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

Optimal. Leaf size=136 \[ \frac{13 c^5 \tan (e+f x)}{2 a^2 f}-\frac{47 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac{112 c^5 \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}+\frac{\tan (e+f x) \left (c^5-c^5 \sec (e+f x)\right )}{2 a^2 f}+\frac{c^5 x}{a^2}-\frac{32 c^5 \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(c^5*x)/a^2 - (47*c^5*ArcTanh[Sin[e + f*x]])/(2*a^2*f) + (13*c^5*Tan[e + f*x])/(2*a^2*f) + (112*c^5*Tan[e + f*
x])/(3*a^2*f*(1 + Sec[e + f*x])) - (32*c^5*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + ((c^5 - c^5*Sec[e + f*
x])*Tan[e + f*x])/(2*a^2*f)

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Rubi [A]  time = 0.402111, antiderivative size = 153, normalized size of antiderivative = 1.12, number of steps used = 26, number of rules used = 14, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3904, 3886, 3473, 8, 2606, 2607, 30, 3767, 2621, 302, 207, 2620, 270, 288} \[ \frac{7 c^5 \tan (e+f x)}{a^2 f}-\frac{64 c^5 \cot ^3(e+f x)}{3 a^2 f}-\frac{48 c^5 \cot (e+f x)}{a^2 f}+\frac{131 c^5 \csc ^3(e+f x)}{6 a^2 f}+\frac{33 c^5 \csc (e+f x)}{2 a^2 f}-\frac{47 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac{c^5 \csc ^3(e+f x) \sec ^2(e+f x)}{2 a^2 f}+\frac{c^5 x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^5*x)/a^2 - (47*c^5*ArcTanh[Sin[e + f*x]])/(2*a^2*f) - (48*c^5*Cot[e + f*x])/(a^2*f) - (64*c^5*Cot[e + f*x]^
3)/(3*a^2*f) + (33*c^5*Csc[e + f*x])/(2*a^2*f) + (131*c^5*Csc[e + f*x]^3)/(6*a^2*f) - (c^5*Csc[e + f*x]^3*Sec[
e + f*x]^2)/(2*a^2*f) + (7*c^5*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]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx &=\frac{\int \cot ^4(e+f x) (c-c \sec (e+f x))^7 \, dx}{a^2 c^2}\\ &=\frac{\int \left (c^7 \cot ^4(e+f x)-7 c^7 \cot ^3(e+f x) \csc (e+f x)+21 c^7 \cot ^2(e+f x) \csc ^2(e+f x)-35 c^7 \cot (e+f x) \csc ^3(e+f x)+35 c^7 \csc ^4(e+f x)-21 c^7 \csc ^4(e+f x) \sec (e+f x)+7 c^7 \csc ^4(e+f x) \sec ^2(e+f x)-c^7 \csc ^4(e+f x) \sec ^3(e+f x)\right ) \, dx}{a^2 c^2}\\ &=\frac{c^5 \int \cot ^4(e+f x) \, dx}{a^2}-\frac{c^5 \int \csc ^4(e+f x) \sec ^3(e+f x) \, dx}{a^2}-\frac{\left (7 c^5\right ) \int \cot ^3(e+f x) \csc (e+f x) \, dx}{a^2}+\frac{\left (7 c^5\right ) \int \csc ^4(e+f x) \sec ^2(e+f x) \, dx}{a^2}+\frac{\left (21 c^5\right ) \int \cot ^2(e+f x) \csc ^2(e+f x) \, dx}{a^2}-\frac{\left (21 c^5\right ) \int \csc ^4(e+f x) \sec (e+f x) \, dx}{a^2}-\frac{\left (35 c^5\right ) \int \cot (e+f x) \csc ^3(e+f x) \, dx}{a^2}+\frac{\left (35 c^5\right ) \int \csc ^4(e+f x) \, dx}{a^2}\\ &=-\frac{c^5 \cot ^3(e+f x)}{3 a^2 f}-\frac{c^5 \int \cot ^2(e+f x) \, dx}{a^2}+\frac{c^5 \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (e+f x)\right )}{a^2 f}+\frac{\left (7 c^5\right ) \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\csc (e+f x)\right )}{a^2 f}+\frac{\left (7 c^5\right ) \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 (21 c^5\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (e+f x)\right )}{a^2 f}+\frac{\left (21 c^5\right ) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{a^2 f}+\frac{\left (35 c^5\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\csc (e+f x)\right )}{a^2 f}-\frac{\left (35 c^5\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (e+f x)\right )}{a^2 f}\\ &=-\frac{34 c^5 \cot (e+f x)}{a^2 f}-\frac{19 c^5 \cot ^3(e+f x)}{a^2 f}-\frac{7 c^5 \csc (e+f x)}{a^2 f}+\frac{14 c^5 \csc ^3(e+f x)}{a^2 f}-\frac{c^5 \csc ^3(e+f x) \sec ^2(e+f x)}{2 a^2 f}+\frac{c^5 \int 1 \, dx}{a^2}+\frac{\left (5 c^5\right ) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{2 a^2 f}+\frac{\left (7 c^5\right ) \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 (21 c^5\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^5 x}{a^2}-\frac{48 c^5 \cot (e+f x)}{a^2 f}-\frac{64 c^5 \cot ^3(e+f x)}{3 a^2 f}+\frac{14 c^5 \csc (e+f x)}{a^2 f}+\frac{21 c^5 \csc ^3(e+f x)}{a^2 f}-\frac{c^5 \csc ^3(e+f x) \sec ^2(e+f x)}{2 a^2 f}+\frac{7 c^5 \tan (e+f x)}{a^2 f}+\frac{\left (5 c^5\right ) \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (e+f x)\right )}{2 a^2 f}+\frac{\left (21 c^5\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{a^2 f}\\ &=\frac{c^5 x}{a^2}-\frac{21 c^5 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac{48 c^5 \cot (e+f x)}{a^2 f}-\frac{64 c^5 \cot ^3(e+f x)}{3 a^2 f}+\frac{33 c^5 \csc (e+f x)}{2 a^2 f}+\frac{131 c^5 \csc ^3(e+f x)}{6 a^2 f}-\frac{c^5 \csc ^3(e+f x) \sec ^2(e+f x)}{2 a^2 f}+\frac{7 c^5 \tan (e+f x)}{a^2 f}+\frac{\left (5 c^5\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{2 a^2 f}\\ &=\frac{c^5 x}{a^2}-\frac{47 c^5 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac{48 c^5 \cot (e+f x)}{a^2 f}-\frac{64 c^5 \cot ^3(e+f x)}{3 a^2 f}+\frac{33 c^5 \csc (e+f x)}{2 a^2 f}+\frac{131 c^5 \csc ^3(e+f x)}{6 a^2 f}-\frac{c^5 \csc ^3(e+f x) \sec ^2(e+f x)}{2 a^2 f}+\frac{7 c^5 \tan (e+f x)}{a^2 f}\\ \end{align*}

