3.1 \(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=105 \[ \frac {a c^4 \tan ^5(e+f x)}{5 f}+\frac {4 a c^4 \tan ^3(e+f x)}{3 f}+\frac {7 a c^4 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a c^4 \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {a c^4 \tan (e+f x) \sec (e+f x)}{8 f} \]

[Out]

7/8*a*c^4*arctanh(sin(f*x+e))/f-1/8*a*c^4*sec(f*x+e)*tan(f*x+e)/f-3/4*a*c^4*sec(f*x+e)^3*tan(f*x+e)/f+4/3*a*c^
4*tan(f*x+e)^3/f+1/5*a*c^4*tan(f*x+e)^5/f

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Rubi [A]  time = 0.20, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3958, 2611, 3770, 2607, 30, 3768, 14} \[ \frac {a c^4 \tan ^5(e+f x)}{5 f}+\frac {4 a c^4 \tan ^3(e+f x)}{3 f}+\frac {7 a c^4 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a c^4 \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {a c^4 \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a*c^4*ArcTanh[Sin[e + f*x]])/(8*f) - (a*c^4*Sec[e + f*x]*Tan[e + f*x])/(8*f) - (3*a*c^4*Sec[e + f*x]^3*Tan[
e + f*x])/(4*f) + (4*a*c^4*Tan[e + f*x]^3)/(3*f) + (a*c^4*Tan[e + f*x]^5)/(5*f)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- m), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^4 \, dx &=-\left ((a c) \int \left (c^3 \sec (e+f x) \tan ^2(e+f x)-3 c^3 \sec ^2(e+f x) \tan ^2(e+f x)+3 c^3 \sec ^3(e+f x) \tan ^2(e+f x)-c^3 \sec ^4(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a c^4\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\right )+\left (a c^4\right ) \int \sec ^4(e+f x) \tan ^2(e+f x) \, dx+\left (3 a c^4\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx-\left (3 a c^4\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a c^4 \sec (e+f x) \tan (e+f x)}{2 f}-\frac {3 a c^4 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {1}{2} \left (a c^4\right ) \int \sec (e+f x) \, dx+\frac {1}{4} \left (3 a c^4\right ) \int \sec ^3(e+f x) \, dx+\frac {\left (a c^4\right ) \operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (3 a c^4\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c^4 \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a c^4 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {3 a c^4 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {a c^4 \tan ^3(e+f x)}{f}+\frac {1}{8} \left (3 a c^4\right ) \int \sec (e+f x) \, dx+\frac {\left (a c^4\right ) \operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {7 a c^4 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a c^4 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {3 a c^4 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {4 a c^4 \tan ^3(e+f x)}{3 f}+\frac {a c^4 \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [B]  time = 1.70, size = 499, normalized size = 4.75 \[ -\frac {a c^4 \sec (e) \sec ^5(e+f x) \left (-1920 \sin (2 e+f x)+780 \sin (e+2 f x)+780 \sin (3 e+2 f x)+640 \sin (2 e+3 f x)-720 \sin (4 e+3 f x)+30 \sin (3 e+4 f x)+30 \sin (5 e+4 f x)+272 \sin (4 e+5 f x)+525 \cos (2 e+3 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+525 \cos (4 e+3 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+105 \cos (4 e+5 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+105 \cos (6 e+5 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+1050 \cos (f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+1050 \cos (2 e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-525 \cos (2 e+3 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-525 \cos (4 e+3 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-105 \cos (4 e+5 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-105 \cos (6 e+5 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+800 \sin (f x)\right )}{3840 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(a*c^4*Sec[e]*Sec[e + f*x]^5*(525*Cos[2*e + 3*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 525*Cos[
4*e + 3*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 105*Cos[4*e + 5*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f
*x)/2]] + 105*Cos[6*e + 5*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 1050*Cos[f*x]*(Log[Cos[(e + f*x)/2]
- Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 1050*Cos[2*e + f*x]*(Log[Cos[(e + f*x)/2] -
Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 525*Cos[2*e + 3*f*x]*Log[Cos[(e + f*x)/2] + Si
n[(e + f*x)/2]] - 525*Cos[4*e + 3*f*x]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 105*Cos[4*e + 5*f*x]*Log[Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]] - 105*Cos[6*e + 5*f*x]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 800*Sin[f*
x] - 1920*Sin[2*e + f*x] + 780*Sin[e + 2*f*x] + 780*Sin[3*e + 2*f*x] + 640*Sin[2*e + 3*f*x] - 720*Sin[4*e + 3*
f*x] + 30*Sin[3*e + 4*f*x] + 30*Sin[5*e + 4*f*x] + 272*Sin[4*e + 5*f*x]))/f

