3.5.58 \(\int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) [458]

Optimal. Leaf size=157 \[ \frac {2 b^4 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d} \]

[Out]

-1/2*b*(a^2+2*b^2)*arctanh(sin(d*x+c))/a^4/d+2*b^4*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^2+3*b^2)*tan(d*x+c)/a^3/d-1/2*b*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*sec(d*x+c)^2*ta
n(d*x+c)/a/d

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Rubi [A]
time = 0.34, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2881, 3134, 3080, 3855, 2738, 211} \begin {gather*} \frac {2 b^4 \text {ArcTan}\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 {b \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 a^3 d}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 2.44, size = 258, normalized size = 1.64 \begin {gather*} \frac {-\frac {24 b^4 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {1}{2} \sec ^3(c+d x) \left (9 b \left (a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 b \left (a^2+2 b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 a \left (4 a^2+3 b^2-3 a b \cos (c+d x)+\left (2 a^2+3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*b^4*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (Sec[c + d*x]^3*(9*b*(a^2 +
2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 3
*b*(a^2 + 2*b^2)*Cos[3*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]]) + 4*a*(4*a^2 + 3*b^2 - 3*a*b*Cos[c + d*x] + (2*a^2 + 3*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/2)/(12*
a^4*d)

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Maple [A]
time = 0.42, size = 252, normalized size = 1.61

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(252\)
default	\(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(252\)
risch	\(\frac {i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}+4 a^{2}+6 b^{2}\right )}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^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 de

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Fricas [A]
time = 0.56, size = 535, normalized size = 3.41 \begin {gather*} \left [-\frac {6 \, \sqrt {-a^{2} + b^{2}} b^{4} \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 (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{5} - 2 \, a^{3} b^{2} + 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - 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 \, \sqrt {a^{2} - b^{2}} b^{4} \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 (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{5} - 2 \, a^{3} b^{2} + 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - 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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(6*sqrt(-a^2 + b^2)*b^4*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*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^
3*log(-sin(d*x + c) + 1) - 2*(2*a^5 - 2*a^3*b^2 + 2*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 - 3*(a^4*b - 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*sqrt(a^2 - b^2)*b^4*arctan(-(a
*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3
*log(sin(d*x + c) + 1) + 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*a^5 - 2*a^3*
b^2 + 2*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 - 3*(a^4*b - 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (140) = 280\).
time = 0.53, size = 286, normalized size = 1.82 \begin {gather*} -\frac {\frac {12 \, {\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 )} b^{4}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {3 \, {\left (a^{2} b + 2 \, 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 (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (6 \, 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} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, 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 ) + 6 \, 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*(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)))*b^4/(sqrt(a^2 - b^2)*a^4) + 3*(a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^
4 - 3*(a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 2*(6*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2
*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3
+ 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)
^2 - 1)^3*a^3))/d

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Mupad [B]
time = 2.68, size = 991, normalized size = 6.31 \begin {gather*} \frac {a^5\,\left (\frac {\sin \left (c+d\,x\right )}{2}+\frac {\sin \left (3\,c+3\,d\,x\right )}{6}\right )-a^4\,\left (\frac {b\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}\right )-a^2\,\left (\frac {3\,b^3\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {b^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{4}\right )-a^3\,\left (\frac {b^2\,\sin \left (c+d\,x\right )}{4}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12}\right )-a\,\left (\frac {b^4\,\sin \left (c+d\,x\right )}{4}+\frac {b^4\,\sin \left (3\,c+3\,d\,x\right )}{4}\right )+\frac {3\,b^5\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {b^5\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,b^4\,\mathrm {atanh}\left (\frac {a^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+8\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-8\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+8\,a^2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^4\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-3\,a^5\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-2\,a^6\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,a^7\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^8\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (a^7+2\,a^5\,b^2-3\,a^3\,b^4\right )}\right )\,\cos \left (c+d\,x\right )\,\sqrt {b^2-a^2}}{2}+\frac {b^4\,\mathrm {atanh}\left (\frac {a^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+8\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-8\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+8\,a^2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^4\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-3\,a^5\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-2\,a^6\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,a^7\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^8\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (a^7+2\,a^5\,b^2-3\,a^3\,b^4\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,\sqrt {b^2-a^2}}{2}}{a^4\,d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )\,\left (a^2-b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^5*(sin(c + d*x)/2 + sin(3*c + 3*d*x)/6) - a^4*((b*sin(2*c + 2*d*x))/4 + (b*atanh(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2))*cos(3*c + 3*d*x))/4 + (3*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4) - a^2*((
3*b^3*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - (b^3*sin(2*c + 2*d*x))/4 + (b^3*atanh(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4) - a^3*((b^2*sin(c + d*x))/4 - (b^2*sin(3*c + 3*d*x))/
12) - a*((b^4*sin(c + d*x))/4 + (b^4*sin(3*c + 3*d*x))/4) + (3*b^5*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2)))/2 + (b^5*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + (3*b^4*atanh((a^9*
sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*b^9*sin(c/2 + (d*x)/2)*(
b^2 - a^2)^(1/2) + 8*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(
1/2) - 3*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*a^7
*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a
*b^2 - a^3)*(a^7 - 3*a^3*b^4 + 2*a^5*b^2)))*cos(c + d*x)*(b^2 - a^2)^(1/2))/2 + (b^4*atanh((a^9*sin(c/2 + (d*x
)/2)*(b^2 - a^2)^(1/2) + 8*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/
2) + 8*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 3*a^5*b
^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*a^7*b^2*sin(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2) - a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(a
^7 - 3*a^3*b^4 + 2*a^5*b^2)))*cos(3*c + 3*d*x)*(b^2 - a^2)^(1/2))/2)/(a^4*d*((3*cos(c + d*x))/4 + cos(3*c + 3*
d*x)/4)*(a^2 - b^2))

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