3.42 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=164 \[ -\frac {7 c^5 \tan ^3(e+f x)}{a^2 f}-\frac {84 c^5 \tan (e+f x)}{a^2 f}+\frac {105 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac {63 c^5 \tan (e+f x) \sec (e+f x)}{2 a^2 f}-\frac {6 c^2 \tan (e+f x) (c-c \sec (e+f x))^3}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^4}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

105/2*c^5*arctanh(sin(f*x+e))/a^2/f-84*c^5*tan(f*x+e)/a^2/f+63/2*c^5*sec(f*x+e)*tan(f*x+e)/a^2/f-6*c^2*(c-c*se
c(f*x+e))^3*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))+2/3*c*(c-c*sec(f*x+e))^4*tan(f*x+e)/f/(a+a*sec(f*x+e))^2-7*c^5*t
an(f*x+e)^3/a^2/f

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Rubi [A]  time = 0.25, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3957, 3791, 3770, 3767, 8, 3768} \[ -\frac {7 c^5 \tan ^3(e+f x)}{a^2 f}-\frac {84 c^5 \tan (e+f x)}{a^2 f}+\frac {105 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac {63 c^5 \tan (e+f x) \sec (e+f x)}{2 a^2 f}-\frac {6 c^2 \tan (e+f x) (c-c \sec (e+f x))^3}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^4}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^2,x]

[Out]

(105*c^5*ArcTanh[Sin[e + f*x]])/(2*a^2*f) - (84*c^5*Tan[e + f*x])/(a^2*f) + (63*c^5*Sec[e + f*x]*Tan[e + f*x])
/(2*a^2*f) - (6*c^2*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a^2 + a^2*Sec[e + f*x])) + (2*c*(c - c*Sec[e + f*
x])^4*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (7*c^5*Tan[e + f*x]^3)/(a^2*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx &=\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {(3 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx}{a}\\ &=-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (21 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^3 \, dx}{a^2}\\ &=-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (21 c^2\right ) \int \left (c^3 \sec (e+f x)-3 c^3 \sec ^2(e+f x)+3 c^3 \sec ^3(e+f x)-c^3 \sec ^4(e+f x)\right ) \, dx}{a^2}\\ &=-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (21 c^5\right ) \int \sec (e+f x) \, dx}{a^2}-\frac {\left (21 c^5\right ) \int \sec ^4(e+f x) \, dx}{a^2}-\frac {\left (63 c^5\right ) \int \sec ^2(e+f x) \, dx}{a^2}+\frac {\left (63 c^5\right ) \int \sec ^3(e+f x) \, dx}{a^2}\\ &=\frac {21 c^5 \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {63 c^5 \sec (e+f x) \tan (e+f x)}{2 a^2 f}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (63 c^5\right ) \int \sec (e+f x) \, dx}{2 a^2}+\frac {\left (21 c^5\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{a^2 f}+\frac {\left (63 c^5\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a^2 f}\\ &=\frac {105 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac {84 c^5 \tan (e+f x)}{a^2 f}+\frac {63 c^5 \sec (e+f x) \tan (e+f x)}{2 a^2 f}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {7 c^5 \tan ^3(e+f x)}{a^2 f}\\ \end {align*}

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Mathematica [B]  time = 1.20, size = 380, normalized size = 2.32 \[ \frac {\cot \left (\frac {1}{2} (e+f x)\right ) \csc ^6\left (\frac {1}{2} (e+f x)\right ) (c-c \sec (e+f x))^5 \left (\sec \left (\frac {e}{2}\right ) \sec (e) \left (-2901 \sin \left (e-\frac {f x}{2}\right )+1197 \sin \left (e+\frac {f x}{2}\right )-3027 \sin \left (2 e+\frac {f x}{2}\right )-273 \sin \left (e+\frac {3 f x}{2}\right )+1827 \sin \left (2 e+\frac {3 f x}{2}\right )-1693 \sin \left (3 e+\frac {3 f x}{2}\right )+1995 \sin \left (e+\frac {5 f x}{2}\right )-117 \sin \left (2 e+\frac {5 f x}{2}\right )+1143 \sin \left (3 e+\frac {5 f x}{2}\right )-969 \sin \left (4 e+\frac {5 f x}{2}\right )+1173 \sin \left (2 e+\frac {7 f x}{2}\right )+117 \sin \left (3 e+\frac {7 f x}{2}\right )+747 \sin \left (4 e+\frac {7 f x}{2}\right )-309 \sin \left (5 e+\frac {7 f x}{2}\right )+494 \sin \left (3 e+\frac {9 f x}{2}\right )+142 \sin \left (4 e+\frac {9 f x}{2}\right )+352 \sin \left (5 e+\frac {9 f x}{2}\right )-1323 \sin \left (\frac {f x}{2}\right )+3247 \sin \left (\frac {3 f x}{2}\right )\right ) \csc ^3\left (\frac {1}{2} (e+f x)\right )+20160 \cos ^3(e+f x) \cot ^3\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{3072 a^2 f (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^2,x]

