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

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

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {2 (8 A-5 B) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(7 A-4 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A - 4*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (2*(8*A - 5*B)*Tan[c + d*x])/(3*a^2*d) + ((7*A - 4*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) - ((8*A - 5*B)*Sec[c + d*x]*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B)
*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 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 ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(a (5 A-2 B)-3 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (3 a^2 (7 A-4 B)-2 a^2 (8 A-5 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a^4}\\ &=-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(2 (8 A-5 B)) \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac {(7 A-4 B) \int \sec ^3(c+d x) \, dx}{a^2}\\ &=\frac {(7 A-4 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(7 A-4 B) \int \sec (c+d x) \, dx}{2 a^2}+\frac {(2 (8 A-5 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=\frac {(7 A-4 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {2 (8 A-5 B) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \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 = 3.43, size = 496, normalized size = 3.26 \[ -\frac {96 (7 A-4 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-14 (A-B) \sin \left (\frac {d x}{2}\right )+(97 A-64 B) \sin \left (\frac {3 d x}{2}\right )-126 A \sin \left (c-\frac {d x}{2}\right )+42 A \sin \left (c+\frac {d x}{2}\right )-98 A \sin \left (2 c+\frac {d x}{2}\right )-3 A \sin \left (c+\frac {3 d x}{2}\right )+37 A \sin \left (2 c+\frac {3 d x}{2}\right )-63 A \sin \left (3 c+\frac {3 d x}{2}\right )+75 A \sin \left (c+\frac {5 d x}{2}\right )+15 A \sin \left (2 c+\frac {5 d x}{2}\right )+39 A \sin \left (3 c+\frac {5 d x}{2}\right )-21 A \sin \left (4 c+\frac {5 d x}{2}\right )+32 A \sin \left (2 c+\frac {7 d x}{2}\right )+12 A \sin \left (3 c+\frac {7 d x}{2}\right )+20 A \sin \left (4 c+\frac {7 d x}{2}\right )+84 B \sin \left (c-\frac {d x}{2}\right )-42 B \sin \left (c+\frac {d x}{2}\right )+56 B \sin \left (2 c+\frac {d x}{2}\right )+6 B \sin \left (c+\frac {3 d x}{2}\right )-34 B \sin \left (2 c+\frac {3 d x}{2}\right )+36 B \sin \left (3 c+\frac {3 d x}{2}\right )-48 B \sin \left (c+\frac {5 d x}{2}\right )-6 B \sin \left (2 c+\frac {5 d x}{2}\right )-30 B \sin \left (3 c+\frac {5 d x}{2}\right )+12 B \sin \left (4 c+\frac {5 d x}{2}\right )-20 B \sin \left (2 c+\frac {7 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )-14 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 228, normalized size = 1.50 \[ \frac {3 \, {\left ({\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (8 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (43 \, A - 28 \, B\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 3 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*((7*A - 4*B)*cos(d*x + c)^4 + 2*(7*A - 4*B)*cos(d*x + c)^3 + (7*A - 4*B)*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) - 3*((7*A - 4*B)*cos(d*x + c)^4 + 2*(7*A - 4*B)*cos(d*x + c)^3 + (7*A - 4*B)*cos(d*x + c)^2)*log(-sin
(d*x + c) + 1) - 2*(4*(8*A - 5*B)*cos(d*x + c)^3 + (43*A - 28*B)*cos(d*x + c)^2 + 6*(A - B)*cos(d*x + c) - 3*A
)*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 = 1.29, size = 198, normalized size = 1.30 \[ \frac {\frac {3 \, {\left (7 \, A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (7 \, A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.18, size = 294, normalized size = 1.93 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{2}}+\frac {5 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{2}}-\frac {A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^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)^3-7/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*
tan(1/2*d*x+1/2*c)-7/2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B+5/2/d/a^2*A/(tan(1/
2*d*x+1/2*c)-1)-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*B+1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^2+5/2/d/a^2*A/(tan(1/2*d*x
+1/2*c)+1)-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B+7/2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-2/d/a^2*ln(tan(1/2*d*x+1/2*c)
+1)*B-1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [B]  time = 0.50, size = 336, normalized size = 2.21 \[ -\frac {A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(A*(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) - B*((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 = 0.30, size = 165, normalized size = 1.09 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A-2\,B\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^2}+\frac {4\,A-2\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A-4\,B\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**3/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)*
*3/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2

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