3.465 \(\int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=118 \[ -\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

[Out]

-2*b*(2*a^2-b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+arctanh(sin(
d*x+c))/a^2/d+b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.22, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2802, 3001, 3770, 2659, 205} \[ -\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/2)*(a + b)^(3/2)*d) +
ArcTanh[Sin[c + d*x]]/(a^2*d) + (b^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*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] + 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*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((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 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 {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (a^2-b^2-a b \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \sec (c+d x) \, dx}{a^2}-\frac {\left (b \left (2 a^2-b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (2 b \left (2 a^2-b^2\right )\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^2 \left (a^2-b^2\right ) d}\\ &=-\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 146, normalized size = 1.24 \[ \frac {\frac {2 b \left (b^2-2 a^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a b^2 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*(-2*a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - Log[Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a*b^2*Sin[c + d*x])/((a - b)*(a + b)*
(a + b*Cos[c + d*x])))/(a^2*d)

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fricas [B]  time = 1.86, size = 592, normalized size = 5.02 \[ \left [-\frac {{\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \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 ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}, -\frac {2 \, {\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\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 ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*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)) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(sin(d*x
+ c) + 1) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^3
*b^2 - a*b^4)*sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d), -1
/2*(2*(2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^
2 - b^2)*sin(d*x + c))) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(sin(d*x + c)
+ 1) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^3*b^2
- a*b^4)*sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d)]

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giac [A]  time = 0.70, size = 198, normalized size = 1.68 \[ \frac {\frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} - \frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2*tan(1/2*d*x + 1/2*c)/((a^3 - a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) - 2*
(2*a^2*b - b^3)*(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)))/((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) + log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - l
og(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2)/d

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maple [B]  time = 0.09, size = 221, normalized size = 1.87 \[ \frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {4 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*cos(d*x+c))^2,x)

[Out]

2/d*b^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)-4/d*b/(a-b)/(a+b)/(
(a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+2/d*b^3/a^2/(a-b)/(a+b)/((a-b)*(a+b))^
(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^2*ln(tan(1/2
*d*x+1/2*c)+1)

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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(sec(d*x+c)/(a+b*cos(d*x+c))^2,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 = 5.98, size = 2886, normalized size = 24.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + b*cos(c + d*x))^2),x)

[Out]

- (atan((((((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*tan(c
/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b
^3 - a^3*b^2)))/a^2 - (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*
b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2 - ((((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^
5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b
^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2 + (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^
5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2)/((((32*(2*a^8*b -
a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*
a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2 - (32*t
an(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^
3 - a^3*b^2))/a^2 - (64*(2*a^4*b - a*b^4 + b^5 - 3*a^2*b^3 + 2*a^3*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (
((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*tan(c/2 + (d*x)/
2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^
2)))/a^2 + (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*
b + a^5 - a^2*b^3 - a^3*b^2))/a^2))*2i)/(a^2*d) - (b*atan(((b*((32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5
+ 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (b*(2*a^2 - b^2)*((32*(2*a^
8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*b*tan(c/2 + (d*x)/2)*(2*a^
2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(
(a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8
- a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4
- 3*a^6*b^2) + (b*((32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^
2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (b*(2*a^2 - b^2)*((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))
/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^
9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^
2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*(2*a^2
- b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))/((64*(2*a^4*b - a*b^4 + b^5
- 3*a^2*b^3 + 2*a^3*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (b*((32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*
b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (b*(2*a^2 - b^2)*((32*(2
*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*b*tan(c/2 + (d*x)/2)*(2
*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)
)/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2))/(
a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4
- 3*a^6*b^2) + (b*((32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^
2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (b*(2*a^2 - b^2)*((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))
/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^
9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^
2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*(2*a^2
- b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(2*a^2 - b^2)*(-(a + b)^3*(a -
b)^3)^(1/2)*2i)/(d*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)) - (2*b^2*tan(c/2 + (d*x)/2))/(d*(a + b)*(a*b - a^
2)*(a + b + tan(c/2 + (d*x)/2)^2*(a - b)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*cos(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)/(a + b*cos(c + d*x))**2, x)

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