Mathematica [B]  time = 3.01918, size = 384, normalized size = 2.82 \[ \frac{\cos ^3(e+f x) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^6\left (\frac{1}{2} (e+f x)\right ) (c-c \sec (e+f x))^5 \left (-\frac{64 \tan \left (\frac{e}{2}\right ) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{f}-\frac{64 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc ^3\left (\frac{1}{2} (e+f x)\right )}{f}+3 \cot ^3\left (\frac{1}{2} (e+f x)\right ) \left (-\frac{28 \sin (f x)}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{1}{f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{1}{f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{94 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{94 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}-4 x\right )-\frac{320 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \cot ^2\left (\frac{1}{2} (e+f x)\right ) \csc \left (\frac{1}{2} (e+f x)\right )}{f}\right )}{96 a^2 (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^3*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^6*(c - c*Sec[e + f*x])^5*((-320*Cot[(e + f*x)/2]^2*Csc[(e +
f*x)/2]*Sec[e/2]*Sin[(f*x)/2])/f - (64*Csc[(e + f*x)/2]^3*Sec[e/2]*Sin[(f*x)/2])/f + 3*Cot[(e + f*x)/2]^3*(-4*
x - (94*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])/f + (94*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])/f + 1/(f*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2) - 1/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2) - (28*Sin[f*x])/(f*(Co
s[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]))) - (64*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*Tan[e/2])/f))/(96*a^2*(1 + Sec[e + f*x])^2)

