3.54 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=164 \[ \frac{7 c^4 \tan (e+f x)}{a^3 f}-\frac{7 c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{14 \tan (e+f x) \left (c^4-c^4 \sec (e+f x)\right )}{3 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{15 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

(-7*c^4*ArcTanh[Sin[e + f*x]])/(a^3*f) + (7*c^4*Tan[e + f*x])/(a^3*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*
x])/(5*f*(a + a*Sec[e + f*x])^3) - (14*(c^2 - c^2*Sec[e + f*x])^2*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2
) + (14*(c^4 - c^4*Sec[e + f*x])*Tan[e + f*x])/(3*f*(a^3 + a^3*Sec[e + f*x]))

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Rubi [A]  time = 0.275221, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3957, 3787, 3770, 3767, 8} \[ \frac{7 c^4 \tan (e+f x)}{a^3 f}-\frac{7 c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{14 \tan (e+f x) \left (c^4-c^4 \sec (e+f x)\right )}{3 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{15 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^3,x]

[Out]

(-7*c^4*ArcTanh[Sin[e + f*x]])/(a^3*f) + (7*c^4*Tan[e + f*x])/(a^3*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*
x])/(5*f*(a + a*Sec[e + f*x])^3) - (14*(c^2 - c^2*Sec[e + f*x])^2*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2
) + (14*(c^4 - c^4*Sec[e + f*x])*Tan[e + f*x])/(3*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 3787

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

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{(7 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{\left (7 c^2\right ) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^2}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{14 \left (c^4-c^4 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac{\left (7 c^3\right ) \int \sec (e+f x) (c-c \sec (e+f x)) \, dx}{a^3}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{14 \left (c^4-c^4 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac{\left (7 c^4\right ) \int \sec (e+f x) \, dx}{a^3}+\frac{\left (7 c^4\right ) \int \sec ^2(e+f x) \, dx}{a^3}\\ &=-\frac{7 c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{14 \left (c^4-c^4 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac{\left (7 c^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a^3 f}\\ &=-\frac{7 c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{7 c^4 \tan (e+f x)}{a^3 f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{14 \left (c^4-c^4 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.31824, size = 826, normalized size = 5.04 \[ \frac{2 \cos (e+f x) \cot \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec \left (\frac{e}{2}\right ) (c-c \sec (e+f x))^4 \sin \left (\frac{f x}{2}\right ) \csc ^7\left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f (\sec (e+f x) a+a)^3}+\frac{2 \cos (e+f x) \cot ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) (c-c \sec (e+f x))^4 \tan \left (\frac{e}{2}\right ) \csc ^6\left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f (\sec (e+f x) a+a)^3}+\frac{8 \cos (e+f x) \cot ^3\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec \left (\frac{e}{2}\right ) (c-c \sec (e+f x))^4 \sin \left (\frac{f x}{2}\right ) \csc ^5\left (\frac{e}{2}+\frac{f x}{2}\right )}{15 f (\sec (e+f x) a+a)^3}+\frac{8 \cos (e+f x) \cot ^4\left (\frac{e}{2}+\frac{f x}{2}\right ) (c-c \sec (e+f x))^4 \tan \left (\frac{e}{2}\right ) \csc ^4\left (\frac{e}{2}+\frac{f x}{2}\right )}{15 f (\sec (e+f x) a+a)^3}+\frac{76 \cos (e+f x) \cot ^5\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec \left (\frac{e}{2}\right ) (c-c \sec (e+f x))^4 \sin \left (\frac{f x}{2}\right ) \csc ^3\left (\frac{e}{2}+\frac{f x}{2}\right )}{15 f (\sec (e+f x) a+a)^3}+\frac{7 \cos (e+f x) \cot ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )-\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right ) (c-c \sec (e+f x))^4 \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f (\sec (e+f x) a+a)^3}-\frac{7 \cos (e+f x) \cot ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )+\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right ) (c-c \sec (e+f x))^4 \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f (\sec (e+f x) a+a)^3}+\frac{\cos (e+f x) \cot ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) (c-c \sec (e+f x))^4 \sin \left (\frac{f x}{2}\right ) \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f (\sec (e+f x) a+a)^3 \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )-\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}+\frac{\cos (e+f x) \cot ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) (c-c \sec (e+f x))^4 \sin \left (\frac{f x}{2}\right ) \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f (\sec (e+f x) a+a)^3 \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )+\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^3,x]

[Out]

