3.22 \(\int \frac{(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=102 \[ \frac{c^4 \tan (e+f x)}{a^2 f}-\frac{32 c^4 \cot ^3(e+f x)}{3 a^2 f}-\frac{16 c^4 \cot (e+f x)}{a^2 f}+\frac{32 c^4 \csc ^3(e+f x)}{3 a^2 f}-\frac{6 c^4 \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac{c^4 x}{a^2} \]

[Out]

(c^4*x)/a^2 - (6*c^4*ArcTanh[Sin[e + f*x]])/(a^2*f) - (16*c^4*Cot[e + f*x])/(a^2*f) - (32*c^4*Cot[e + f*x]^3)/
(3*a^2*f) + (32*c^4*Csc[e + f*x]^3)/(3*a^2*f) + (c^4*Tan[e + f*x])/(a^2*f)

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Rubi [A]  time = 0.308967, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3904, 3886, 3473, 8, 2606, 2607, 30, 3767, 2621, 302, 207, 2620, 270} \[ \frac{c^4 \tan (e+f x)}{a^2 f}-\frac{32 c^4 \cot ^3(e+f x)}{3 a^2 f}-\frac{16 c^4 \cot (e+f x)}{a^2 f}+\frac{32 c^4 \csc ^3(e+f x)}{3 a^2 f}-\frac{6 c^4 \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac{c^4 x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^2,x]

[Out]

(c^4*x)/a^2 - (6*c^4*ArcTanh[Sin[e + f*x]])/(a^2*f) - (16*c^4*Cot[e + f*x])/(a^2*f) - (32*c^4*Cot[e + f*x]^3)/
(3*a^2*f) + (32*c^4*Csc[e + f*x]^3)/(3*a^2*f) + (c^4*Tan[e + f*x])/(a^2*f)

Rule 3904

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

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx &=\frac{\int \cot ^4(e+f x) (c-c \sec (e+f x))^6 \, dx}{a^2 c^2}\\ &=\frac{\int \left (c^6 \cot ^4(e+f x)-6 c^6 \cot ^3(e+f x) \csc (e+f x)+15 c^6 \cot ^2(e+f x) \csc ^2(e+f x)-20 c^6 \cot (e+f x) \csc ^3(e+f x)+15 c^6 \csc ^4(e+f x)-6 c^6 \csc ^4(e+f x) \sec (e+f x)+c^6 \csc ^4(e+f x) \sec ^2(e+f x)\right ) \, dx}{a^2 c^2}\\ &=\frac{c^4 \int \cot ^4(e+f x) \, dx}{a^2}+\frac{c^4 \int \csc ^4(e+f x) \sec ^2(e+f x) \, dx}{a^2}-\frac{\left (6 c^4\right ) \int \cot ^3(e+f x) \csc (e+f x) \, dx}{a^2}-\frac{\left (6 c^4\right ) \int \csc ^4(e+f x) \sec (e+f x) \, dx}{a^2}+\frac{\left (15 c^4\right ) \int \cot ^2(e+f x) \csc ^2(e+f x) \, dx}{a^2}+\frac{\left (15 c^4\right ) \int \csc ^4(e+f x) \, dx}{a^2}-\frac{\left (20 c^4\right ) \int \cot (e+f x) \csc ^3(e+f x) \, dx}{a^2}\\ &=-\frac{c^4 \cot ^3(e+f x)}{3 a^2 f}-\frac{c^4 \int \cot ^2(e+f x) \, dx}{a^2}+\frac{c^4 \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^4} \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{\left (6 c^4\right ) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{a^2 f}+\frac{\left (6 c^4\right ) \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\csc (e+f x)\right )}{a^2 f}+\frac{\left (15 c^4\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (e+f x)\right )}{a^2 f}-\frac{\left (15 c^4\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (e+f x)\right )}{a^2 f}+\frac{\left (20 c^4\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\csc (e+f x)\right )}{a^2 f}\\ &=-\frac{14 c^4 \cot (e+f x)}{a^2 f}-\frac{31 c^4 \cot ^3(e+f x)}{3 a^2 f}-\frac{6 c^4 \csc (e+f x)}{a^2 f}+\frac{26 c^4 \csc ^3(e+f x)}{3 a^2 f}+\frac{c^4 \int 1 \, dx}{a^2}+\frac{c^4 \operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}+\frac{2}{x^2}\right ) \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{\left (6 c^4\right ) \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (e+f x)\right )}{a^2 f}\\ &=\frac{c^4 x}{a^2}-\frac{16 c^4 \cot (e+f x)}{a^2 f}-\frac{32 c^4 \cot ^3(e+f x)}{3 a^2 f}+\frac{32 c^4 \csc ^3(e+f x)}{3 a^2 f}+\frac{c^4 \tan (e+f x)}{a^2 f}+\frac{\left (6 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{a^2 f}\\ &=\frac{c^4 x}{a^2}-\frac{6 c^4 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac{16 c^4 \cot (e+f x)}{a^2 f}-\frac{32 c^4 \cot ^3(e+f x)}{3 a^2 f}+\frac{32 c^4 \csc ^3(e+f x)}{3 a^2 f}+\frac{c^4 \tan (e+f x)}{a^2 f}\\ \end{align*}

