3.5.75 \(\int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [475]

Optimal. Leaf size=182 \[ -\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]

[Out]

-b*(6*a^4-5*a^2*b^2+2*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+ar
ctanh(sin(d*x+c))/a^3/d+1/2*b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/2*b^2*(5*a^2-2*b^2)*sin(d*x+c)/a
^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))

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Rubi [A]
time = 0.31, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2881, 3134, 3080, 3855, 2738, 211} \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

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 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-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 3080

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 3134

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] + D
ist[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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

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))^3} \, dx &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\int \frac {\left (2 \left (a^2-b^2\right )-2 a b \cos (c+d x)+b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {\left (2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \sec (c+d x) \, dx}{a^3}-\frac {\left (b \left (6 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\left (b \left (6 a^4-5 a^2 b^2+2 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^2 d}\\ &=-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 1.17, size = 192, normalized size = 1.05 \begin {gather*} \frac {\frac {2 b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^2 \left (6 a^3-3 a b^2+b \left (5 a^2-2 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{2 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 251, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {2 b \left (\frac {-\frac {\left (6 a^{2}+a b -2 b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-a b -2 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )^{2}}+\frac {\left (6 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}}{d}\)	\(251\)
default	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {2 b \left (\frac {-\frac {\left (6 a^{2}+a b -2 b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-a b -2 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )^{2}}+\frac {\left (6 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}}{d}\)	\(251\)
risch	\(\frac {i b \left (4 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+16 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-7 a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+5 a^{2} b^{2}-2 b^{4}\right )}{d \,a^{2} \left (a^{2}-b^{2}\right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}+\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}\)	\(722\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*cos(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/a^3*ln(tan(1/2*d*x+1/2*c)+1)-1/a^3*ln(tan(1/2*d*x+1/2*c)-1)-2*b/a^3*((-1/2*(6*a^2+a*b-2*b^2)*a*b/(a-b)/
(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(6*a^2-a*b-2*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(a*ta
n(1/2*d*x+1/2*c)^2-b*tan(1/2*d*x+1/2*c)^2+a+b)^2+1/2*(6*a^4-5*a^2*b^2+2*b^4)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))
^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*cos(d*x+c))^3,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 de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (169) = 338\).
time = 1.01, size = 1142, normalized size = 6.27 \begin {gather*} \left [-\frac {{\left (6 \, a^{6} b - 5 \, a^{4} b^{3} + 2 \, a^{2} b^{5} + {\left (6 \, a^{4} b^{3} - 5 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\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 ) - 2 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (6 \, a^{6} b^{2} - 9 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + {\left (5 \, a^{5} b^{3} - 7 \, a^{3} b^{5} + 2 \, a b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{9} b^{2} - 3 \, a^{7} b^{4} + 3 \, a^{5} b^{6} - a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{10} b - 3 \, a^{8} b^{3} + 3 \, a^{6} b^{5} - a^{4} b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 3 \, a^{9} b^{2} + 3 \, a^{7} b^{4} - a^{5} b^{6}\right )} d\right )}}, -\frac {{\left (6 \, a^{6} b - 5 \, a^{4} b^{3} + 2 \, a^{2} b^{5} + {\left (6 \, a^{4} b^{3} - 5 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\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^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, a^{6} b^{2} - 9 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + {\left (5 \, a^{5} b^{3} - 7 \, a^{3} b^{5} + 2 \, a b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{9} b^{2} - 3 \, a^{7} b^{4} + 3 \, a^{5} b^{6} - a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{10} b - 3 \, a^{8} b^{3} + 3 \, a^{6} b^{5} - a^{4} b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 3 \, a^{9} b^{2} + 3 \, a^{7} b^{4} - a^{5} b^{6}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5 + (6*a^4*b^3 - 5*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + 2*(6*a^5*b^2 - 5*a^
3*b^4 + 2*a*b^6)*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*sqr
t(-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)
) - 2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^2 + 2*(a^7
*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2
*b^6 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(
d*x + c))*log(-sin(d*x + c) + 1) - 2*(6*a^6*b^2 - 9*a^4*b^4 + 3*a^2*b^6 + (5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*co
s(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^
3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d), -1/2*((6*a^6*b - 5*a^4*
b^3 + 2*a^2*b^5 + (6*a^4*b^3 - 5*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + 2*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*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^8 - 3*a^6*b^2 + 3*a^4
*b^4 - a^2*b^6 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a
*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 + (a^6*b^2 - 3*a^4*b^4 + 3*
a^2*b^6 - b^8)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1)
- (6*a^6*b^2 - 9*a^4*b^4 + 3*a^2*b^6 + (5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^
2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x
+ c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*cos(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)/(a + b*cos(c + d*x))**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (169) = 338\).
time = 0.49, size = 344, normalized size = 1.89 \begin {gather*} \frac {\frac {{\left (6 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\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} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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 )}^{2}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 9.06, size = 2500, normalized size = 13.74 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + b*cos(c + d*x))^3),x)

[Out]

