3.468 \(\int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=270 \[ \frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {b \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac {b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a^3 d \left (a^2-b^2\right )} \]

[Out]

2*b^4*(5*a^2-4*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^5/(a-b)^(3/2)/(a+b)^(3/2)/d-b*(a^2+4*
b^2)*arctanh(sin(d*x+c))/a^5/d+1/3*(2*a^4+7*a^2*b^2-12*b^4)*tan(d*x+c)/a^4/(a^2-b^2)/d-b*(a^2-2*b^2)*sec(d*x+c
)*tan(d*x+c)/a^3/(a^2-b^2)/d+1/3*(a^2-4*b^2)*sec(d*x+c)^2*tan(d*x+c)/a^2/(a^2-b^2)/d+b^2*sec(d*x+c)^2*tan(d*x+
c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.97, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2802, 3055, 3001, 3770, 2659, 205} \[ \frac {2 b^4 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (7 a^2 b^2+2 a^4-12 b^4\right ) \tan (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac {b \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac {b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^4*(5*a^2 - 4*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a + b)^(3/2)*d)
- (b*(a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(a^5*d) + ((2*a^4 + 7*a^2*b^2 - 12*b^4)*Tan[c + d*x])/(3*a^4*(a^2 -
b^2)*d) - (b*(a^2 - 2*b^2)*Sec[c + d*x]*Tan[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 4*b^2)*Sec[c + d*x]^2*Tan
[c + d*x])/(3*a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]^2*Tan[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 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 {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (a^2-4 b^2-a b \cos (c+d x)+3 b^2 \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-6 b \left (a^2-2 b^2\right )+a \left (2 a^2+b^2\right ) \cos (c+d x)+2 b \left (a^2-4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (2 \left (2 a^4+7 a^2 b^2-12 b^4\right )-2 a b \left (a^2+2 b^2\right ) \cos (c+d x)-6 b^2 \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-6 b \left (a^4+3 a^2 b^2-4 b^4\right )-6 a b^2 \left (a^2-2 b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (b^4 \left (5 a^2-4 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}-\frac {\left (b \left (a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx}{a^5}\\ &=-\frac {b \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (2 b^4 \left (5 a^2-4 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^5 \left (a^2-b^2\right ) d}\\ &=\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {b \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (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 = 6.15, size = 499, normalized size = 1.85 \[ -\frac {b^5 \sin (c+d x)}{a^4 d (a-b) (a+b) (a+b \cos (c+d x))}+\frac {a-6 b}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {6 b-a}{12 a^3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\left (a^2 b+4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (-a^2 b-4 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}-\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{a^5 d \left (a^2-b^2\right ) \sqrt {b^2-a^2}}+\frac {2 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^4*(5*a^2 - 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(a^5*(a^2 - b^2)*Sqrt[-a^2 + b^2
]*d) + ((a^2*b + 4*b^3)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(a^5*d) + ((-(a^2*b) - 4*b^3)*Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]])/(a^5*d) + (a - 6*b)/(12*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c +
d*x)/2]/(6*a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + Sin[(c + d*x)/2]/(6*a^2*d*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^3) + (-a + 6*b)/(12*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*a^2*Sin[(c + d*x)/2] + 9
*b^2*Sin[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*a^2*Sin[(c + d*x)/2] + 9*b^2*Sin[(
c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (b^5*Sin[c + d*x])/(a^4*(a - b)*(a + b)*d*(a +
b*Cos[c + d*x]))

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fricas [A]  time = 2.77, size = 1001, normalized size = 3.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*((5*a^2*b^5 - 4*b^7)*cos(d*x + c)^4 + (5*a^3*b^4 - 4*a*b^6)*cos(d*x + c)^3)*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)) + 3*((a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x
+ c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((a^6*b^2 + 2*a^
4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(-sin
(d*x + c) + 1) - 2*(a^8 - 2*a^6*b^2 + a^4*b^4 + (2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7)*cos(d*x + c)^3 +
2*(a^8 + a^6*b^2 - 5*a^4*b^4 + 3*a^2*b^6)*cos(d*x + c)^2 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c))*sin(
d*x + c))/((a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^4 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)^3), 1/
6*(6*((5*a^2*b^5 - 4*b^7)*cos(d*x + c)^4 + (5*a^3*b^4 - 4*a*b^6)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*co
s(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - 3*((a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4
+ (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) + 3*((a^6*b^2 + 2*a^4*b^4 -
7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c
) + 1) + 2*(a^8 - 2*a^6*b^2 + a^4*b^4 + (2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7)*cos(d*x + c)^3 + 2*(a^8
+ a^6*b^2 - 5*a^4*b^4 + 3*a^2*b^6)*cos(d*x + c)^2 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c))*sin(d*x + c)
)/((a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^4 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)^3)]

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giac [A]  time = 1.39, size = 368, normalized size = 1.36 \[ -\frac {\frac {6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - a^{4} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {6 \, {\left (5 \, a^{2} b^{4} - 4 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - a^{5} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {3 \, {\left (a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*b^5*tan(1/2*d*x + 1/2*c)/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b
)) + 6*(5*a^2*b^4 - 4*b^6)*(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^7 - a^5*b^2)*sqrt(a^2 - b^2)) + 3*(a^2*b + 4*b^3)*log(abs(tan
(1/2*d*x + 1/2*c) + 1))/a^5 - 3*(a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 + 2*(3*a^2*tan(1/2*d*x
+ 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*a^2*tan(1/2*d*x + 1/2*c)^3 - 18*b
^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*b^2*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^4))/d

