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*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]))

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Rubi [A]  time = 0.309145, antiderivative size = 193, normalized size of antiderivative = 1., 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 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 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 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 8

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

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]

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 (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*}

Mathematica [A]  time = 1.37743, size = 380, normalized size = 1.97 \[ \frac{\cot \left (\frac{1}{2} (e+f x)\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right ) (c-c \sec (e+f x))^5 \left (\sec \left (\frac{e}{2}\right ) \sec (e) \left (7351 \sin \left (e-\frac{f x}{2}\right )-5271 \sin \left (e+\frac{f x}{2}\right )+5545 \sin \left (2 e+\frac{f x}{2}\right )+2205 \sin \left (e+\frac{3 f x}{2}\right )-4515 \sin \left (2 e+\frac{3 f x}{2}\right )+3805 \sin \left (3 e+\frac{3 f x}{2}\right )-4407 \sin \left (e+\frac{5 f x}{2}\right )+585 \sin \left (2 e+\frac{5 f x}{2}\right )-3447 \sin \left (3 e+\frac{5 f x}{2}\right )+1545 \sin \left (4 e+\frac{5 f x}{2}\right )-2155 \sin \left (2 e+\frac{7 f x}{2}\right )-75 \sin \left (3 e+\frac{7 f x}{2}\right )-1755 \sin \left (4 e+\frac{7 f x}{2}\right )+325 \sin \left (5 e+\frac{7 f x}{2}\right )-496 \sin \left (3 e+\frac{9 f x}{2}\right )-80 \sin \left (4 e+\frac{9 f x}{2}\right )-416 \sin \left (5 e+\frac{9 f x}{2}\right )+3465 \sin \left (\frac{f x}{2}\right )-6115 \sin \left (\frac{3 f x}{2}\right )\right ) \csc ^5\left (\frac{1}{2} (e+f x)\right )-40320 \cos ^2(e+f x) \cot ^5\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{5120 a^3 f (\sec (e+f x)+1)^3} \]

Warning: Unable to verify antiderivative.

[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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Maple [A]  time = 0.103, size = 208, normalized size = 1.1 \begin{align*}{\frac{8\,{c}^{5}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+8\,{\frac{{c}^{5} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{f{a}^{3}}}+48\,{\frac{{c}^{5}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}+{\frac{{c}^{5}}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-2}}-{\frac{17\,{c}^{5}}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{63\,{c}^{5}}{2\,f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{c}^{5}}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-2}}-{\frac{17\,{c}^{5}}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{63\,{c}^{5}}{2\,f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}

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

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Maxima [B]  time = 1.06249, size = 918, normalized size = 4.76 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [A]  time = 0.503679, size = 620, normalized size = 3.21 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{c^{5} \left (\int - \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}\, 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 - \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}\, 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 - \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}\, 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}} \end{align*}

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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Giac [A]  time = 1.35371, size = 225, normalized size = 1.17 \begin{align*} -\frac{\frac{315 \, c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{315 \, c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} + \frac{10 \,{\left (17 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 15 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3}} - \frac{16 \,{\left (a^{12} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 5 \, a^{12} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, a^{12} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{15}}}{10 \, f} \end{align*}

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]

-1/10*(315*c^5*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 315*c^5*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 + 10*(1
7*c^5*tan(1/2*f*x + 1/2*e)^3 - 15*c^5*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^3) - 16*(a^12*c^
5*tan(1/2*f*x + 1/2*e)^5 + 5*a^12*c^5*tan(1/2*f*x + 1/2*e)^3 + 30*a^12*c^5*tan(1/2*f*x + 1/2*e))/a^15)/f