3.53 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=193 \[ \frac {42 c^5 \tan (e+f x)}{a^3 f}-\frac {63 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^3 f}-\frac {21 c^5 \tan (e+f x) \sec (e+f x)}{2 a^3 f}+\frac {42 c \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac {6 c^2 \tan (e+f x) (c-c \sec (e+f x))^3}{5 a f (a \sec (e+f x)+a)^2}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^4}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

-63/2*c^5*arctanh(sin(f*x+e))/a^3/f+42*c^5*tan(f*x+e)/a^3/f-21/2*c^5*sec(f*x+e)*tan(f*x+e)/a^3/f-6/5*c^2*(c-c*
sec(f*x+e))^3*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2+2/5*c*(c-c*sec(f*x+e))^4*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+42/5*
c*(c^2-c^2*sec(f*x+e))^2*tan(f*x+e)/f/(a^3+a^3*sec(f*x+e))

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Rubi [A]  time = 0.31, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3957, 3788, 3767, 8, 4046, 3770} \[ \frac {42 c^5 \tan (e+f x)}{a^3 f}-\frac {63 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^3 f}-\frac {21 c^5 \tan (e+f x) \sec (e+f x)}{2 a^3 f}+\frac {42 c \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac {6 c^2 \tan (e+f x) (c-c \sec (e+f x))^3}{5 a f (a \sec (e+f x)+a)^2}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^4}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^3,x]

[Out]

(-63*c^5*ArcTanh[Sin[e + f*x]])/(2*a^3*f) + (42*c^5*Tan[e + f*x])/(a^3*f) - (21*c^5*Sec[e + f*x]*Tan[e + f*x])
/(2*a^3*f) - (6*c^2*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(5*a*f*(a + a*Sec[e + f*x])^2) + (2*c*(c - c*Sec[e +
f*x])^4*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (42*c*(c^2 - c^2*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(a^3
+ a^3*Sec[e + f*x]))

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 3770

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

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]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^3} \, dx &=\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {(9 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (21 c^2\right ) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx}{5 a^2}\\ &=\frac {42 c^3 (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\left (21 c^3\right ) \int \sec (e+f x) (c-c \sec (e+f x))^2 \, dx}{a^3}\\ &=\frac {42 c^3 (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\left (21 c^3\right ) \int \sec (e+f x) \left (c^2+c^2 \sec ^2(e+f x)\right ) \, dx}{a^3}+\frac {\left (42 c^5\right ) \int \sec ^2(e+f x) \, dx}{a^3}\\ &=-\frac {21 c^5 \sec (e+f x) \tan (e+f x)}{2 a^3 f}+\frac {42 c^3 (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\left (63 c^5\right ) \int \sec (e+f x) \, dx}{2 a^3}-\frac {\left (42 c^5\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a^3 f}\\ &=-\frac {63 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^3 f}+\frac {42 c^5 \tan (e+f x)}{a^3 f}-\frac {21 c^5 \sec (e+f x) \tan (e+f x)}{2 a^3 f}+\frac {42 c^3 (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {6 c^2 (c-c \sec (e+f x))^3 \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 1.41, size = 380, normalized size = 1.97 \[ \frac {\cot \left (\frac {1}{2} (e+f x)\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right ) (c-c \sec (e+f x))^5 \left (\sec \left (\frac {e}{2}\right ) \sec (e) \left (7351 \sin \left (e-\frac {f x}{2}\right )-5271 \sin \left (e+\frac {f x}{2}\right )+5545 \sin \left (2 e+\frac {f x}{2}\right )+2205 \sin \left (e+\frac {3 f x}{2}\right )-4515 \sin \left (2 e+\frac {3 f x}{2}\right )+3805 \sin \left (3 e+\frac {3 f x}{2}\right )-4407 \sin \left (e+\frac {5 f x}{2}\right )+585 \sin \left (2 e+\frac {5 f x}{2}\right )-3447 \sin \left (3 e+\frac {5 f x}{2}\right )+1545 \sin \left (4 e+\frac {5 f x}{2}\right )-2155 \sin \left (2 e+\frac {7 f x}{2}\right )-75 \sin \left (3 e+\frac {7 f x}{2}\right )-1755 \sin \left (4 e+\frac {7 f x}{2}\right )+325 \sin \left (5 e+\frac {7 f x}{2}\right )-496 \sin \left (3 e+\frac {9 f x}{2}\right )-80 \sin \left (4 e+\frac {9 f x}{2}\right )-416 \sin \left (5 e+\frac {9 f x}{2}\right )+3465 \sin \left (\frac {f x}{2}\right )-6115 \sin \left (\frac {3 f x}{2}\right )\right ) \csc ^5\left (\frac {1}{2} (e+f x)\right )-40320 \cos ^2(e+f x) \cot ^5\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{5120 a^3 f (\sec (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^3,x]