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maple [B]  time = 0.12, size = 535, normalized size = 1.98 \[ -\frac {2 b^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \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 {10 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 )}{d \,a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {8 b^{6} \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^{5} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 b^{2}}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {4 b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{5}}-\frac {1}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b^{2}}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {4 b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^5/a^4/(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)+10/d*b^4/a^3/(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))-8/d*b^6/a^5/(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))-1/3/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/
a^2/(tan(1/2*d*x+1/2*c)-1)^2-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2*b-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)-1/d/a^3/(tan(1/
2*d*x+1/2*c)-1)*b-3/d/a^4/(tan(1/2*d*x+1/2*c)-1)*b^2+1/d*b/a^3*ln(tan(1/2*d*x+1/2*c)-1)+4/d*b^3/a^5*ln(tan(1/2
*d*x+1/2*c)-1)-1/3/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2+1/d/a^3/(tan(1/2*d*x+1/2*
c)+1)^2*b-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*b-3/d/a^4/(tan(1/2*d*x+1/2*c)+1)*b^2-1
/d*b/a^3*ln(tan(1/2*d*x+1/2*c)+1)-4/d*b^3/a^5*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)^4/(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 = 7.09, size = 3843, normalized size = 14.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b*cos(c + d*x))^2),x)

[Out]

((2*tan(c/2 + (d*x)/2)^7*(a^5 - 2*a*b^4 + 4*b^5 - 3*a^2*b^3 + a^3*b^2))/(a^4*(a + b)*(a - b)) + (2*tan(c/2 + (
d*x)/2)^3*(6*a*b^4 - 8*a^4*b + a^5 + 36*b^5 - 19*a^2*b^3 - 7*a^3*b^2))/(3*a^4*(a + b)*(a - b)) + (2*tan(c/2 +
(d*x)/2)^5*(6*a*b^4 + 8*a^4*b + a^5 - 36*b^5 + 19*a^2*b^3 - 7*a^3*b^2))/(3*a^4*(a + b)*(a - b)) + (2*tan(c/2 +
(d*x)/2)*(a^5 - 2*a*b^4 - 4*b^5 + 3*a^2*b^3 + a^3*b^2))/(a^4*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)
^8*(a - b) - tan(c/2 + (d*x)/2)^2*(2*a + 4*b) + tan(c/2 + (d*x)/2)^6*(2*a - 4*b) + 6*b*tan(c/2 + (d*x)/2)^4))
+ (b*atan(((b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^
8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2
) + (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3)
)/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (32*b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a^10*b^6 + 2*a^
11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2))))/a^5)*1i)/a^5 + (b*
(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7
+ 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b*(a^2 +
4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^
15 - a^12*b^3 - a^13*b^2) - (32*b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^1
2*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2))))/a^5)*1i)/a^5)/((64*(64*b^14 - 32
*a*b^13 - 112*a^2*b^12 + 48*a^3*b^11 + 12*a^4*b^10 - 6*a^5*b^9 + 31*a^6*b^8 - 5*a^7*b^7 + 5*a^8*b^6))/(a^14*b
+ a^15 - a^12*b^3 - a^13*b^2) + (b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 +
48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^1
1 - a^8*b^3 - a^9*b^2) + (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5
*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (32*b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*
b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2))
))/a^5))/a^5 - (b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^
4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9
*b^2) - (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*
b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a^10*b^6 +
2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2))))/a^5))/a^5))*(a
^2 + 4*b^2)*2i)/(a^5*d) + (b^4*atan(((b^4*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)
*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4
- 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) + (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9
*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (32*b^4*tan(c/2 + (d*
x)/2)*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b
^3 - 2*a^14*b^2))/((a^10*b + a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*b
^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))*1i)/(a^11 - a^5*b^6 + 3*a^7*b^4 -
3*a^9*b^2) + (b^4*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 -
48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))
/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 -
5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b^4*tan(c/2 + (d*x)/2)*(5*a^2 - 4*b^2)*(-
(a + b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/((a^10
*b + a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^
3)^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))*1i)/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))/((64*(64*b^1
4 - 32*a*b^13 - 112*a^2*b^12 + 48*a^3*b^11 + 12*a^4*b^10 - 6*a^5*b^9 + 31*a^6*b^8 - 5*a^7*b^7 + 5*a^8*b^6))/(a
^14*b + a^15 - a^12*b^3 - a^13*b^2) + (b^4*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2
)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4
- 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) + (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 +
9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (32*b^4*tan(c/2 + (d
*x)/2)*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*
b^3 - 2*a^14*b^2))/((a^10*b + a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*
b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*
a^9*b^2) - (b^4*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48
*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(
a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5
*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b^4*tan(c/2 + (d*x)/2)*(5*a^2 - 4*b^2)*(-(a
+ b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/((a^10*b
+ a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)
^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*b^2)*
(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(d*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*cos(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*cos(c + d*x))**2, x)

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