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

Optimal. Leaf size=148 \[ \frac {(4 A-3 B+3 C) \tan ^3(c+d x)}{3 a d}+\frac {(4 A-3 B+3 C) \tan (c+d x)}{a d}-\frac {(3 A-3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(3 A-3 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3041, 2748, 3767, 3768, 3770} \[ \frac {(4 A-3 B+3 C) \tan ^3(c+d x)}{3 a d}+\frac {(4 A-3 B+3 C) \tan (c+d x)}{a d}-\frac {(3 A-3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(3 A-3 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int (a (4 A-3 B+3 C)-a (3 A-3 B+2 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A-3 B+2 C) \int \sec ^3(c+d x) \, dx}{a}+\frac {(4 A-3 B+3 C) \int \sec ^4(c+d x) \, dx}{a}\\ &=-\frac {(3 A-3 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A-3 B+2 C) \int \sec (c+d x) \, dx}{2 a}-\frac {(4 A-3 B+3 C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac {(3 A-3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(4 A-3 B+3 C) \tan (c+d x)}{a d}-\frac {(3 A-3 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(4 A-3 B+3 C) \tan ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B]  time = 3.78, size = 351, normalized size = 2.37 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (12 (A-B+C) \tan \left (\frac {1}{2} (c+d x)\right )+\frac {4 (5 A-3 B+3 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 (5 A-3 B+3 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+6 (3 A-3 B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 (3 A-3 B+2 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 A-3 B}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {3 B-2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{6 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*(6*(3*A - 3*B + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*(3*A - 3*B + 2*C)*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]] + (-2*A + 3*B)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*A*Sin[(c + d*x)
/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*(5*A - 3*B + 3*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]) + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (2*A - 3*B)/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2 + (4*(5*A - 3*B + 3*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*(A
- B + C)*Tan[(c + d*x)/2]))/(6*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.48, size = 194, normalized size = 1.31 \[ -\frac {3 \, {\left ({\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (4 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 3 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.51, size = 243, normalized size = 1.64 \[ -\frac {\frac {3 \, {\left (3 \, A - 3 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {3 \, {\left (3 \, A - 3 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(3*A - 3*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(3*A - 3*B + 2*C)*log(abs(tan(1/2*d*x + 1/2
*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(15*A*tan(1/2
*d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 16*A*tan(1/2*d*x + 1/2*c)^3 + 12*B
*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c) +
6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d

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maple [B]  time = 0.25, size = 442, normalized size = 2.99 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{a d}-\frac {5 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{a d}-\frac {5 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 485, normalized size = 3.28 \[ \frac {A {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \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 - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\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 )} - 6 \, C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\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 )}{{\left (a - \frac {a \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 {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(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*si
n(d*x + c)/(a*(cos(d*x + c) + 1))) - 6*C*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*
x + c)/(a*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 1.77, size = 187, normalized size = 1.26 \[ \frac {\left (5\,A-3\,B+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,B-\frac {16\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}-\frac {2\,\mathrm {atanh}\left (\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2}-\frac {3\,B}{2}+C\right )}{3\,A-3\,B+2\,C}\right )\,\left (\frac {3\,A}{2}-\frac {3\,B}{2}+C\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*A - B + 2*C) + tan(c/2 + (d*x)/2)^5*(5*A - 3*B + 2*C) - tan(c/2 + (d*x)/2)^3*((16*A)/3
- 4*B + 4*C))/(d*(a - 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 - a*tan(c/2 + (d*x)/2)^6)) + (tan(c/
2 + (d*x)/2)*(A - B + C))/(a*d) - (2*atanh((2*tan(c/2 + (d*x)/2)*((3*A)/2 - (3*B)/2 + C))/(3*A - 3*B + 2*C))*(
(3*A)/2 - (3*B)/2 + C))/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)**4/(cos(c + d*x) + 1
), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**4/(cos(c + d*x) + 1), x))/a

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