- (atan((((((8*(12*a^14*b - 4*a^15 + 4*a^6*b^9 - 2*a^7*b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*b^4
- 34*a^12*b^3 + 8*a^13*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11
*b^2) - (8*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^10*b^6 + 48*
a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/(a^3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^
7*b^4 - 3*a^8*b^3 - 3*a^9*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*a^10 - 8*a^9*b - 8*a*b^9 + 8*b^10 - 32*a^2*b^8
+ 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4 + 32*a^7*b^3 + 24*a^8*b^2))/(a^10*b + a^11 - a^4*b^7 - a^
5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2))*1i)/a^3 - ((((8*(12*a^14*b - 4*a^15 + 4*a^6*b^9 - 2*a^
7*b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*b^4 - 34*a^12*b^3 + 8*a^13*b^2))/(a^12*b + a^13 - a^6*b^
7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + (8*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10
+ 8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/(
a^3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2)))/a^3 + (8*tan(c/2 + (
d*x)/2)*(4*a^10 - 8*a^9*b - 8*a*b^9 + 8*b^10 - 32*a^2*b^8 + 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4
+ 32*a^7*b^3 + 24*a^8*b^2))/(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2
))*1i)/a^3)/((((8*(12*a^14*b - 4*a^15 + 4*a^6*b^9 - 2*a^7*b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*
b^4 - 34*a^12*b^3 + 8*a^13*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a
^11*b^2) - (8*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^10*b^6 +
48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/(a^3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3
*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*a^10 - 8*a^9*b - 8*a*b^9 + 8*b^10 - 32*a^2*
b^8 + 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4 + 32*a^7*b^3 + 24*a^8*b^2))/(a^10*b + a^11 - a^4*b^7 -
a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2))/a^3 + (((8*(12*a^14*b - 4*a^15 + 4*a^6*b^9 - 2*a^7*
b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*b^4 - 34*a^12*b^3 + 8*a^13*b^2))/(a^12*b + a^13 - a^6*b^7
- a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + (8*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10 +
8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/(a^
3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2)))/a^3 + (8*tan(c/2 + (d*
x)/2)*(4*a^10 - 8*a^9*b - 8*a*b^9 + 8*b^10 - 32*a^2*b^8 + 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4 +
32*a^7*b^3 + 24*a^8*b^2))/(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2))
/a^3 - (16*(12*a^8*b - 2*a*b^8 + 4*b^9 - 18*a^2*b^7 + 13*a^3*b^6 + 36*a^4*b^5 - 26*a^5*b^4 - 34*a^6*b^3 + 24*a
^7*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2)))*2i)/(a^3*d) -
((tan(c/2 + (d*x)/2)^3*(a*b^3 - 2*b^4 + 6*a^2*b^2))/((a^2*b - a^3)*(a + b)^2) + (tan(c/2 + (d*x)/2)*(a*b^3 +
2*b^4 - 6*a^2*b^2))/((a + b)*(a^4 - 2*a^3*b + a^2*b^2)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + ta
n(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) - (b*atan(((b*((8*tan(c/2 + (d*x)/2)*(4*a^10 - 8*a^9*b -
8*a*b^9 + 8*b^10 - 32*a^2*b^8 + 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4 + 32*a^7*b^3 + 24*a^8*b^2))/
(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) - (b*((8*(12*a^14*b - 4*a^
15 + 4*a^6*b^9 - 2*a^7*b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*b^4 - 34*a^12*b^3 + 8*a^13*b^2))/(a
^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) - (4*b*tan(c/2 + (d*x)/2)*
(-(a + b)^5*(a - b)^5)^(1/2)*(6*a^4 + 2*b^4 - 5*a^2*b^2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^8 - 32*
a^9*b^7 - 48*a^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/((a^13 - a^3*b^10 + 5*a^5*b^8 -
10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2)*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3
- 3*a^9*b^2)))*(-(a + b)^5*(a - b)^5)^(1/2)*(6*a^4 + 2*b^4 - 5*a^2*b^2))/(2*(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*
a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2)))*(-(a + b)^5*(a - b)^5)^(1/2)*(6*a^4 + 2*b^4 - 5*a^2*b^2)*1i)/(2*(a^13 - a
^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2)) + (b*((8*tan(c/2 + (d*x)/2)*(4*a^10 - 8*a^9*b - 8
*a*b^9 + 8*b^10 - 32*a^2*b^8 + 32*a^3*b^7 + 57*a^4*b^6 - 48*a^5*b^5 - 52*a^6*b^4 + 32*a^7*b^3 + 24*a^8*b^2))/(
a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) + (b*((8*(12*a^14*b - 4*a^1
5 + 4*a^6*b^9 - 2*a^7*b^8 - 18*a^8*b^7 + 4*a^9*b^6 + 36*a^10*b^5 - 6*a^11*b^4 - 34*a^12*b^3 + 8*a^13*b^2))/(a^
12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + (4*b*tan(c/2 + (d*x)/2)*(
-(a + b)^5*(a - b)^5)^(1/2)*(6*a^4 + 2*b^4 - 5*...

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