3.11 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=150 \[ -\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac {7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {a^2 c^4 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}-\frac {a^2 c^4 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac {a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {5 a^2 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

[Out]

7/16*a^2*c^4*arctanh(sin(f*x+e))/f-5/16*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f-1/8*a^2*c^4*sec(f*x+e)^3*tan(f*x+e)/f+
1/4*a^2*c^4*sec(f*x+e)*tan(f*x+e)^3/f+1/6*a^2*c^4*sec(f*x+e)^3*tan(f*x+e)^3/f-2/5*a^2*c^4*tan(f*x+e)^5/f

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3958, 2611, 3770, 2607, 30, 3768} \[ -\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac {7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {a^2 c^4 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}-\frac {a^2 c^4 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac {a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {5 a^2 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^2*c^4*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^2*c^4*Sec[e + f*x]*Tan[e + f*x])/(16*f) - (a^2*c^4*Sec[e + f*x
]^3*Tan[e + f*x])/(8*f) + (a^2*c^4*Sec[e + f*x]*Tan[e + f*x]^3)/(4*f) + (a^2*c^4*Sec[e + f*x]^3*Tan[e + f*x]^3
)/(6*f) - (2*a^2*c^4*Tan[e + f*x]^5)/(5*f)

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))^2 (c-c \sec (e+f x))^4 \, dx &=\left (a^2 c^2\right ) \int \left (c^2 \sec (e+f x) \tan ^4(e+f x)-2 c^2 \sec ^2(e+f x) \tan ^4(e+f x)+c^2 \sec ^3(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^4\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\left (a^2 c^4\right ) \int \sec ^3(e+f x) \tan ^4(e+f x) \, dx-\left (2 a^2 c^4\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac {1}{2} \left (a^2 c^4\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\frac {1}{4} \left (3 a^2 c^4\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx-\frac {\left (2 a^2 c^4\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {3 a^2 c^4 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac {1}{8} \left (a^2 c^4\right ) \int \sec ^3(e+f x) \, dx+\frac {1}{8} \left (3 a^2 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac {3 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {5 a^2 c^4 \sec (e+f x) \tan (e+f x)}{16 f}-\frac {a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac {1}{16} \left (a^2 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac {7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {5 a^2 c^4 \sec (e+f x) \tan (e+f x)}{16 f}-\frac {a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac {a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac {2 a^2 c^4 \tan ^5(e+f x)}{5 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.40, size = 91, normalized size = 0.61 \[ \frac {a^2 c^4 \left (1680 \tanh ^{-1}(\sin (e+f x))+(330 \sin (e+f x)-240 \sin (2 (e+f x))-445 \sin (3 (e+f x))+192 \sin (4 (e+f x))-135 \sin (5 (e+f x))-48 \sin (6 (e+f x))) \sec ^6(e+f x)\right )}{3840 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^4*(1680*ArcTanh[Sin[e + f*x]] + Sec[e + f*x]^6*(330*Sin[e + f*x] - 240*Sin[2*(e + f*x)] - 445*Sin[3*(e
+ f*x)] + 192*Sin[4*(e + f*x)] - 135*Sin[5*(e + f*x)] - 48*Sin[6*(e + f*x)])))/(3840*f)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 161, normalized size = 1.07 \[ \frac {105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (96 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} + 135 \, a^{2} c^{4} \cos \left (f x + e\right )^{4} - 192 \, a^{2} c^{4} \cos \left (f x + e\right )^{3} + 10 \, a^{2} c^{4} \cos \left (f x + e\right )^{2} + 96 \, a^{2} c^{4} \cos \left (f x + e\right ) - 40 \, a^{2} c^{4}\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(105*a^2*c^4*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 105*a^2*c^4*cos(f*x + e)^6*log(-sin(f*x + e) + 1) -
2*(96*a^2*c^4*cos(f*x + e)^5 + 135*a^2*c^4*cos(f*x + e)^4 - 192*a^2*c^4*cos(f*x + e)^3 + 10*a^2*c^4*cos(f*x +
e)^2 + 96*a^2*c^4*cos(f*x + e) - 40*a^2*c^4)*sin(f*x + e))/(f*cos(f*x + e)^6)

