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

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

[Out]

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

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Rubi [A]  time = 0.484825, antiderivative size = 157, normalized size of antiderivative = 1., 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 = {3851, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 b^3 d}-\frac{a \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{\tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*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 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 4092

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]

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]

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 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 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 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]

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 ) \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^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 = 2.35536, size = 258, normalized size = 1.64 \[ \frac{\frac{1}{2} \sec ^3(c+d x) \left (4 b \sin (c+d x) \left (\left (3 a^2+2 b^2\right ) \cos (2 (c+d x))+3 a^2-3 a b \cos (c+d x)+4 b^2\right )+9 a \left (2 a^2+b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+3 a \left (2 a^2+b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )-\frac{24 a^4 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{12 b^4 d} \]

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)

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Maple [B]  time = 0.056, size = 400, normalized size = 2.6 \begin{align*} 2\,{\frac{{a}^{4}}{d{b}^{4}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{3\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{a}{2\,d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{{a}^{2}}{d{b}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{a}{2\,d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{3\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{a}{2\,d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{a}^{2}}{d{b}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{a}{2\,d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.92305, size = 1243, normalized size = 7.92 \begin{align*} \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{align*}

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)]

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

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)

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Giac [B]  time = 1.3424, size = 386, normalized size = 2.46 \begin{align*} \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{align*}

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