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

Optimal. Leaf size=152 \[ -\frac {2 (8 B-5 C) \tan (c+d x)}{3 a^2 d}+\frac {(7 B-4 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(7 B-4 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(8 B-5 C) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

1/2*(7*B-4*C)*arctanh(sin(d*x+c))/a^2/d-2/3*(8*B-5*C)*tan(d*x+c)/a^2/d+1/2*(7*B-4*C)*sec(d*x+c)*tan(d*x+c)/a^2
/d-1/3*(8*B-5*C)*sec(d*x+c)*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(B-C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))
^2

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Rubi [A]  time = 0.40, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {2 (8 B-5 C) \tan (c+d x)}{3 a^2 d}+\frac {(7 B-4 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(7 B-4 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(8 B-5 C) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

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])^2,x]

[Out]

((7*B - 4*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (2*(8*B - 5*C)*Tan[c + d*x])/(3*a^2*d) + ((7*B - 4*C)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) - ((8*B - 5*C)*Sec[c + d*x]*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((B - C)
*Sec[c + d*x]*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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

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Mathematica [B]  time = 3.19, size = 496, normalized size = 3.26 \[ -\frac {96 (7 B-4 C) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-126 B \sin \left (c-\frac {d x}{2}\right )+42 B \sin \left (c+\frac {d x}{2}\right )-98 B \sin \left (2 c+\frac {d x}{2}\right )-3 B \sin \left (c+\frac {3 d x}{2}\right )+37 B \sin \left (2 c+\frac {3 d x}{2}\right )-63 B \sin \left (3 c+\frac {3 d x}{2}\right )+75 B \sin \left (c+\frac {5 d x}{2}\right )+15 B \sin \left (2 c+\frac {5 d x}{2}\right )+39 B \sin \left (3 c+\frac {5 d x}{2}\right )-21 B \sin \left (4 c+\frac {5 d x}{2}\right )+32 B \sin \left (2 c+\frac {7 d x}{2}\right )+12 B \sin \left (3 c+\frac {7 d x}{2}\right )+20 B \sin \left (4 c+\frac {7 d x}{2}\right )-14 (B-C) \sin \left (\frac {d x}{2}\right )+(97 B-64 C) \sin \left (\frac {3 d x}{2}\right )+84 C \sin \left (c-\frac {d x}{2}\right )-42 C \sin \left (c+\frac {d x}{2}\right )+56 C \sin \left (2 c+\frac {d x}{2}\right )+6 C \sin \left (c+\frac {3 d x}{2}\right )-34 C \sin \left (2 c+\frac {3 d x}{2}\right )+36 C \sin \left (3 c+\frac {3 d x}{2}\right )-48 C \sin \left (c+\frac {5 d x}{2}\right )-6 C \sin \left (2 c+\frac {5 d x}{2}\right )-30 C \sin \left (3 c+\frac {5 d x}{2}\right )+12 C \sin \left (4 c+\frac {5 d x}{2}\right )-20 C \sin \left (2 c+\frac {7 d x}{2}\right )-6 C \sin \left (3 c+\frac {7 d x}{2}\right )-14 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(96*(7*B - 4*C)*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-14*(B - C)*Sin[(d*x)/2] + (97*B - 64*C)*S
in[(3*d*x)/2] - 126*B*Sin[c - (d*x)/2] + 84*C*Sin[c - (d*x)/2] + 42*B*Sin[c + (d*x)/2] - 42*C*Sin[c + (d*x)/2]
- 98*B*Sin[2*c + (d*x)/2] + 56*C*Sin[2*c + (d*x)/2] - 3*B*Sin[c + (3*d*x)/2] + 6*C*Sin[c + (3*d*x)/2] + 37*B*
Sin[2*c + (3*d*x)/2] - 34*C*Sin[2*c + (3*d*x)/2] - 63*B*Sin[3*c + (3*d*x)/2] + 36*C*Sin[3*c + (3*d*x)/2] + 75*
B*Sin[c + (5*d*x)/2] - 48*C*Sin[c + (5*d*x)/2] + 15*B*Sin[2*c + (5*d*x)/2] - 6*C*Sin[2*c + (5*d*x)/2] + 39*B*S
in[3*c + (5*d*x)/2] - 30*C*Sin[3*c + (5*d*x)/2] - 21*B*Sin[4*c + (5*d*x)/2] + 12*C*Sin[4*c + (5*d*x)/2] + 32*B
*Sin[2*c + (7*d*x)/2] - 20*C*Sin[2*c + (7*d*x)/2] + 12*B*Sin[3*c + (7*d*x)/2] - 6*C*Sin[3*c + (7*d*x)/2] + 20*
B*Sin[4*c + (7*d*x)/2] - 14*C*Sin[4*c + (7*d*x)/2]))/(a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.49, size = 228, normalized size = 1.50 \[ \frac {3 \, {\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (8 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (43 \, B - 28 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (B - C\right )} \cos \left (d x + c\right ) - 3 \, B\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/12*(3*((7*B - 4*C)*cos(d*x + c)^4 + 2*(7*B - 4*C)*cos(d*x + c)^3 + (7*B - 4*C)*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) - 3*((7*B - 4*C)*cos(d*x + c)^4 + 2*(7*B - 4*C)*cos(d*x + c)^3 + (7*B - 4*C)*cos(d*x + c)^2)*log(-sin
(d*x + c) + 1) - 2*(4*(8*B - 5*C)*cos(d*x + c)^3 + (43*B - 28*C)*cos(d*x + c)^2 + 6*(B - C)*cos(d*x + c) - 3*B
)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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giac [A]  time = 0.39, size = 198, normalized size = 1.30 \[ \frac {\frac {3 \, {\left (7 \, B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (7 \, B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/6*(3*(7*B - 4*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(7*B - 4*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^2 + 6*(5*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 - 3*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^2) - (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 +
21*B*a^4*tan(1/2*d*x + 1/2*c) - 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.25, size = 294, normalized size = 1.93 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 d \,a^{2}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 d \,a^{2}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}}-\frac {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

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))^2,x)

[Out]

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

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maxima [B]  time = 0.42, size = 336, normalized size = 2.21 \[ -\frac {B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

-1/6*(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) + s
in(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) - C*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 1.17, size = 165, normalized size = 1.09 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,B-2\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,B-2\,C\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^2}+\frac {4\,B-2\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,B-4\,C\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*(5*B - 2*C) - tan(c/2 + (d*x)/2)*(3*B - 2*C))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c
/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^2) + (4*B - 2*C)/(2*a^2)))/d - (tan(c/2 + (d*x
)/2)^3*(B - C))/(6*a^2*d) + (atanh(tan(c/2 + (d*x)/2))*(7*B - 4*C))/(a^2*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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**2,x)

[Out]

Timed out

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