(7*Cos[e + f*x]*Cot[e/2 + (f*x)/2]^6*Csc[e/2 + (f*x)/2]^2*Log[Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2]]*(c - c*
Sec[e + f*x])^4)/(2*f*(a + a*Sec[e + f*x])^3) - (7*Cos[e + f*x]*Cot[e/2 + (f*x)/2]^6*Csc[e/2 + (f*x)/2]^2*Log[
Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]]*(c - c*Sec[e + f*x])^4)/(2*f*(a + a*Sec[e + f*x])^3) + (76*Cos[e + f*
x]*Cot[e/2 + (f*x)/2]^5*Csc[e/2 + (f*x)/2]^3*Sec[e/2]*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(15*f*(a + a*Sec[e
+ f*x])^3) + (8*Cos[e + f*x]*Cot[e/2 + (f*x)/2]^3*Csc[e/2 + (f*x)/2]^5*Sec[e/2]*(c - c*Sec[e + f*x])^4*Sin[(f*
x)/2])/(15*f*(a + a*Sec[e + f*x])^3) + (2*Cos[e + f*x]*Cot[e/2 + (f*x)/2]*Csc[e/2 + (f*x)/2]^7*Sec[e/2]*(c - c
*Sec[e + f*x])^4*Sin[(f*x)/2])/(5*f*(a + a*Sec[e + f*x])^3) + (Cos[e + f*x]*Cot[e/2 + (f*x)/2]^6*Csc[e/2 + (f*
x)/2]^2*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(2*f*(a + a*Sec[e + f*x])^3*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x
)/2] - Sin[e/2 + (f*x)/2])) + (Cos[e + f*x]*Cot[e/2 + (f*x)/2]^6*Csc[e/2 + (f*x)/2]^2*(c - c*Sec[e + f*x])^4*S
in[(f*x)/2])/(2*f*(a + a*Sec[e + f*x])^3*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])) + (8
*Cos[e + f*x]*Cot[e/2 + (f*x)/2]^4*Csc[e/2 + (f*x)/2]^4*(c - c*Sec[e + f*x])^4*Tan[e/2])/(15*f*(a + a*Sec[e +
f*x])^3) + (2*Cos[e + f*x]*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*Tan[e/2])/(5*f*(a
+ a*Sec[e + f*x])^3)

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Maple [A]  time = 0.085, size = 160, normalized size = 1. \begin{align*}{\frac{4\,{c}^{4}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{8\,{c}^{4}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+12\,{\frac{{c}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}-{\frac{{c}^{4}}{f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-7\,{\frac{{c}^{4}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{f{a}^{3}}}-{\frac{{c}^{4}}{f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+7\,{\frac{{c}^{4}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^3,x)

[Out]

4/5/f*c^4/a^3*tan(1/2*f*x+1/2*e)^5+8/3/f*c^4/a^3*tan(1/2*f*x+1/2*e)^3+12/f*c^4/a^3*tan(1/2*f*x+1/2*e)-1/f*c^4/
a^3/(tan(1/2*f*x+1/2*e)+1)-7/f*c^4/a^3*ln(tan(1/2*f*x+1/2*e)+1)-1/f*c^4/a^3/(tan(1/2*f*x+1/2*e)-1)+7/f*c^4/a^3
*ln(tan(1/2*f*x+1/2*e)-1)

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Maxima [B]  time = 1.0405, size = 635, normalized size = 3.87 \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))^4/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(3*c^4*(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*sin(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) + 4*c^4*(
(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) + 6*c^4*(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/(c
os(f*x + e) + 1)^5)/a^3 + c^4*(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 - 12*c^4*(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 = 0.489546, size = 579, normalized size = 3.53 \begin{align*} -\frac{105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + 3 \, c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + 3 \, c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (167 \, c^{4} \cos \left (f x + e\right )^{3} + 381 \, c^{4} \cos \left (f x + e\right )^{2} + 277 \, c^{4} \cos \left (f x + e\right ) + 15 \, c^{4}\right )} \sin \left (f x + e\right )}{30 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f \cos \left (f x + e\right )\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))^4/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/30*(105*(c^4*cos(f*x + e)^4 + 3*c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + c^4*cos(f*x + e))*log(sin(f*x +
e) + 1) - 105*(c^4*cos(f*x + e)^4 + 3*c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + c^4*cos(f*x + e))*log(-sin(
f*x + e) + 1) - 2*(167*c^4*cos(f*x + e)^3 + 381*c^4*cos(f*x + e)^2 + 277*c^4*cos(f*x + e) + 15*c^4)*sin(f*x +
e))/(a^3*f*cos(f*x + e)^4 + 3*a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + a^3*f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{4} \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{4 \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{6 \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{4 \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{\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\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))**4/(a+a*sec(f*x+e))**3,x)

[Out]

c**4*(Integral(sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(-4*sec(e
+ f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(6*sec(e + f*x)**3/(sec(e
+ f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(-4*sec(e + f*x)**4/(sec(e + f*x)**3 + 3*sec
(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**5/(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 = 1.37334, size = 200, normalized size = 1.22 \begin{align*} -\frac{\frac{105 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{105 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} + \frac{30 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a^{3}} - \frac{4 \,{\left (3 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 10 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 45 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{15}}}{15 \, 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))^4/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(105*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 105*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 + 30*c^
4*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2*e)^2 - 1)*a^3) - 4*(3*a^12*c^4*tan(1/2*f*x + 1/2*e)^5 + 10*a^12*c^4
*tan(1/2*f*x + 1/2*e)^3 + 45*a^12*c^4*tan(1/2*f*x + 1/2*e))/a^15)/f