Mathematica [B]  time = 6.01563, size = 448, normalized size = 4.39 \[ \frac{4 c^4 \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \left (-\frac{1}{16} \sec ^3\left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (-64 \cos (e+f x)-16 \cos (2 (e+f x))+90 \cos (2 e+f x)+27 \cos (e+2 f x)+21 \cos (3 e+2 f x)+16 \cos (e)+102 \cos (f x)-48) \csc ^5\left (\frac{1}{2} (e+f x)\right )-22 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \cot ^4\left (\frac{1}{2} (e+f x)\right ) \csc \left (\frac{1}{2} (e+f x)\right )+4 \left (\sin \left (\frac{3 e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \sec ^3\left (\frac{e}{2}\right ) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )-3 \cos (e) \sec ^2\left (\frac{e}{2}\right ) \cot ^5\left (\frac{1}{2} (e+f x)\right ) \left (6 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-6 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+f x\right )-\left (\tan ^2\left (\frac{e}{2}\right )-1\right ) \cot ^3\left (\frac{1}{2} (e+f x)\right ) \left (3 \left (6 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-6 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+f x\right )-8 \tan \left (\frac{e}{2}\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 a^2 f \left (\tan \left (\frac{e}{2}\right )-1\right ) \left (\tan \left (\frac{e}{2}\right )+1\right ) (\cos (e+f x)+1)^2 \left (\cot \left (\frac{1}{2} (e+f x)\right )-1\right ) \left (\cot \left (\frac{1}{2} (e+f x)\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^2,x]

[Out]

(4*c^4*Cos[(e + f*x)/2]*Sin[(e + f*x)/2]^3*(-3*Cos[e]*Cot[(e + f*x)/2]^5*(f*x + 6*Log[Cos[(e + f*x)/2] - Sin[(
e + f*x)/2]] - 6*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sec[e/2]^2 + 4*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*
Sec[e/2]^3*(-Sin[e/2] + Sin[(3*e)/2]) - 22*Cot[(e + f*x)/2]^4*Csc[(e + f*x)/2]*Sec[e/2]*Sin[(f*x)/2] - ((-48 +
16*Cos[e] + 102*Cos[f*x] - 64*Cos[e + f*x] - 16*Cos[2*(e + f*x)] + 90*Cos[2*e + f*x] + 27*Cos[e + 2*f*x] + 21
*Cos[3*e + 2*f*x])*Csc[(e + f*x)/2]^5*Sec[e/2]^3*Sin[(f*x)/2])/16 - Cot[(e + f*x)/2]^3*(3*(f*x + 6*Log[Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2]] - 6*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 8*Csc[(e + f*x)/2]^2*Tan[e/2])*(
-1 + Tan[e/2]^2)))/(3*a^2*f*(1 + Cos[e + f*x])^2*(-1 + Cot[(e + f*x)/2])*(1 + Cot[(e + f*x)/2])*(-1 + Tan[e/2]
)*(1 + Tan[e/2]))