[Out]

(Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^4*(c - c*Sec[e + f*x])^5*(-40320*Cos[e + f*x]^2*Cot[(e + f*x)/2]^5*(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]^5*Sec[e/2]*Se
c[e]*(3465*Sin[(f*x)/2] - 6115*Sin[(3*f*x)/2] + 7351*Sin[e - (f*x)/2] - 5271*Sin[e + (f*x)/2] + 5545*Sin[2*e +
(f*x)/2] + 2205*Sin[e + (3*f*x)/2] - 4515*Sin[2*e + (3*f*x)/2] + 3805*Sin[3*e + (3*f*x)/2] - 4407*Sin[e + (5*
f*x)/2] + 585*Sin[2*e + (5*f*x)/2] - 3447*Sin[3*e + (5*f*x)/2] + 1545*Sin[4*e + (5*f*x)/2] - 2155*Sin[2*e + (7
*f*x)/2] - 75*Sin[3*e + (7*f*x)/2] - 1755*Sin[4*e + (7*f*x)/2] + 325*Sin[5*e + (7*f*x)/2] - 496*Sin[3*e + (9*f
*x)/2] - 80*Sin[4*e + (9*f*x)/2] - 416*Sin[5*e + (9*f*x)/2])))/(5120*a^3*f*(1 + Sec[e + f*x])^3)

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fricas [A]  time = 0.47, size = 250, normalized size = 1.30 \[ -\frac {315 \, {\left (c^{5} \cos \left (f x + e\right )^{5} + 3 \, c^{5} \cos \left (f x + e\right )^{4} + 3 \, c^{5} \cos \left (f x + e\right )^{3} + c^{5} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 315 \, {\left (c^{5} \cos \left (f x + e\right )^{5} + 3 \, c^{5} \cos \left (f x + e\right )^{4} + 3 \, c^{5} \cos \left (f x + e\right )^{3} + c^{5} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (496 \, c^{5} \cos \left (f x + e\right )^{4} + 1163 \, c^{5} \cos \left (f x + e\right )^{3} + 801 \, c^{5} \cos \left (f x + e\right )^{2} + 65 \, c^{5} \cos \left (f x + e\right ) - 5 \, c^{5}\right )} \sin \left (f x + e\right )}{20 \, {\left (a^{3} f \cos \left (f x + e\right )^{5} + 3 \, a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + a^{3} f \cos \left (f x + e\right )^{2}\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

-1/20*(315*(c^5*cos(f*x + e)^5 + 3*c^5*cos(f*x + e)^4 + 3*c^5*cos(f*x + e)^3 + c^5*cos(f*x + e)^2)*log(sin(f*x
+ e) + 1) - 315*(c^5*cos(f*x + e)^5 + 3*c^5*cos(f*x + e)^4 + 3*c^5*cos(f*x + e)^3 + c^5*cos(f*x + e)^2)*log(-
sin(f*x + e) + 1) - 2*(496*c^5*cos(f*x + e)^4 + 1163*c^5*cos(f*x + e)^3 + 801*c^5*cos(f*x + e)^2 + 65*c^5*cos(
f*x + e) - 5*c^5)*sin(f*x + e))/(a^3*f*cos(f*x + e)^5 + 3*a^3*f*cos(f*x + e)^4 + 3*a^3*f*cos(f*x + e)^3 + a^3*
f*cos(f*x + e)^2)