________________________________________________________________________________________

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))^2*(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^2*c^4/32*ln(abs(tan((f*x+exp(1))/2)-1))+7*a^2*c^4/32
*ln(abs(tan((f*x+exp(1))/2)+1))-(105*tan((f*x+exp(1))/2)^11*a^2*c^4-595*tan((f*x+exp(1))/2)^9*a^2*c^4-1686*tan
((f*x+exp(1))/2)^7*a^2*c^4+1386*tan((f*x+exp(1))/2)^5*a^2*c^4-595*tan((f*x+exp(1))/2)^3*a^2*c^4+105*tan((f*x+e
xp(1))/2)*a^2*c^4)*1/240/(tan((f*x+exp(1))/2)^2-1)^6)

________________________________________________________________________________________

maple [A]  time = 1.66, size = 167, normalized size = 1.11 \[ -\frac {a^{2} c^{4} \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{24 f}-\frac {9 a^{2} c^{4} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}+\frac {7 a^{2} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f}-\frac {2 a^{2} c^{4} \tan \left (f x +e \right )}{5 f}+\frac {4 a^{2} c^{4} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{5 f}-\frac {2 a^{2} c^{4} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f}+\frac {a^{2} c^{4} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{6 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*a^2*c^4*sec(f*x+e)^3*tan(f*x+e)/f-9/16*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f+7/16/f*a^2*c^4*ln(sec(f*x+e)+tan(
f*x+e))-2/5*a^2*c^4*tan(f*x+e)/f+4/5/f*a^2*c^4*tan(f*x+e)*sec(f*x+e)^2-2/5/f*a^2*c^4*tan(f*x+e)*sec(f*x+e)^4+1
/6/f*a^2*c^4*tan(f*x+e)*sec(f*x+e)^5

________________________________________________________________________________________

maxima [B]  time = 0.56, size = 321, normalized size = 2.14 \[ -\frac {64 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} - 640 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} + 5 \, a^{2} c^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 30 \, a^{2} c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{2} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 480 \, a^{2} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 960 \, a^{2} c^{4} \tan \left (f x + e\right )}{480 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(64*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^4 - 640*(tan(f*x + e)^3 + 3*tan(f*x
+ e))*a^2*c^4 + 5*a^2*c^4*(2*(15*sin(f*x + e)^5 - 40*sin(f*x + e)^3 + 33*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin
(f*x + e)^4 + 3*sin(f*x + e)^2 - 1) - 15*log(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 30*a^2*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^2*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)) - 480*a^2*c^4*log(sec(f*x + e) + tan(f*x + e)) + 960*a^2*c^4*tan(f*x + e))/f

________________________________________________________________________________________

mupad [B]  time = 5.61, size = 219, normalized size = 1.46 \[ \frac {-\frac {7\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}+\frac {119\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}+\frac {281\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{20}-\frac {231\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20}+\frac {119\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24}-\frac {7\,a^2\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {7\,a^2\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((119*a^2*c^4*tan(e/2 + (f*x)/2)^3)/24 - (231*a^2*c^4*tan(e/2 + (f*x)/2)^5)/20 + (281*a^2*c^4*tan(e/2 + (f*x)/
2)^7)/20 + (119*a^2*c^4*tan(e/2 + (f*x)/2)^9)/24 - (7*a^2*c^4*tan(e/2 + (f*x)/2)^11)/8 - (7*a^2*c^4*tan(e/2 +
(f*x)/2))/8)/(f*(15*tan(e/2 + (f*x)/2)^4 - 6*tan(e/2 + (f*x)/2)^2 - 20*tan(e/2 + (f*x)/2)^6 + 15*tan(e/2 + (f*
x)/2)^8 - 6*tan(e/2 + (f*x)/2)^10 + tan(e/2 + (f*x)/2)^12 + 1)) + (7*a^2*c^4*atanh(tan(e/2 + (f*x)/2)))/(8*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} c^{4} \left (\int \sec {\left (e + f x \right )}\, dx + \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 4 \sec ^{4}{\left (e + f x \right )}\, dx + \int \left (- \sec ^{5}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{6}{\left (e + f x \right )}\right )\, dx + \int \sec ^{7}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c**4*(Integral(sec(e + f*x), x) + Integral(-2*sec(e + f*x)**2, x) + Integral(-sec(e + f*x)**3, x) + Integ
ral(4*sec(e + f*x)**4, x) + Integral(-sec(e + f*x)**5, x) + Integral(-2*sec(e + f*x)**6, x) + Integral(sec(e +
f*x)**7, x))

________________________________________________________________________________________