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fricas [A]  time = 0.46, size = 131, normalized size = 1.25 \[ \frac {105 \, a c^{4} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a c^{4} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (136 \, a c^{4} \cos \left (f x + e\right )^{4} + 15 \, a c^{4} \cos \left (f x + e\right )^{3} - 112 \, a c^{4} \cos \left (f x + e\right )^{2} + 90 \, a c^{4} \cos \left (f x + e\right ) - 24 \, a c^{4}\right )} \sin \left (f x + e\right )}{240 \, f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(105*a*c^4*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 105*a*c^4*cos(f*x + e)^5*log(-sin(f*x + e) + 1) - 2*(1
36*a*c^4*cos(f*x + e)^4 + 15*a*c^4*cos(f*x + e)^3 - 112*a*c^4*cos(f*x + e)^2 + 90*a*c^4*cos(f*x + e) - 24*a*c^
4)*sin(f*x + e))/(f*cos(f*x + e)^5)

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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)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^4,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*(-7*a*c^4/16*ln(abs(tan((f*x+exp(1))/2)-1))+7*a*c^4/16*ln(
abs(tan((f*x+exp(1))/2)+1))-(105*tan((f*x+exp(1))/2)^9*a*c^4+790*tan((f*x+exp(1))/2)^7*a*c^4-896*tan((f*x+exp(
1))/2)^5*a*c^4+490*tan((f*x+exp(1))/2)^3*a*c^4-105*tan((f*x+exp(1))/2)*a*c^4)*1/120/(tan((f*x+exp(1))/2)^2-1)^
5)

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maple [A]  time = 1.72, size = 130, normalized size = 1.24 \[ -\frac {3 a \,c^{4} \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{4 f}-\frac {a \,c^{4} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{8 f}+\frac {7 a \,c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8 f}-\frac {17 a \,c^{4} \tan \left (f x +e \right )}{15 f}+\frac {14 a \,c^{4} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{15 f}+\frac {a \,c^{4} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*a*c^4*sec(f*x+e)^3*tan(f*x+e)/f-1/8*a*c^4*sec(f*x+e)*tan(f*x+e)/f+7/8/f*a*c^4*ln(sec(f*x+e)+tan(f*x+e))-1
7/15/f*a*c^4*tan(f*x+e)+14/15/f*a*c^4*tan(f*x+e)*sec(f*x+e)^2+1/5/f*a*c^4*tan(f*x+e)*sec(f*x+e)^4

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maxima [B]  time = 0.65, size = 215, normalized size = 2.05 \[ \frac {16 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a c^{4} + 160 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a c^{4} + 45 \, a c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 240 \, a c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 720 \, a c^{4} \tan \left (f x + e\right )}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a*c^4 + 160*(tan(f*x + e)^3 + 3*tan(f*x + e
))*a*c^4 + 45*a*c^4*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin
(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 120*a*c^4*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e)
+ 1) + log(sin(f*x + e) - 1)) + 240*a*c^4*log(sec(f*x + e) + tan(f*x + e)) - 720*a*c^4*tan(f*x + e))/f

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mupad [B]  time = 6.64, size = 176, normalized size = 1.68 \[ \frac {7\,a\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {7\,a\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{4}+\frac {79\,a\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{6}-\frac {224\,a\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{15}+\frac {49\,a\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{6}-\frac {7\,a\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a*c^4*atanh(tan(e/2 + (f*x)/2)))/(4*f) - ((49*a*c^4*tan(e/2 + (f*x)/2)^3)/6 - (7*a*c^4*tan(e/2 + (f*x)/2))/
4 - (224*a*c^4*tan(e/2 + (f*x)/2)^5)/15 + (79*a*c^4*tan(e/2 + (f*x)/2)^7)/6 + (7*a*c^4*tan(e/2 + (f*x)/2)^9)/4
)/(f*(5*tan(e/2 + (f*x)/2)^2 - 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + ta
n(e/2 + (f*x)/2)^10 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**4*(Integral(sec(e + f*x), x) + Integral(-3*sec(e + f*x)**2, x) + Integral(2*sec(e + f*x)**3, x) + Integra
l(2*sec(e + f*x)**4, x) + Integral(-3*sec(e + f*x)**5, x) + Integral(sec(e + f*x)**6, x))

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