3.549 \(\int \frac {\sec ^5(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=140 \[ -\frac {a \left (2 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {\cos (c+d x) (b-a \tan (c+d x))}{\sqrt {a^2+b^2}}\right )}{b^4 d}+\frac {\left (a^2+b^2\right ) \sec (c+d x)}{b^3 d}-\frac {a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {\sec ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 152, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3510, 3486, 3768, 3770, 3509, 206} \[ \frac {\left (a^2+b^2\right ) \sec (c+d x)}{b^3 d}-\frac {a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac {\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {\cos (c+d x) (b-a \tan (c+d x))}{\sqrt {a^2+b^2}}\right )}{b^4 d}-\frac {a \tanh ^{-1}(\sin (c+d x))}{2 b^2 d}-\frac {a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {\sec ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

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

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

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Mathematica [B]  time = 2.00, size = 321, normalized size = 2.29 \[ \frac {48 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )+\sec ^3(c+d x) \left (6 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 b \left (a^2+b^2\right ) \cos (2 (c+d x))+9 a \left (2 a^2+3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+12 a^2 b-6 a b^2 \sin (2 (c+d x))+9 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+20 b^3\right )}{24 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*(a^2 + b^2)^(3/2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sec[c + d*x]^3*(12*a^2*b + 20*b^3 +
12*b*(a^2 + b^2)*Cos[2*(c + d*x)] + 6*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*a*b^2
*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*a*(2*a^2 + 3*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]]) - 6*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] - 9*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 6*a*b^2*Sin[2*
(c + d*x)]))/(24*b^4*d)

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fricas [A]  time = 1.89, size = 259, normalized size = 1.85 \[ \frac {6 \, {\left (a^{2} + b^{2}\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b^{3} + 12 \, {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, b^{4} d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(6*(a^2 + b^2)^(3/2)*cos(d*x + c)^3*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 -
2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b
^2)*cos(d*x + c)^2 + b^2)) - 3*(2*a^3 + 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^3 + 3*a*b^2)*co
s(d*x + c)^3*log(-sin(d*x + c) + 1) - 6*a*b^2*cos(d*x + c)*sin(d*x + c) + 4*b^3 + 12*(a^2*b + b^3)*cos(d*x + c
)^2)/(b^4*d*cos(d*x + c)^3)

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giac [B]  time = 0.81, size = 278, normalized size = 1.99 \[ -\frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac {6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} + 8 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1
/2*c) - 1))/b^4 + 6*(a^4 + 2*a^2*b^2 + b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*
a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6
*a^2*tan(1/2*d*x + 1/2*c)^4 + 12*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d
*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 + 8*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^3))/d

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maple [B]  time = 0.47, size = 488, normalized size = 3.49 \[ \frac {2 \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) a^{4}}{d \,b^{4} \sqrt {a^{2}+b^{2}}}+\frac {4 \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) a^{2}}{d \,b^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}}-\frac {1}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2}}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{4}}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,b^{2}}+\frac {1}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a^{2}}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{4}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*tan(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.45, size = 361, normalized size = 2.58 \[ \frac {\frac {2 \, {\left (6 \, a^{2} + 8 \, b^{2} - \frac {3 \, a b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a b \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {12 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, {\left (a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )}}{b^{3} - \frac {3 \, b^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, b^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {b^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{4}} - \frac {6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*(6*a^2 + 8*b^2 - 3*a*b*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a*b*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 12
*(a^2 + b^2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*(a^2 + 2*b^2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/(b^3 -
3*b^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*b^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - b^3*sin(d*x + c)^6/(c
os(d*x + c) + 1)^6) - 3*(2*a^3 + 3*a*b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^4 + 3*(2*a^3 + 3*a*b^2)*l
og(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^4 - 6*(a^4 + 2*a^2*b^2 + b^4)*log((b - a*sin(d*x + c)/(cos(d*x + c)
+ 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4))/d

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mupad [B]  time = 5.40, size = 724, normalized size = 5.17 \[ \frac {b^3\,\left (\cos \left (c+d\,x\right )+\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {\cos \left (3\,c+3\,d\,x\right )}{3}+\frac {5}{6}\right )-b^2\,\left (\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,a\,\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 {9\,a\,\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 )+b\,\left (\frac {3\,a^2\,\cos \left (c+d\,x\right )}{4}+\frac {a^2}{2}+\frac {a^2\,\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{4}\right )+\frac {\mathrm {atanh}\left (\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\cos \left (3\,c+3\,d\,x\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{2}-\frac {3\,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 )}{2}-\frac {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 )}{2}+\frac {3\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{2}}{b^4\,d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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