3.5.87 \(\int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) [487]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.33, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3936, 4177, 4167, 4083, 3855, 3916, 2738, 214} \begin {gather*} \frac {2 a^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 b^3 d}-\frac {a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

-1/2*(a*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(b^4*d) + (2*a^4*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a +
b]])/(Sqrt[a - b]*b^4*Sqrt[a + b]*d) + ((3*a^2 + 2*b^2)*Tan[c + d*x])/(3*b^3*d) - (a*Sec[c + d*x]*Tan[c + d*x]
)/(2*b^2*d) + (Sec[c + d*x]^2*Tan[c + d*x])/(3*b*d)

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3936

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-d^3)*Cot[
e + f*x]*((d*Csc[e + f*x])^(n - 3)/(b*f*(n - 2))), x] + Dist[d^3/(b*(n - 2)), Int[(d*Csc[e + f*x])^(n - 3)*(Si
mp[a*(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*Csc[e + f*x]^2, x]/(a + b*Csc[e + f*x])), x], x] /; FreeQ[{a
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3]

Rule 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4177

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^
(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m +
2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C,
m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 2.79, size = 258, normalized size = 1.64 \begin {gather*} \frac {-\frac {24 a^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 a \left (2 a^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 a \left (2 a^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 b \left (3 a^2+4 b^2-3 a b \cos (c+d x)+\left (3 a^2+2 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 b^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

((-24*a^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*a*(2*a^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*
a*(2*a^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*b*(3*a^2 + 4*b^2 - 3*a*b*Cos[c + d*x] + (3*a^2 + 2*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/2)/(12*b
^4*d)

________________________________________________________________________________________

Maple [A]
time = 0.26, size = 252, normalized size = 1.61

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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)^5/(a+b*sec(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*a^2-4*b^2>0)', see `assume?`
for more de

________________________________________________________________________________________

Fricas [A]
time = 3.41, size = 557, normalized size = 3.55 \begin {gather*} \left [\frac {6 \, \sqrt {a^{2} - b^{2}} a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {12 \, \sqrt {-a^{2} + b^{2}} a^{4} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\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)^5/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(a^2 - b^2)*a^4*cos(d*x + c)^3*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^
2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 3
*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*l
og(-sin(d*x + c) + 1) + 2*(2*a^2*b^3 - 2*b^5 + 2*(3*a^4*b - a^2*b^3 - 2*b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b
^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c)^3), 1/12*(12*sqrt(-a^2 + b^2)*a^4*arctan(-sqrt
(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 3*(2*a^5 - a^3*b^2 - a*b^4)*cos
(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*a
^2*b^3 - 2*b^5 + 2*(3*a^4*b - a^2*b^3 - 2*b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c)
)/((a^2*b^4 - b^6)*d*cos(d*x + c)^3)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*sec(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**5/(a + b*sec(c + d*x)), x)

________________________________________________________________________________________

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 )} a^{4}}{\sqrt {-a^{2} + b^{2}} b^{4}} - \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{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} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, 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} b^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Mupad [B]
time = 2.45, size = 1021, normalized size = 6.50 \begin {gather*} -\frac {\frac {9\,a^4\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {8\,a^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^7\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^5\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,a^2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^3\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^4\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (4\,a^4\,\left (a^2-b^2\right )+b^4\,\left (a^2-b^2\right )+2\,a^5\,b-4\,a^6+2\,a^3\,b^3+4\,a^2\,b^2\,\left (a^2-b^2\right )-a\,b^3\,\left (a^2-b^2\right )-2\,a^3\,b\,\left (a^2-b^2\right )\right )}\right )}{2}-\frac {3\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {a^2-b^2}}{2}-\frac {b^3\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a^2-b^2}}{2}+\frac {3\,a^4\,\cos \left (3\,c+3\,d\,x\right )\,\mathrm {atanh}\left (\frac {8\,a^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^7\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^5\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,a^2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^3\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^4\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (4\,a^4\,\left (a^2-b^2\right )+b^4\,\left (a^2-b^2\right )+2\,a^5\,b-4\,a^6+2\,a^3\,b^3+4\,a^2\,b^2\,\left (a^2-b^2\right )-a\,b^3\,\left (a^2-b^2\right )-2\,a^3\,b\,\left (a^2-b^2\right )\right )}\right )}{2}+\frac {3\,a^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 )\,\sqrt {a^2-b^2}}{2}-\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )\,\sqrt {a^2-b^2}}{4}+\frac {9\,a^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 )\,\sqrt {a^2-b^2}}{2}+\frac {3\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {a^2-b^2}}{4}-\frac {3\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a^2-b^2}}{4}+\frac {9\,a\,b^2\,\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 )\,\sqrt {a^2-b^2}}{4}+\frac {3\,a\,b^2\,\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 )\,\sqrt {a^2-b^2}}{4}}{3\,b^4\,d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )\,\sqrt {a^2-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))),x)

[Out]

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

________________________________________________________________________________________