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

Optimal. Leaf size=119 \[ -\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 119, 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 = {3851, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3770

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

Rule 3831

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

Rule 3851

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)*Simp
[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 3998

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 4082

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

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Mathematica [A]  time = 1.14, size = 238, normalized size = 2.00 \[ \frac {-4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 a^3 \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 a b \tan (c+d x)+\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.67, size = 485, normalized size = 4.08 \[ \left [\frac {2 \, \sqrt {a^{2} - b^{2}} a^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, \sqrt {-a^{2} + b^{2}} a^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 211, normalized size = 1.77 \[ -\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3}}{\sqrt {-a^{2} + b^{2}} b^{3}} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.42, size = 262, normalized size = 2.20 \[ -\frac {2 a^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}{d \,b^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d b}-\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}}{d \,b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.65, size = 1002, normalized size = 8.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(c + d*x)/(2*b*d*(cos(2*c + 2*d*x)/2 + 1/2)) + atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(2*b*d*(cos(2*c
+ 2*d*x)/2 + 1/2)) + (a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^3*d*(cos(2*c + 2*d*x)/2 + 1/2)) +
(atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/(2*b*d*(cos(2*c + 2*d*x)/2 + 1/2)) - (a*sin(2*
c + 2*d*x))/(2*b^2*d*(cos(2*c + 2*d*x)/2 + 1/2)) - (a^3*atan(((8*a^6*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^8*si
n(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))*1i)/(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))))*1i)/(b^3*d*(a^2 - b^2)^(1/2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (a^2*atanh(sin(c
/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/(b^3*d*(cos(2*c + 2*d*x)/2 + 1/2)) - (a^3*cos(2*c + 2*d*x)
*atan(((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))*1i)/(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))))*1i)/(b^3*d*(a^2 - b^
2)^(1/2)*(cos(2*c + 2*d*x)/2 + 1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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