3.257 \(\int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\)
Optimal. Leaf size=187 \[ \frac {2 b^3 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {(A b-a B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2 A-3 a b B+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
[Out]
-1/2*(a^2+2*b^2)*(A*b-B*a)*arctanh(sin(d*x+c))/a^4/d+2*b^3*(A*b-B*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+
b)^(1/2))/a^4/d/(a-b)^(1/2)/(a+b)^(1/2)+1/3*(2*A*a^2+3*A*b^2-3*B*a*b)*tan(d*x+c)/a^3/d-1/2*(A*b-B*a)*sec(d*x+c
)*tan(d*x+c)/a^2/d+1/3*A*sec(d*x+c)^2*tan(d*x+c)/a/d
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Rubi [A] time = 0.77, antiderivative size = 187, 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 =
{3000, 3055, 3001, 3770, 2659, 205} \[ \frac {2 b^3 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2 A-3 a b B+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {(A b-a B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]
[Out]
(2*b^3*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sqrt[a + b]*d) - ((a^2
+ 2*b^2)*(A*b - a*B)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) + ((2*a^2*A + 3*A*b^2 - 3*a*b*B)*Tan[c + d*x])/(3*a^3*d
) - ((A*b - a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 2659
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]
Rule 3000
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3001
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3055
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*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
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+b \cos (c+d x)} \, dx &=\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (-3 (A b-a B)+2 a A \cos (c+d x)+2 A b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a}\\ &=-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (2 \left (2 a^2 A+3 A b^2-3 a b B\right )+a (A b+3 a B) \cos (c+d x)-3 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^2}\\ &=\frac {\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (-3 \left (a^2+2 b^2\right ) (A b-a B)-3 a b (A b-a B) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3}\\ &=\frac {\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (b^3 (A b-a B)\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^4}-\frac {\left (\left (a^2+2 b^2\right ) (A b-a B)\right ) \int \sec (c+d x) \, dx}{2 a^4}\\ &=-\frac {\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (2 b^3 (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac {2 b^3 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}-\frac {\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d}\\ \end {align*}
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Mathematica [B] time = 2.27, size = 422, normalized size = 2.26 \[ \frac {\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 a \left (2 a^2 A-3 a b B+3 A b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a \left (2 a^2 A-3 a b B+3 A b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-6 \left (a^2+2 b^2\right ) (a B-A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (a^2+2 b^2\right ) (a B-A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {24 b^3 (a B-A b) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {a^2 (a (A+3 B)-3 A b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2 (a (A+3 B)-3 A b)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}}{12 a^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]
[Out]
((24*b^3*(-(A*b) + a*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 6*(a^2 + 2*b^
2)*(-(A*b) + a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(a^2 + 2*b^2)*(-(A*b) + a*B)*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + (a^2*(-3*A*b + a*(A + 3*B)))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*a^3*A*Sin[(
c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*a*(2*a^2*A + 3*A*b^2 - 3*a*b*B)*Sin[(c + d*x)/2])/(C
os[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (a^
2*(-3*A*b + a*(A + 3*B)))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(2*a^2*A + 3*A*b^2 - 3*a*b*B)*Sin[(c
+ d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(12*a^4*d)
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fricas [A] time = 2.72, size = 729, normalized size = 3.90 \[ \left [\frac {6 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{5} - 2 \, A a^{3} b^{2} + 2 \, {\left (2 \, A a^{5} - 3 \, B a^{4} b + A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {12 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, A a^{5} - 2 \, A a^{3} b^{2} + 2 \, {\left (2 \, A a^{5} - 3 \, B a^{4} b + A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} 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+b*cos(d*x+c)),x, algorithm="fricas")
[Out]
[1/12*(6*(B*a*b^3 - A*b^4)*sqrt(-a^2 + b^2)*cos(d*x + c)^3*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c
)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x
+ c) + a^2)) + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x +
c) + 1) - 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) +
1) + 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*(2*A*a^5 - 3*B*a^4*b + A*a^3*b^2 + 3*B*a^2*b^3 - 3*A*a*b^4)*cos(d*x + c)^2