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Maple [A]  time = 0.088, size = 159, normalized size = 1.6 \begin{align*}{\frac{8\,{c}^{4}}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+8\,{\frac{{c}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{2}}}+2\,{\frac{{c}^{4}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{2}}}-{\frac{{c}^{4}}{f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-6\,{\frac{{c}^{4}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{f{a}^{2}}}-{\frac{{c}^{4}}{f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+6\,{\frac{{c}^{4}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{f{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x)

[Out]

8/3/f*c^4/a^2*tan(1/2*f*x+1/2*e)^3+8/f*c^4/a^2*tan(1/2*f*x+1/2*e)+2/f*c^4/a^2*arctan(tan(1/2*f*x+1/2*e))-1/f*c
^4/a^2/(tan(1/2*f*x+1/2*e)+1)-6/f*c^4/a^2*ln(tan(1/2*f*x+1/2*e)+1)-1/f*c^4/a^2/(tan(1/2*f*x+1/2*e)-1)+6/f*c^4/
a^2*ln(tan(1/2*f*x+1/2*e)-1)

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Maxima [B]  time = 1.54283, size = 558, normalized size = 5.47 \begin{align*} \frac{c^{4}{\left (\frac{\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac{12 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac{12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac{a^{2} \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 )}}\right )} + 4 \, c^{4}{\left (\frac{\frac{9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - c^{4}{\left (\frac{\frac{9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac{6 \, c^{4}{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac{4 \, c^{4}{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(c^4*((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*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 4*c^4*((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) - c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 -
12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 6*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(c
os(f*x + e) + 1)^3)/a^2 - 4*c^4*(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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Fricas [B]  time = 1.13238, size = 541, normalized size = 5.3 \begin{align*} \frac{3 \, c^{4} f x \cos \left (f x + e\right )^{3} + 6 \, c^{4} f x \cos \left (f x + e\right )^{2} + 3 \, c^{4} f x \cos \left (f x + e\right ) - 9 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, 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 ) + 9 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, 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 ) +{\left (19 \, c^{4} \cos \left (f x + e\right )^{2} + 38 \, c^{4} \cos \left (f x + e\right ) + 3 \, c^{4}\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/3*(3*c^4*f*x*cos(f*x + e)^3 + 6*c^4*f*x*cos(f*x + e)^2 + 3*c^4*f*x*cos(f*x + e) - 9*(c^4*cos(f*x + e)^3 + 2*
c^4*cos(f*x + e)^2 + c^4*cos(f*x + e))*log(sin(f*x + e) + 1) + 9*(c^4*cos(f*x + e)^3 + 2*c^4*cos(f*x + e)^2 +
c^4*cos(f*x + e))*log(-sin(f*x + e) + 1) + (19*c^4*cos(f*x + e)^2 + 38*c^4*cos(f*x + e) + 3*c^4)*sin(f*x + e))
/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x + e)^2 + a^2*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{4 \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{4 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{1}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**4/(a+a*sec(f*x+e))**2,x)

[Out]

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

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Giac [A]  time = 1.32759, size = 190, normalized size = 1.86 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} c^{4}}{a^{2}} - \frac{18 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac{18 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{6 \, 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^{2}} + \frac{8 \,{\left (a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 \, a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{6}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

1/3*(3*(f*x + e)*c^4/a^2 - 18*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 + 18*c^4*log(abs(tan(1/2*f*x + 1/2*e)
- 1))/a^2 - 6*c^4*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2*e)^2 - 1)*a^2) + 8*(a^4*c^4*tan(1/2*f*x + 1/2*e)^3
+ 3*a^4*c^4*tan(1/2*f*x + 1/2*e))/a^6)/f