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Maple [A]  time = 0.105, size = 207, normalized size = 1.5 \begin{align*}{\frac{16\,{c}^{5}}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+32\,{\frac{{c}^{5}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{2}}}+2\,{\frac{{c}^{5}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{2}}}+{\frac{{c}^{5}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-2}}-{\frac{15\,{c}^{5}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{47\,{c}^{5}}{2\,f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{c}^{5}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-2}}-{\frac{15\,{c}^{5}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{47\,{c}^{5}}{2\,f{a}^{2}}\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((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*f*x+1/2*e)^3+32/f*c^5/a^2*tan(1/2*f*x+1/2*e)+2/f*c^5/a^2*arctan(tan(1/2*f*x+1/2*e))+1/2
/f*c^5/a^2/(tan(1/2*f*x+1/2*e)+1)^2-15/2/f*c^5/a^2/(tan(1/2*f*x+1/2*e)+1)-47/2/f*c^5/a^2*ln(tan(1/2*f*x+1/2*e)
+1)-1/2/f*c^5/a^2/(tan(1/2*f*x+1/2*e)-1)^2-15/2/f*c^5/a^2/(tan(1/2*f*x+1/2*e)-1)+47/2/f*c^5/a^2*ln(tan(1/2*f*x
+1/2*e)-1)

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Maxima [B]  time = 1.59072, size = 814, normalized size = 5.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(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*sin(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(sin(f*x +
e)/(cos(f*x + e) + 1) - 1)/a^2) + 5*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) - 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 - 12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 10*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 - 5*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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Fricas [A]  time = 1.15021, size = 601, normalized size = 4.42 \begin{align*} \frac{12 \, c^{5} f x \cos \left (f x + e\right )^{4} + 24 \, c^{5} f x \cos \left (f x + e\right )^{3} + 12 \, c^{5} f x \cos \left (f x + e\right )^{2} - 141 \,{\left (c^{5} \cos \left (f x + e\right )^{4} + 2 \, 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 ) + 141 \,{\left (c^{5} \cos \left (f x + e\right )^{4} + 2 \, 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 (202 \, c^{5} \cos \left (f x + e\right )^{3} + 305 \, c^{5} \cos \left (f x + e\right )^{2} + 36 \, c^{5} \cos \left (f x + e\right ) - 3 \, c^{5}\right )} \sin \left (f x + e\right )}{12 \,{\left (a^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*c^5*f*x*cos(f*x + e)^4 + 24*c^5*f*x*cos(f*x + e)^3 + 12*c^5*f*x*cos(f*x + e)^2 - 141*(c^5*cos(f*x + e
)^4 + 2*c^5*cos(f*x + e)^3 + c^5*cos(f*x + e)^2)*log(sin(f*x + e) + 1) + 141*(c^5*cos(f*x + e)^4 + 2*c^5*cos(f
*x + e)^3 + c^5*cos(f*x + e)^2)*log(-sin(f*x + e) + 1) + 2*(202*c^5*cos(f*x + e)^3 + 305*c^5*cos(f*x + e)^2 +
36*c^5*cos(f*x + e) - 3*c^5)*sin(f*x + e))/(a^2*f*cos(f*x + e)^4 + 2*a^2*f*cos(f*x + e)^3 + a^2*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{5 \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{10 \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{10 \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{5 \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{\sec ^{5}{\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))**5/(a+a*sec(f*x+e))**2,x)

[Out]

-c**5*(Integral(5*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-10*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(-5*sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**5/(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 = 1.46604, size = 217, normalized size = 1.6 \begin{align*} \frac{\frac{6 \,{\left (f x + e\right )} c^{5}}{a^{2}} - \frac{141 \, c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac{141 \, c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (15 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 13 \, 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^{2}} + \frac{32 \,{\left (a^{4} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, a^{4} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{6}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(f*x + e)*c^5/a^2 - 141*c^5*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 + 141*c^5*log(abs(tan(1/2*f*x + 1/2*
e) - 1))/a^2 - 6*(15*c^5*tan(1/2*f*x + 1/2*e)^3 - 13*c^5*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2
*a^2) + 32*(a^4*c^5*tan(1/2*f*x + 1/2*e)^3 + 6*a^4*c^5*tan(1/2*f*x + 1/2*e))/a^6)/f