+ 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3),
-1/12*(12*(B*a*b^3 - A*b^4)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d
*x + c)^3 - 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c)
+ 1) + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1)
- 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*(2*A*a^5 - 3*B*a^4*b + A*a^3*b^2 + 3*B*a^2*b^3 - 3*A*a*b^4)*cos(d*x + c)^2 + 3
*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3)]
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giac [B] time = 0.62, size = 412, normalized size = 2.20 \[ \frac {\frac {3 \, {\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, {\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {12 \, {\left (B a b^{3} - A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
1/6*(3*(B*a^3 - A*a^2*b + 2*B*a*b^2 - 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*(B*a^3 - A*a^2*b + 2
*B*a*b^2 - 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 12*(B*a*b^3 - A*b^4)*(pi*floor(1/2*(d*x + c)/pi +
1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2
- b^2)*a^4) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c
)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*B*
a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*
d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))/((t
an(1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d
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maple [B] time = 0.18, size = 688, normalized size = 3.68 \[ \frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A \,b^{3}}{d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B \,b^{2}}{d \,a^{3}}-\frac {A \,b^{2}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A b}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A \,b^{3}}{d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B \,b^{2}}{d \,a^{3}}-\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}{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 {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A b}{2 d \,a^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 a d}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 a d}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A b}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {A \,b^{2}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {B b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A}{2 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 )}-\frac {A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A b}{2 d \,a^{2}}-\frac {A b}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A b}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \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+b*cos(d*x+c)),x)
[Out]
1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A*b^3-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B*b^2-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*A
*b^2+1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*B*b+1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*A*b-1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1
)*A*b^3+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B*b^2-1/3/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-1/3/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+1/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1
)*A*b-1/2/a/d*ln(tan(1/2*d*x+1/2*c)-1)*B-1/a/d*A/(tan(1/2*d*x+1/2*c)+1)+1/2/a/d*ln(tan(1/2*d*x+1/2*c)+1)*B-1/a
/d*A/(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2*A*b-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*A*b^2+1/d/a^
2/(tan(1/2*d*x+1/2*c)+1)*B*b+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)*B+1/2/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)*B-1/2/a/d*A/(tan(1/2*d*x+1/2*c)-1)^2-1/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*A*b-1/2/d/a^2/(tan(1
/2*d*x+1/2*c)+1)*A*b-1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)*A*b+2/d*b^4/a^4/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1
/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A-2/d*b^3/a^3/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b
))^(1/2))*B
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?
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mupad [B] time = 4.90, size = 4696, normalized size = 25.11 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B*cos(c + d*x))/(cos(c + d*x)^4*(a + b*cos(c + d*x))),x)
[Out]
(atan(((((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3
*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*
a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A
*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 + (((8*(2*B*a^13 - 4*A*a^8
*b^5 + 6*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*
B*a^12*b))/a^9 - (4*tan(c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*
b^2))/a^10)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/(2*a^4))*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2)*1i)/(2*a
^4) + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*
b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a
^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*
B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 - (((8*(2*B*a^13 - 4*A*a^8*
