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

Optimal. Leaf size=179 \[ \frac{4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d}+\frac{4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(10 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-((10*A - 7*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + (4*(3*A - 2*B)*Tan[c + d*x])/(a^2*d) - ((10*A - 7*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) - ((10*A - 7*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A
- B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(3*A - 2*B)*Tan[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.364972, antiderivative size = 179, normalized size of antiderivative = 1., 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 = {2978, 2748, 3767, 3768, 3770} \[ \frac{4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d}+\frac{4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(10 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((10*A - 7*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + (4*(3*A - 2*B)*Tan[c + d*x])/(a^2*d) - ((10*A - 7*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) - ((10*A - 7*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A
- B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(3*A - 2*B)*Tan[c + d*x]^3)/(3*a^2*d)

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 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{(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(3 a (2 A-B)-4 a (A-B) \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (12 a^2 (3 A-2 B)-3 a^2 (10 A-7 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4}\\ &=-\frac{(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(10 A-7 B) \int \sec ^3(c+d x) \, dx}{a^2}+\frac{(4 (3 A-2 B)) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac{(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(10 A-7 B) \int \sec (c+d x) \, dx}{2 a^2}-\frac{(4 (3 A-2 B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac{(10 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac{(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 4.84746, size = 609, normalized size = 3.4 \[ \frac{192 (10 A-7 B) \cos ^4\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 ^3(c+d x) \left ((45 B-6 A) \sin \left (\frac{d x}{2}\right )+(310 A-201 B) \sin \left (\frac{3 d x}{2}\right )-306 A \sin \left (c-\frac{d x}{2}\right )+42 A \sin \left (c+\frac{d x}{2}\right )-270 A \sin \left (2 c+\frac{d x}{2}\right )+50 A \sin \left (c+\frac{3 d x}{2}\right )+90 A \sin \left (2 c+\frac{3 d x}{2}\right )-170 A \sin \left (3 c+\frac{3 d x}{2}\right )+198 A \sin \left (c+\frac{5 d x}{2}\right )+42 A \sin \left (2 c+\frac{5 d x}{2}\right )+66 A \sin \left (3 c+\frac{5 d x}{2}\right )-90 A \sin \left (4 c+\frac{5 d x}{2}\right )+114 A \sin \left (2 c+\frac{7 d x}{2}\right )+36 A \sin \left (3 c+\frac{7 d x}{2}\right )+48 A \sin \left (4 c+\frac{7 d x}{2}\right )-30 A \sin \left (5 c+\frac{7 d x}{2}\right )+48 A \sin \left (3 c+\frac{9 d x}{2}\right )+22 A \sin \left (4 c+\frac{9 d x}{2}\right )+26 A \sin \left (5 c+\frac{9 d x}{2}\right )+195 B \sin \left (c-\frac{d x}{2}\right )-51 B \sin \left (c+\frac{d x}{2}\right )+189 B \sin \left (2 c+\frac{d x}{2}\right )-B \sin \left (c+\frac{3 d x}{2}\right )-81 B \sin \left (2 c+\frac{3 d x}{2}\right )+119 B \sin \left (3 c+\frac{3 d x}{2}\right )-129 B \sin \left (c+\frac{5 d x}{2}\right )-9 B \sin \left (2 c+\frac{5 d x}{2}\right )-57 B \sin \left (3 c+\frac{5 d x}{2}\right )+63 B \sin \left (4 c+\frac{5 d x}{2}\right )-75 B \sin \left (2 c+\frac{7 d x}{2}\right )-15 B \sin \left (3 c+\frac{7 d x}{2}\right )-39 B \sin \left (4 c+\frac{7 d x}{2}\right )+21 B \sin \left (5 c+\frac{7 d x}{2}\right )-32 B \sin \left (3 c+\frac{9 d x}{2}\right )-12 B \sin \left (4 c+\frac{9 d x}{2}\right )-20 B \sin \left (5 c+\frac{9 d x}{2}\right )\right )}{96 a^2 d (\cos (c+d x)+1)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(192*(10*A - 7*B)*Cos[(c + d*x)/2]^4*(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]^3*((-6*A + 45*B)*Sin[(d*x)/2] + (310*A - 201*B)*S
in[(3*d*x)/2] - 306*A*Sin[c - (d*x)/2] + 195*B*Sin[c - (d*x)/2] + 42*A*Sin[c + (d*x)/2] - 51*B*Sin[c + (d*x)/2
] - 270*A*Sin[2*c + (d*x)/2] + 189*B*Sin[2*c + (d*x)/2] + 50*A*Sin[c + (3*d*x)/2] - B*Sin[c + (3*d*x)/2] + 90*
A*Sin[2*c + (3*d*x)/2] - 81*B*Sin[2*c + (3*d*x)/2] - 170*A*Sin[3*c + (3*d*x)/2] + 119*B*Sin[3*c + (3*d*x)/2] +
