3.63 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=192 \[ -\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {(13 A+2 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(76 A+11 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 A+C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

1/2*(13*A+2*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(76*A+11*C)*tan(d*x+c)/a^3/d+1/2*(13*A+2*C)*sec(d*x+c)*tan(d*x+c
)/a^3/d-1/5*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(11*A+C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(
d*x+c))^2-1/15*(76*A+11*C)*sec(d*x+c)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.51, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {(13 A+2 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(76 A+11 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 A+C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]

[Out]

((13*A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(76*A + 11*C)*Tan[c + d*x])/(15*a^3*d) + ((13*A + 2*C)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A + C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((11*A
+ C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((76*A + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(1
5*d*(a^3 + a^3*Cos[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

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 {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (7 A+2 C)-a (4 A-C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (43 A+8 C)-3 a^2 (11 A+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (15 a^3 (13 A+2 C)-2 a^3 (76 A+11 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 A+2 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac {(2 (76 A+11 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 A+2 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(2 (76 A+11 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 4.76, size = 597, normalized size = 3.11 \[ -\frac {1920 (13 A+2 C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-4329 A \sin \left (c-\frac {d x}{2}\right )+1989 A \sin \left (c+\frac {d x}{2}\right )-3575 A \sin \left (2 c+\frac {d x}{2}\right )-475 A \sin \left (c+\frac {3 d x}{2}\right )+2005 A \sin \left (2 c+\frac {3 d x}{2}\right )-2275 A \sin \left (3 c+\frac {3 d x}{2}\right )+2673 A \sin \left (c+\frac {5 d x}{2}\right )+105 A \sin \left (2 c+\frac {5 d x}{2}\right )+1593 A \sin \left (3 c+\frac {5 d x}{2}\right )-975 A \sin \left (4 c+\frac {5 d x}{2}\right )+1325 A \sin \left (2 c+\frac {7 d x}{2}\right )+255 A \sin \left (3 c+\frac {7 d x}{2}\right )+875 A \sin \left (4 c+\frac {7 d x}{2}\right )-195 A \sin \left (5 c+\frac {7 d x}{2}\right )+304 A \sin \left (3 c+\frac {9 d x}{2}\right )+90 A \sin \left (4 c+\frac {9 d x}{2}\right )+214 A \sin \left (5 c+\frac {9 d x}{2}\right )-5 (247 A+98 C) \sin \left (\frac {d x}{2}\right )+5 (761 A+106 C) \sin \left (\frac {3 d x}{2}\right )-654 C \sin \left (c-\frac {d x}{2}\right )+654 C \sin \left (c+\frac {d x}{2}\right )-490 C \sin \left (2 c+\frac {d x}{2}\right )-350 C \sin \left (c+\frac {3 d x}{2}\right )+530 C \sin \left (2 c+\frac {3 d x}{2}\right )-350 C \sin \left (3 c+\frac {3 d x}{2}\right )+378 C \sin \left (c+\frac {5 d x}{2}\right )-150 C \sin \left (2 c+\frac {5 d x}{2}\right )+378 C \sin \left (3 c+\frac {5 d x}{2}\right )-150 C \sin \left (4 c+\frac {5 d x}{2}\right )+190 C \sin \left (2 c+\frac {7 d x}{2}\right )-30 C \sin \left (3 c+\frac {7 d x}{2}\right )+190 C \sin \left (4 c+\frac {7 d x}{2}\right )-30 C \sin \left (5 c+\frac {7 d x}{2}\right )+44 C \sin \left (3 c+\frac {9 d x}{2}\right )+44 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/480*(1920*(13*A + 2*C)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-5*(247*A + 98*C)*Sin[(d*x)/2] + 5*(76
1*A + 106*C)*Sin[(3*d*x)/2] - 4329*A*Sin[c - (d*x)/2] - 654*C*Sin[c - (d*x)/2] + 1989*A*Sin[c + (d*x)/2] + 654
*C*Sin[c + (d*x)/2] - 3575*A*Sin[2*c + (d*x)/2] - 490*C*Sin[2*c + (d*x)/2] - 475*A*Sin[c + (3*d*x)/2] - 350*C*
Sin[c + (3*d*x)/2] + 2005*A*Sin[2*c + (3*d*x)/2] + 530*C*Sin[2*c + (3*d*x)/2] - 2275*A*Sin[3*c + (3*d*x)/2] -
350*C*Sin[3*c + (3*d*x)/2] + 2673*A*Sin[c + (5*d*x)/2] + 378*C*Sin[c + (5*d*x)/2] + 105*A*Sin[2*c + (5*d*x)/2]
- 150*C*Sin[2*c + (5*d*x)/2] + 1593*A*Sin[3*c + (5*d*x)/2] + 378*C*Sin[3*c + (5*d*x)/2] - 975*A*Sin[4*c + (5*
d*x)/2] - 150*C*Sin[4*c + (5*d*x)/2] + 1325*A*Sin[2*c + (7*d*x)/2] + 190*C*Sin[2*c + (7*d*x)/2] + 255*A*Sin[3*
c + (7*d*x)/2] - 30*C*Sin[3*c + (7*d*x)/2] + 875*A*Sin[4*c + (7*d*x)/2] + 190*C*Sin[4*c + (7*d*x)/2] - 195*A*S
in[5*c + (7*d*x)/2] - 30*C*Sin[5*c + (7*d*x)/2] + 304*A*Sin[3*c + (9*d*x)/2] + 44*C*Sin[3*c + (9*d*x)/2] + 90*
A*Sin[4*c + (9*d*x)/2] + 214*A*Sin[5*c + (9*d*x)/2] + 44*C*Sin[5*c + (9*d*x)/2]))/(a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.63, size = 289, normalized size = 1.51 \[ \frac {15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (76 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A + 34 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A + 64 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, A \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4 + 3*(13*A + 2*C)*cos(d*x + c)^3 + (13*A
+ 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4
+ 3*(13*A + 2*C)*cos(d*x + c)^3 + (13*A + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(76*A + 11*C)*co
s(d*x + c)^4 + 3*(239*A + 34*C)*cos(d*x + c)^3 + (479*A + 64*C)*cos(d*x + c)^2 + 45*A*cos(d*x + c) - 15*A)*sin
(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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giac [A]  time = 0.54, size = 207, normalized size = 1.08 \[ \frac {\frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(7*A*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3
*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*C*a^1
2*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 0.23, size = 289, normalized size = 1.51 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {13 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{3}}+\frac {A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7 A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {13 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{3}}-\frac {A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7 A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(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*C*tan(1/2*d*x+1/2*c)^5-2/3/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/3/d/a^
3*C*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*A*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)-13/2/d/a^3*A*ln(tan(1/
2*d*x+1/2*c)-1)-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^2+7/2/d/a^3*A/(tan(1/2*d
*x+1/2*c)-1)+13/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^3*A/(tan(1/2*d*x
+1/2*c)+1)^2+7/2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)

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maxima [A]  time = 0.35, size = 330, normalized size = 1.72 \[ -\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(A*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*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 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + C*((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 - 60*log(sin(
d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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mupad [B]  time = 0.91, size = 195, normalized size = 1.02 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {13\,A}{2}+C\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (5\,A+C\right )}{4\,a^3}+\frac {10\,A-2\,C}{4\,a^3}\right )}{d}-\frac {5\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {5\,A+C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^3),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*((13*A)/2 + C))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^3) + (3*(5*A + C)
)/(4*a^3) + (10*A - 2*C)/(4*a^3)))/d - (5*A*tan(c/2 + (d*x)/2) - 7*A*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (
d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (5*A + C)/(12*a^3)))
/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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