3.5.98 \(\int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [498]

Optimal. Leaf size=164 \[ -\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3930, 4167, 4083, 3855, 3916, 2738, 214} \begin {gather*} \frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3930

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)
*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist
[d^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b
*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

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]

Rubi steps

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

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Mathematica [A]
time = 1.12, size = 162, normalized size = 0.99 \begin {gather*} \frac {-\frac {2 a^2 \left (2 a^2-3 b^2\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 )^{3/2}}+2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 b \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+b \tan (c+d x)}{b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.28, size = 209, normalized size = 1.27

method	result	size
derivativedivides	\(\frac {-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}-\frac {2 a^{2} \left (\frac {b a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (2 a^{2}-3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(209\)
default	\(\frac {-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}-\frac {2 a^{2} \left (\frac {b a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (2 a^{2}-3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(209\)
risch	\(\frac {2 i \left (-b \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}-2 a^{3}+b^{2} a \right )}{\left (-a^{2}+b^{2}\right ) d \,b^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{3} d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{3} d}\)	\(533\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/b^2/(tan(1/2*d*x+1/2*c)+1)-2*a/b^3*ln(tan(1/2*d*x+1/2*c)+1)-2/b^3*a^2*(b*a/(a^2-b^2)*tan(1/2*d*x+1/2*c
)/(a*tan(1/2*d*x+1/2*c)^2-b*tan(1/2*d*x+1/2*c)^2-a-b)-(2*a^2-3*b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))+2*a/b^3*ln(tan(1/2*d*x+1/2*c)-1)-1/b^2/(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 \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*sec(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*a^2-4*b^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. 350 vs. \(2 (155) = 310\).
time = 3.98, size = 760, normalized size = 4.63 \begin {gather*} \left [\frac {{\left ({\left (2 \, a^{5} - 3 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\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 (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 ) - 2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (2 \, a^{5} - 3 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \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 ) - {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (155) = 310\).
time = 0.49, size = 331, normalized size = 2.02 \begin {gather*} \frac {2 \, {\left (\frac {{\left (2 \, a^{4} - 3 \, a^{2} b^{2}\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^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.81, size = 3159, normalized size = 19.26 \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)^4*(a + b/cos(c + d*x))^2),x)

[Out]

((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + a^2*b - 2*a^3 - b^3))/(b^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(a*b^2 -
a^2*b - 2*a^3 + b^3))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) - 2*a*tan(c/2 + (d*x)/2
)^2)) + (a*atan(((a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^5*b^3
- 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (2*a*((32*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 +
a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (64*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*
a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3)*2i)/b^3 + (a*((32
*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 +
b^7 - a^2*b^5 - a^3*b^4) + (2*a*((32*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a
*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (64*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*
a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3)*2i)/b^3)/((64*(8*a^8 - 4*a^7*b + 12*a^4*b
^4 + 6*a^5*b^3 - 20*a^6*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (2*a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*
b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (2*a*((3
2*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (
64*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 +
b^7 - a^2*b^5 - a^3*b^4))))/b^3))/b^3 + (2*a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5
+ 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (2*a*((32*(2*a*b^11 - 3*a^2*b^10
- 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (64*a*tan(c/2 + (d*x)/2)*(
2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)
)))/b^3))/b^3))*4i)/(b^3*d) + (a^2*atan(((a^2*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5
+ 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a^2*(2*a^2 - 3*b^2)*((32*(2*a*b^
11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*a^2*ta
n(c/2 + (d*x)/2)*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 +
2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^
3*(a - b)^3)^(1/2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*1i)/(
b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (a^2*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5
+ 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (a^2*(2*a^2 - 3*b^2)*((32*(2*a*b^
11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (32*a^2*ta
n(c/2 + (d*x)/2)*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 +
2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^
3*(a - b)^3)^(1/2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*1i)/(
b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))/((64*(8*a^8 - 4*a^7*b + 12*a^4*b^4 + 6*a^5*b^3 - 20*a^6*b^2))/(a*b^8 +
b^9 - a^2*b^7 - a^3*b^6) + (a^2*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4
+ 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a^2*(2*a^2 - 3*b^2)*((32*(2*a*b^11 - 3*a^2*b^
10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*a^2*tan(c/2 + (d*x)
/2)*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2
*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^
(1/2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(b^9 - 3*a^2*b^7
+ 3*a^4*b^5 - a^6*b^3) - (a^2*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 1
6*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (a^2*(2*a^2 - 3*b^2)*((32*(2*a*b^11 - 3*a^2*b^10
- 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (32*a^2*tan(c/2 + (d*x)/2)
*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^
6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^(1/
2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(b^9 - 3*a^2*b^7 + 3
*a^4*b^5 - a^6*b^3)))*(2*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*2i)/(d*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^
3))

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