198*A*Sin[c + (5*d*x)/2] - 129*B*Sin[c + (5*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] - 9*B*Sin[2*c + (5*d*x)/2] +
66*A*Sin[3*c + (5*d*x)/2] - 57*B*Sin[3*c + (5*d*x)/2] - 90*A*Sin[4*c + (5*d*x)/2] + 63*B*Sin[4*c + (5*d*x)/2]
+ 114*A*Sin[2*c + (7*d*x)/2] - 75*B*Sin[2*c + (7*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] - 15*B*Sin[3*c + (7*d*x)/
2] + 48*A*Sin[4*c + (7*d*x)/2] - 39*B*Sin[4*c + (7*d*x)/2] - 30*A*Sin[5*c + (7*d*x)/2] + 21*B*Sin[5*c + (7*d*x
)/2] + 48*A*Sin[3*c + (9*d*x)/2] - 32*B*Sin[3*c + (9*d*x)/2] + 22*A*Sin[4*c + (9*d*x)/2] - 12*B*Sin[4*c + (9*d
*x)/2] + 26*A*Sin[5*c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2]))/(96*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [B]  time = 0.125, size = 382, normalized size = 2.1 \begin{align*}{\frac{A}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{9\,A}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,B}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,A}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{B}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+5\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{{a}^{2}d}}-{\frac{7\,B}{2\,{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-5\,{\frac{A}{{a}^{2}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+{\frac{5\,B}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{3\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-5\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{{a}^{2}d}}+{\frac{7\,B}{2\,{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,A}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{B}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-5\,{\frac{A}{{a}^{2}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{5\,B}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{3\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*A*tan(1/2*d*x+1/2*c)-7/2/d/a^2*B*t
an(1/2*d*x+1/2*c)-3/2/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2*B+5/d/a^2*A*ln(tan(1
/2*d*x+1/2*c)-1)-7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B-5/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)+5/2/d/a^2/(tan(1/2*d*x+
1/2*c)-1)*B-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^3-5/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)+7/2/d/a^2*ln(tan(1/2*d*x+1
/2*c)+1)*B+3/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2-1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*B-5/d/a^2*A/(tan(1/2*d*x+1/
2*c)+1)+5/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^3

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Maxima [B]  time = 1.04846, size = 574, normalized size = 3.21 \begin{align*} \frac{A{\left (\frac{4 \,{\left (\frac{9 \, \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{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) +
sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/a^2))/d

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Fricas [A]  time = 1.47871, size = 617, normalized size = 3.45 \begin{align*} -\frac{3 \,{\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (66 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} -{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (10*A - 7*B)*cos(d*x + c)^3)*log(sin(d
*x + c) + 1) - 3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (10*A - 7*B)*cos(d*x + c)^3)*l
og(-sin(d*x + c) + 1) - 2*(16*(3*A - 2*B)*cos(d*x + c)^4 + (66*A - 43*B)*cos(d*x + c)^3 + 6*(2*A - B)*cos(d*x
+ c)^2 - (2*A - 3*B)*cos(d*x + c) + 2*A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*
cos(d*x + c)^3)

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

[Out]

Timed out

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Giac [A]  time = 1.20286, size = 305, normalized size = 1.7 \begin{align*} -\frac{\frac{3 \,{\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (30 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, 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 )}^{3} a^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(10*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(10*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1)
)/a^2 + 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 - 40*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(
1/2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2
) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*ta
n(1/2*d*x + 1/2*c))/a^6)/d