3.276 \(\int \frac{(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=196 \[ -\frac{8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{(13 B-6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{4 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(11 B-6 C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

((13*B - 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (8*(19*B - 9*C)*Tan[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Sec[
c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((B - C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((11*B -
6*C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (4*(19*B - 9*C)*Sec[c + d*x]*Tan[c + d*x])/
(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.565793, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{(13 B-6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{4 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(11 B-6 C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

((13*B - 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (8*(19*B - 9*C)*Tan[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Sec[
c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((B - C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((11*B -
6*C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (4*(19*B - 9*C)*Sec[c + d*x]*Tan[c + d*x])/
(15*d*(a^3 + a^3*Cos[c + d*x]))

Rule 3029

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

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 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 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]

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 8

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

Rubi steps

\begin{align*} \int \frac{\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=\int \frac{(B+C \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (7 B-2 C)-4 a (B-C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 B-6 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 B-18 C)-3 a^2 (11 B-6 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{4 (19 B-9 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 B-6 C)-8 a^3 (19 B-9 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6}\\ &=-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(8 (19 B-9 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac{(13 B-6 C) \int \sec ^3(c+d x) \, dx}{a^3}\\ &=\frac{(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(13 B-6 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac{(8 (19 B-9 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{(13 B-6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 4.72756, size = 610, normalized size = 3.11 \[ -\frac{1920 (13 B-6 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 B \sin \left (c-\frac{d x}{2}\right )+1989 B \sin \left (c+\frac{d x}{2}\right )-3575 B \sin \left (2 c+\frac{d x}{2}\right )-475 B \sin \left (c+\frac{3 d x}{2}\right )+2005 B \sin \left (2 c+\frac{3 d x}{2}\right )-2275 B \sin \left (3 c+\frac{3 d x}{2}\right )+2673 B \sin \left (c+\frac{5 d x}{2}\right )+105 B \sin \left (2 c+\frac{5 d x}{2}\right )+1593 B \sin \left (3 c+\frac{5 d x}{2}\right )-975 B \sin \left (4 c+\frac{5 d x}{2}\right )+1325 B \sin \left (2 c+\frac{7 d x}{2}\right )+255 B \sin \left (3 c+\frac{7 d x}{2}\right )+875 B \sin \left (4 c+\frac{7 d x}{2}\right )-195 B \sin \left (5 c+\frac{7 d x}{2}\right )+304 B \sin \left (3 c+\frac{9 d x}{2}\right )+90 B \sin \left (4 c+\frac{9 d x}{2}\right )+214 B \sin \left (5 c+\frac{9 d x}{2}\right )+(870 C-1235 B) \sin \left (\frac{d x}{2}\right )+5 (761 B-366 C) \sin \left (\frac{3 d x}{2}\right )+2094 C \sin \left (c-\frac{d x}{2}\right )-1314 C \sin \left (c+\frac{d x}{2}\right )+1650 C \sin \left (2 c+\frac{d x}{2}\right )+450 C \sin \left (c+\frac{3 d x}{2}\right )-1230 C \sin \left (2 c+\frac{3 d x}{2}\right )+1050 C \sin \left (3 c+\frac{3 d x}{2}\right )-1278 C \sin \left (c+\frac{5 d x}{2}\right )+90 C \sin \left (2 c+\frac{5 d x}{2}\right )-918 C \sin \left (3 c+\frac{5 d x}{2}\right )+450 C \sin \left (4 c+\frac{5 d x}{2}\right )-630 C \sin \left (2 c+\frac{7 d x}{2}\right )-60 C \sin \left (3 c+\frac{7 d x}{2}\right )-480 C \sin \left (4 c+\frac{7 d x}{2}\right )+90 C \sin \left (5 c+\frac{7 d x}{2}\right )-144 C \sin \left (3 c+\frac{9 d x}{2}\right )-30 C \sin \left (4 c+\frac{9 d x}{2}\right )-114 C \sin \left (5 c+\frac{9 d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