[Out]

(Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^6*(c - c*Sec[e + f*x])^5*(20160*Cos[e + f*x]^3*Cot[(e + f*x)/2]^3*(Log[Cos[
(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + Csc[(e + f*x)/2]^3*Sec[e/2]*Sec
[e]*(-1323*Sin[(f*x)/2] + 3247*Sin[(3*f*x)/2] - 2901*Sin[e - (f*x)/2] + 1197*Sin[e + (f*x)/2] - 3027*Sin[2*e +
(f*x)/2] - 273*Sin[e + (3*f*x)/2] + 1827*Sin[2*e + (3*f*x)/2] - 1693*Sin[3*e + (3*f*x)/2] + 1995*Sin[e + (5*f
*x)/2] - 117*Sin[2*e + (5*f*x)/2] + 1143*Sin[3*e + (5*f*x)/2] - 969*Sin[4*e + (5*f*x)/2] + 1173*Sin[2*e + (7*f
*x)/2] + 117*Sin[3*e + (7*f*x)/2] + 747*Sin[4*e + (7*f*x)/2] - 309*Sin[5*e + (7*f*x)/2] + 494*Sin[3*e + (9*f*x
)/2] + 142*Sin[4*e + (9*f*x)/2] + 352*Sin[5*e + (9*f*x)/2])))/(3072*a^2*f*(1 + Sec[e + f*x])^2)

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fricas [A]  time = 0.47, size = 210, normalized size = 1.28 \[ \frac {315 \, {\left (c^{5} \cos \left (f x + e\right )^{5} + 2 \, c^{5} \cos \left (f x + e\right )^{4} + c^{5} \cos \left (f x + e\right )^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 315 \, {\left (c^{5} \cos \left (f x + e\right )^{5} + 2 \, c^{5} \cos \left (f x + e\right )^{4} + c^{5} \cos \left (f x + e\right )^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (494 \, c^{5} \cos \left (f x + e\right )^{4} + 679 \, c^{5} \cos \left (f x + e\right )^{3} + 102 \, c^{5} \cos \left (f x + e\right )^{2} - 17 \, c^{5} \cos \left (f x + e\right ) + 2 \, c^{5}\right )} \sin \left (f x + e\right )}{12 \, {\left (a^{2} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} f \cos \left (f x + e\right )^{4} + a^{2} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^5/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/12*(315*(c^5*cos(f*x + e)^5 + 2*c^5*cos(f*x + e)^4 + c^5*cos(f*x + e)^3)*log(sin(f*x + e) + 1) - 315*(c^5*co
s(f*x + e)^5 + 2*c^5*cos(f*x + e)^4 + c^5*cos(f*x + e)^3)*log(-sin(f*x + e) + 1) - 2*(494*c^5*cos(f*x + e)^4 +
679*c^5*cos(f*x + e)^3 + 102*c^5*cos(f*x + e)^2 - 17*c^5*cos(f*x + e) + 2*c^5)*sin(f*x + e))/(a^2*f*cos(f*x +
e)^5 + 2*a^2*f*cos(f*x + e)^4 + a^2*f*cos(f*x + e)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^5/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*((8/3*tan((f*x+exp(1))/2)^3*c^5*a^4+32*tan((f*x+exp(1))/2
)*c^5*a^4)/a^6+(-165*tan((f*x+exp(1))/2)^5*c^5+280*tan((f*x+exp(1))/2)^3*c^5-123*tan((f*x+exp(1))/2)*c^5)*1/6/
a^2/(tan((f*x+exp(1))/2)^2-1)^3+105*c^5*1/4/a^2*ln(abs(tan((f*x+exp(1))/2)-1))-105*c^5*1/4/a^2*ln(abs(tan((f*x
+exp(1))/2)+1)))

