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

Optimal. Leaf size=131 \[ \frac {(4 A-3 B) \tan ^3(c+d x)}{3 a d}+\frac {(4 A-3 B) \tan (c+d x)}{a d}-\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {3 (A-B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {(4 A-3 B) \tan ^3(c+d x)}{3 a d}+\frac {(4 A-3 B) \tan (c+d x)}{a d}-\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {3 (A-B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(A - B)*ArcTanh[Sin[c + d*x]])/(2*a*d) + ((4*A - 3*B)*Tan[c + d*x])/(a*d) - (3*(A - B)*Sec[c + d*x]*Tan[c
+ d*x])/(2*a*d) - ((A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((4*A - 3*B)*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 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 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)} \, dx &=-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int (a (4 A-3 B)-3 a (A-B) \cos (c+d x)) \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(4 A-3 B) \int \sec ^4(c+d x) \, dx}{a}-\frac {(3 (A-B)) \int \sec ^3(c+d x) \, dx}{a}\\ &=-\frac {3 (A-B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 (A-B)) \int \sec (c+d x) \, dx}{2 a}-\frac {(4 A-3 B) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(4 A-3 B) \tan (c+d x)}{a d}-\frac {3 (A-B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(4 A-3 B) \tan ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B]  time = 4.61, size = 490, normalized size = 3.74 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (6 (A+B) \sin \left (\frac {d x}{2}\right )+3 (13 A-9 B) \sin \left (\frac {3 d x}{2}\right )-24 A \sin \left (c-\frac {d x}{2}\right )-6 A \sin \left (c+\frac {d x}{2}\right )-24 A \sin \left (2 c+\frac {d x}{2}\right )+21 A \sin \left (c+\frac {3 d x}{2}\right )+9 A \sin \left (2 c+\frac {3 d x}{2}\right )-9 A \sin \left (3 c+\frac {3 d x}{2}\right )+7 A \sin \left (c+\frac {5 d x}{2}\right )+A \sin \left (2 c+\frac {5 d x}{2}\right )-3 A \sin \left (3 c+\frac {5 d x}{2}\right )-9 A \sin \left (4 c+\frac {5 d x}{2}\right )+16 A \sin \left (2 c+\frac {7 d x}{2}\right )+10 A \sin \left (3 c+\frac {7 d x}{2}\right )+6 A \sin \left (4 c+\frac {7 d x}{2}\right )+12 B \sin \left (c-\frac {d x}{2}\right )+6 B \sin \left (c+\frac {d x}{2}\right )+24 B \sin \left (2 c+\frac {d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )+9 B \sin \left (3 c+\frac {3 d x}{2}\right )-3 B \sin \left (c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )+9 B \sin \left (4 c+\frac {5 d x}{2}\right )-12 B \sin \left (2 c+\frac {7 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )-6 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )+144 (A-B) \cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{48 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(144*(A - B)*Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(6*(A + B)*Sin[(d*x)/2] + 3*(13*A - 9*B)*Sin[(3*d*x)/
2] - 24*A*Sin[c - (d*x)/2] + 12*B*Sin[c - (d*x)/2] - 6*A*Sin[c + (d*x)/2] + 6*B*Sin[c + (d*x)/2] - 24*A*Sin[2*
c + (d*x)/2] + 24*B*Sin[2*c + (d*x)/2] + 21*A*Sin[c + (3*d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 9*A*Sin[2*c + (3*d
*x)/2] - 9*B*Sin[2*c + (3*d*x)/2] - 9*A*Sin[3*c + (3*d*x)/2] + 9*B*Sin[3*c + (3*d*x)/2] + 7*A*Sin[c + (5*d*x)/
2] - 3*B*Sin[c + (5*d*x)/2] + A*Sin[2*c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] - 3*A*Sin[3*c + (5*d*x)/2] + 3
*B*Sin[3*c + (5*d*x)/2] - 9*A*Sin[4*c + (5*d*x)/2] + 9*B*Sin[4*c + (5*d*x)/2] + 16*A*Sin[2*c + (7*d*x)/2] - 12
*B*Sin[2*c + (7*d*x)/2] + 10*A*Sin[3*c + (7*d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] + 6*A*Sin[4*c + (7*d*x)/2] - 6*
B*Sin[4*c + (7*d*x)/2])))/(48*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.55, size = 168, normalized size = 1.28 \[ -\frac {9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (4 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(9*((A - B)*cos(d*x + c)^4 + (A - B)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*((A - B)*cos(d*x + c)^4 +
(A - B)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(4*A - 3*B)*cos(d*x + c)^3 + (7*A - 3*B)*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 = 0.40, size = 182, normalized size = 1.39 \[ -\frac {\frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.17, size = 340, normalized size = 2.60 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 a d}-\frac {5 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 a d}-\frac {5 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*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)-1/3/a/d*A/(tan(1/2*d*x+1/2*c)-1)^3+1/2/a/d/(tan(1/2*d*x+
1/2*c)-1)^2*B-1/a/d*A/(tan(1/2*d*x+1/2*c)-1)^2+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-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-1/3/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*l
n(tan(1/2*d*x+1/2*c)+1)*B-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

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maxima [B]  time = 0.47, size = 368, normalized size = 2.81 \[ \frac {A {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

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)),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))))/d

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mupad [B]  time = 0.64, size = 152, normalized size = 1.16 \[ \frac {\left (5\,A-3\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,B-\frac {16\,A}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)),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))/a

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