-(1920*(13*B - 6*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*((-1235*B + 870*C)*Sin[(d*x)/2] + 5*(761*B -
366*C)*Sin[(3*d*x)/2] - 4329*B*Sin[c - (d*x)/2] + 2094*C*Sin[c - (d*x)/2] + 1989*B*Sin[c + (d*x)/2] - 1314*C*S
in[c + (d*x)/2] - 3575*B*Sin[2*c + (d*x)/2] + 1650*C*Sin[2*c + (d*x)/2] - 475*B*Sin[c + (3*d*x)/2] + 450*C*Sin
[c + (3*d*x)/2] + 2005*B*Sin[2*c + (3*d*x)/2] - 1230*C*Sin[2*c + (3*d*x)/2] - 2275*B*Sin[3*c + (3*d*x)/2] + 10
50*C*Sin[3*c + (3*d*x)/2] + 2673*B*Sin[c + (5*d*x)/2] - 1278*C*Sin[c + (5*d*x)/2] + 105*B*Sin[2*c + (5*d*x)/2]
+ 90*C*Sin[2*c + (5*d*x)/2] + 1593*B*Sin[3*c + (5*d*x)/2] - 918*C*Sin[3*c + (5*d*x)/2] - 975*B*Sin[4*c + (5*d
*x)/2] + 450*C*Sin[4*c + (5*d*x)/2] + 1325*B*Sin[2*c + (7*d*x)/2] - 630*C*Sin[2*c + (7*d*x)/2] + 255*B*Sin[3*c
+ (7*d*x)/2] - 60*C*Sin[3*c + (7*d*x)/2] + 875*B*Sin[4*c + (7*d*x)/2] - 480*C*Sin[4*c + (7*d*x)/2] - 195*B*Si
n[5*c + (7*d*x)/2] + 90*C*Sin[5*c + (7*d*x)/2] + 304*B*Sin[3*c + (9*d*x)/2] - 144*C*Sin[3*c + (9*d*x)/2] + 90*
B*Sin[4*c + (9*d*x)/2] - 30*C*Sin[4*c + (9*d*x)/2] + 214*B*Sin[5*c + (9*d*x)/2] - 114*C*Sin[5*c + (9*d*x)/2]))
/(480*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.07, size = 334, normalized size = 1.7 \begin{align*} -{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{13\,B}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{3}}}+{\frac{7\,B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7\,B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{13\,B}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{3}}}-{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x)

[Out]

-1/20/d/a^3*B*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*B+1/2/d/a^
3*C*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*B*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)-13/2/d/a^3*B*ln(tan(1
/2*d*x+1/2*c)-1)+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C+7/2/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)-1/d/a^3/(tan(1/2*d*x+1/
2*c)-1)*C+1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)^2+7/2/d/a^3*B/(tan(1/2*d*x+1/2*c)+1)-1/d/a^3/(tan(1/2*d*x+1/2*c)+
1)*C+13/2/d/a^3*B*ln(tan(1/2*d*x+1/2*c)+1)-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)+
1)^2

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Maxima [B]  time = 1.05891, size = 509, normalized size = 2.6 \begin{align*} -\frac{B{\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 )} - 3 \, C{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \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{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(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) - 3*C*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^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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Fricas [A]  time = 1.70459, size = 756, normalized size = 3.86 \begin{align*} \frac{15 \,{\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (19 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) - 15 \, B\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4 + 3*(13*B - 6*C)*cos(d*x + c)^3 + (13*B
- 6*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4
+ 3*(13*B - 6*C)*cos(d*x + c)^3 + (13*B - 6*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(19*B - 9*C)*co
s(d*x + c)^4 + 3*(239*B - 114*C)*cos(d*x + c)^3 + (479*B - 234*C)*cos(d*x + c)^2 + 15*(3*B - 2*C)*cos(d*x + c)
- 15*B)*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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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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.442, size = 315, normalized size = 1.61 \begin{align*} \frac{\frac{30 \,{\left (13 \, B - 6 \, 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 \, B - 6 \, 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 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, 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 )}^{2} a^{3}} - \frac{3 \, B 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 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(7*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 - 5*B*tan(1/2*d*x + 1/2*c) + 2*C*tan(1/2
*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*B*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*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*B*a^12*tan(1/2*d*x + 1/
2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d