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maple [A]  time = 0.80, size = 234, normalized size = 1.43 \[ -\frac {16 c^{5} \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 f \,a^{2}}-\frac {64 c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{2}}+\frac {c^{5}}{3 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )^{3}}+\frac {4 c^{5}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )^{2}}+\frac {55 c^{5}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {105 c^{5} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{2 f \,a^{2}}+\frac {c^{5}}{3 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )^{3}}-\frac {4 c^{5}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )^{2}}+\frac {55 c^{5}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {105 c^{5} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{2 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^5/(a+a*sec(f*x+e))^2,x)

[Out]

-16/3/f*c^5/a^2*tan(1/2*e+1/2*f*x)^3-64/f*c^5/a^2*tan(1/2*e+1/2*f*x)+1/3/f*c^5/a^2/(tan(1/2*e+1/2*f*x)-1)^3+4/
f*c^5/a^2/(tan(1/2*e+1/2*f*x)-1)^2+55/2/f*c^5/a^2/(tan(1/2*e+1/2*f*x)-1)-105/2/f*c^5/a^2*ln(tan(1/2*e+1/2*f*x)
-1)+1/3/f*c^5/a^2/(tan(1/2*e+1/2*f*x)+1)^3-4/f*c^5/a^2/(tan(1/2*e+1/2*f*x)+1)^2+55/2/f*c^5/a^2/(tan(1/2*e+1/2*
f*x)+1)+105/2/f*c^5/a^2*ln(tan(1/2*e+1/2*f*x)+1)

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maxima [B]  time = 0.35, size = 765, normalized size = 4.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^5/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/6*(c^5*(4*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^5/(
cos(f*x + e) + 1)^5)/(a^2 - 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1
)^4 - a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) + (27*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3)/a^2 - 30*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 30*log(sin(f*x + e)/(cos(f*x + e) + 1) -
1)/a^2) + 5*c^5*(6*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2 - 2*a^2*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) + (21*sin(f*x + e)/(cos(f*x + e)
+ 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 21*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 21*log(si
n(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + 10*c^5*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*
x + e) + 1)^3)/a^2 - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1)
- 1)/a^2 + 12*sin(f*x + e)/((a^2 - a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 10*c^5*((9*
sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(cos(f*x + e)
+ 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + 5*c^5*(3*sin(f*x + e)/(cos(f*x + e) + 1) + s
in(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^5*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e
) + 1)^3)/a^2)/f

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mupad [B]  time = 1.73, size = 170, normalized size = 1.04 \[ \frac {55\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-\frac {280\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+41\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^2\right )}-\frac {64\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}-\frac {16\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}+\frac {105\,c^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^5/(cos(e + f*x)*(a + a/cos(e + f*x))^2),x)

[Out]

(55*c^5*tan(e/2 + (f*x)/2)^5 - (280*c^5*tan(e/2 + (f*x)/2)^3)/3 + 41*c^5*tan(e/2 + (f*x)/2))/(f*(3*a^2*tan(e/2
+ (f*x)/2)^2 - 3*a^2*tan(e/2 + (f*x)/2)^4 + a^2*tan(e/2 + (f*x)/2)^6 - a^2)) - (64*c^5*tan(e/2 + (f*x)/2))/(a
^2*f) - (16*c^5*tan(e/2 + (f*x)/2)^3)/(3*a^2*f) + (105*c^5*atanh(tan(e/2 + (f*x)/2)))/(a^2*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {c^{5} \left (\int \left (- \frac {\sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {5 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {10 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {10 \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {5 \sec ^{5}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{6}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**5/(a+a*sec(f*x+e))**2,x)

[Out]

-c**5*(Integral(-sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(5*sec(e + f*x)**2/(sec(e +
f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-10*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) +
Integral(10*sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-5*sec(e + f*x)**5/(sec(e +
f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**6/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2

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