b^5 + 6*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B
*a^12*b))/a^9 + (4*tan(c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b
^2))/a^10)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/(2*a^4))*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2)*1i)/(2*a^
4))/((16*(4*A^3*b^11 - 6*A^3*a*b^10 + 6*A^3*a^2*b^9 - 5*A^3*a^3*b^8 + 2*A^3*a^4*b^7 - A^3*a^5*b^6 - 4*B^3*a^3*
b^8 + 6*B^3*a^4*b^7 - 6*B^3*a^5*b^6 + 5*B^3*a^6*b^5 - 2*B^3*a^7*b^4 + B^3*a^8*b^3 - 12*A^2*B*a*b^10 + 12*A*B^2
*a^2*b^9 - 18*A*B^2*a^3*b^8 + 18*A*B^2*a^4*b^7 - 15*A*B^2*a^5*b^6 + 6*A*B^2*a^6*b^5 - 3*A*B^2*a^7*b^4 + 18*A^2
*B*a^2*b^9 - 18*A^2*B*a^3*b^8 + 15*A^2*B*a^4*b^7 - 6*A^2*B*a^5*b^6 + 3*A^2*B*a^6*b^5))/a^9 + (((8*tan(c/2 + (d
*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 -
7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4
+ 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*
b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 + (((8*(2*B*a^13 - 4*A*a^8*b^5 + 6*A*a^9*b^4 - 2*A
*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b))/a^9 - (4*tan(
c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/a^10)*(2*A*b^3 - B
*a^3 + A*a^2*b - 2*B*a*b^2))/(2*a^4))*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/(2*a^4) - (((8*tan(c/2 + (d*x)/
2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^
2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 1
3*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5
- 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 - (((8*(2*B*a^13 - 4*A*a^8*b^5 + 6*A*a^9*b^4 - 2*A*a^1
0*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b))/a^9 + (4*tan(c/2
+ (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/a^10)*(2*A*b^3 - B*a^3
+ A*a^2*b - 2*B*a*b^2))/(2*a^4))*(2*A*b^3 - B*a^3 + A*a^2*b - 2*B*a*b^2))/(2*a^4)))*(2*A*b^3 - B*a^3 + A*a^2*
b - 2*B*a*b^2)*1i)/(a^4*d) - ((tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2 - A*a*b - 2*B*a*b))/a^3 + (tan(c/
2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 - B*a^2 + A*a*b - 2*B*a*b))/a^3 - (4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 -
3*B*a*b))/(3*a^3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) + (b^3*at
an(((b^3*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B
^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*
B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8
+ 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^
7*b^2))/a^6 + (b^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 6*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^1
1*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b))/a^9 - (8*b^3*tan(c/2 + (d*x)/2)*
(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/
(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) + (b^3*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*A^2
*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4
+ 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6
*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*
a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 - (b^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 6
*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b
))/a^9 + (8*b^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(
a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2))/((16*(4*A^3*b^11 - 6*A^3*a*b^10 + 6*A
^3*a^2*b^9 - 5*A^3*a^3*b^8 + 2*A^3*a^4*b^7 - A^3*a^5*b^6 - 4*B^3*a^3*b^8 + 6*B^3*a^4*b^7 - 6*B^3*a^5*b^6 + 5*B
^3*a^6*b^5 - 2*B^3*a^7*b^4 + B^3*a^8*b^3 - 12*A^2*B*a*b^10 + 12*A*B^2*a^2*b^9 - 18*A*B^2*a^3*b^8 + 18*A*B^2*a^
4*b^7 - 15*A*B^2*a^5*b^6 + 6*A*B^2*a^6*b^5 - 3*A*B^2*a^7*b^4 + 18*A^2*B*a^2*b^9 - 18*A^2*B*a^3*b^8 + 15*A^2*B*
a^4*b^7 - 6*A^2*B*a^5*b^6 + 3*A^2*B*a^6*b^5))/a^9 + (b^3*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d
*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 -
7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4
+ 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*
b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 + (b^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4*
A*a^8*b^5 + 6*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b
- 2*B*a^12*b))/a^9 - (8*b^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 1
6*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2) - (b^3*(-(a + b)*(a - b))^(1
/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16
*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 +
16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^
7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 - (b^3*(-(a + b)*(
a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 6*A*a^9*b^4 - 2*A*a^10*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10
*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b))/a^9 + (8*b^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b
- B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2)))/(a^6 - a^4*b
^2)))*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*2i)/(d*(a^6 - a^4*b^2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)**4/(a+b*cos(d*x+c)),x)
[Out]
Integral((A + B*cos(c + d*x))*sec(c + d*x)**4/(a + b*cos(c + d*x)), x)
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