}{8\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{3\,A}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{15\,C}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{A}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{25\,C}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,A}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-1/4/a/d*C/(tan(1/2*d*x+1/2*c)+1)^4+5/6/a/d*C/(tan(1/2*d
*x+1/2*c)+1)^3-15/8/a/d/(tan(1/2*d*x+1/2*c)+1)^2*C-1/2/a/d/(tan(1/2*d*x+1/2*c)+1)^2*A+15/8/a/d*ln(tan(1/2*d*x+
1/2*c)+1)*C+3/2/a/d*ln(tan(1/2*d*x+1/2*c)+1)*A+25/8/a/d/(tan(1/2*d*x+1/2*c)+1)*C+3/2/a/d/(tan(1/2*d*x+1/2*c)+1
)*A+1/4/a/d*C/(tan(1/2*d*x+1/2*c)-1)^4+5/6/a/d*C/(tan(1/2*d*x+1/2*c)-1)^3-15/8/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C-
3/2/a/d*ln(tan(1/2*d*x+1/2*c)-1)*A+15/8/a/d/(tan(1/2*d*x+1/2*c)-1)^2*C+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)^2*A+25/8
/a/d/(tan(1/2*d*x+1/2*c)-1)*C+3/2/a/d/(tan(1/2*d*x+1/2*c)-1)*A

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Maxima [B]  time = 0.956555, size = 551, normalized size = 3.34 \begin{align*} -\frac{C{\left (\frac{2 \,{\left (\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a - \frac{4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{45 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{45 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{24 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 12 \, A{\left (\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(C*(2*(21*sin(d*x + c)/(cos(d*x + c) + 1) - 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8) - 45*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 45*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a
+ 24*sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 12*A*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*log(
sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos
(d*x + c) + 1))))/d

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Fricas [A]  time = 0.517481, size = 477, normalized size = 2.89 \begin{align*} \frac{9 \,{\left ({\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} +{\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \,{\left ({\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} +{\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (12 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (12 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) - 6 \, C\right )} \sin \left (d x + c\right )}{48 \,{\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(9*((4*A + 5*C)*cos(d*x + c)^5 + (4*A + 5*C)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 9*((4*A + 5*C)*cos(d
*x + c)^5 + (4*A + 5*C)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(16*(3*A + 4*C)*cos(d*x + c)^4 + (12*A + 19
*C)*cos(d*x + c)^3 - (12*A + 13*C)*cos(d*x + c)^2 + 2*C*cos(d*x + c) - 6*C)*sin(d*x + c))/(a*d*cos(d*x + c)^5
+ a*d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{6}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c + d*x) + 1), x))/a

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Giac [A]  time = 1.21083, size = 288, normalized size = 1.75 \begin{align*} \frac{\frac{9 \,{\left (4 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{9 \,{\left (4 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{24 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (36 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 75 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 84 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 115 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 60 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 109 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, C \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 )}^{4} a}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(4*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(4*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
- 24*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(36*A*tan(1/2*d*x + 1/2*c)^7 + 75*C*tan(1/2*d*x +
1/2*c)^7 - 84*A*tan(1/2*d*x + 1/2*c)^5 - 115*C*tan(1/2*d*x + 1/2*c)^5 + 60*A*tan(1/2*d*x + 1/2*c)^3 + 109*C*t
an(1/2*d*x + 1/2*c)^3 - 12*A*tan(1/2*d*x + 1/2*c) - 21*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^4
*a))/d