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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))^3,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*((-4/5*tan((f*x+exp(1))/2)^5*c^5*a^12-4*tan((f*x+exp(1))/
2)^3*c^5*a^12-24*tan((f*x+exp(1))/2)*c^5*a^12)/a^15-(-17*tan((f*x+exp(1))/2)^3*c^5+15*tan((f*x+exp(1))/2)*c^5)
*1/2/a^3/(tan((f*x+exp(1))/2)^2-1)^2-63*c^5*1/4/a^3*ln(abs(tan((f*x+exp(1))/2)-1))+63*c^5*1/4/a^3*ln(abs(tan((
f*x+exp(1))/2)+1)))

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maple [A]  time = 0.80, size = 208, normalized size = 1.08 \[ \frac {8 c^{5} \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 f \,a^{3}}+\frac {8 c^{5} \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{3}}+\frac {48 c^{5} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{3}}-\frac {c^{5}}{2 f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )^{2}}-\frac {17 c^{5}}{2 f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}+\frac {63 c^{5} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{2 f \,a^{3}}+\frac {c^{5}}{2 f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )^{2}}-\frac {17 c^{5}}{2 f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}-\frac {63 c^{5} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{2 f \,a^{3}} \]

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))^3,x)

[Out]

8/5/f*c^5/a^3*tan(1/2*e+1/2*f*x)^5+8/f*c^5/a^3*tan(1/2*e+1/2*f*x)^3+48/f*c^5/a^3*tan(1/2*e+1/2*f*x)-1/2/f*c^5/
a^3/(tan(1/2*e+1/2*f*x)-1)^2-17/2/f*c^5/a^3/(tan(1/2*e+1/2*f*x)-1)+63/2/f*c^5/a^3*ln(tan(1/2*e+1/2*f*x)-1)+1/2
/f*c^5/a^3/(tan(1/2*e+1/2*f*x)+1)^2-17/2/f*c^5/a^3/(tan(1/2*e+1/2*f*x)+1)-63/2/f*c^5/a^3*ln(tan(1/2*e+1/2*f*x)
+1)

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maxima [B]  time = 0.37, size = 680, normalized size = 3.52 \[ \frac {c^{5} {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {7 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + 15 \, c^{5} {\left (\frac {40 \, \sin \left (f x + e\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + 10 \, c^{5} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {10 \, c^{5} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{5} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {15 \, c^{5} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

1/60*(c^5*(60*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 7*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^3 - 2*a^3*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 + a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) + (465*sin(f*x + e)/(cos(f*x + e) + 1)
+ 40*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 390*log(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)/a^3 + 390*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) + 15*c^5*(40*sin(f*x + e)/((a^
3 - a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)) + (85*sin(f*x + e)/(cos(f*x + e) + 1) + 10*si
n(f*x + e)^3/(cos(f*x + e) + 1)^3 + sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x +
e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) + 10*c^5*((105*sin(f*x + e)/(cos(f*x + e)
+ 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x +
e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) + 10*c^5*(15*sin(f*x + e)/(c
os(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + c^5*(
15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) +
1)^5)/a^3 - 15*c^5*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

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mupad [B]  time = 1.64, size = 159, normalized size = 0.82 \[ \frac {48\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^3\,f}-\frac {17\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-15\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a^3\right )}+\frac {8\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{a^3\,f}+\frac {8\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5\,a^3\,f}-\frac {63\,c^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^3\,f} \]

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))^3),x)

[Out]

(48*c^5*tan(e/2 + (f*x)/2))/(a^3*f) - (17*c^5*tan(e/2 + (f*x)/2)^3 - 15*c^5*tan(e/2 + (f*x)/2))/(f*(a^3*tan(e/
2 + (f*x)/2)^4 - 2*a^3*tan(e/2 + (f*x)/2)^2 + a^3)) + (8*c^5*tan(e/2 + (f*x)/2)^3)/(a^3*f) + (8*c^5*tan(e/2 +
(f*x)/2)^5)/(5*a^3*f) - (63*c^5*atanh(tan(e/2 + (f*x)/2)))/(a^3*f)

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

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))**3,x)

[Out]

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

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