3.23 \(\int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=121 \[ \frac {a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac {5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^3 c^4 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac {5 a^3 c^4 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac {5 a^3 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3958, 2611, 3770, 2607, 30} \[ \frac {a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac {5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^3 c^4 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac {5 a^3 c^4 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac {5 a^3 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*c^4*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^3*c^4*Sec[e + f*x]*Tan[e + f*x])/(16*f) + (5*a^3*c^4*Sec[e + f
*x]*Tan[e + f*x]^3)/(24*f) - (a^3*c^4*Sec[e + f*x]*Tan[e + f*x]^5)/(6*f) + (a^3*c^4*Tan[e + f*x]^7)/(7*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 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))^3 (c-c \sec (e+f x))^4 \, dx &=-\left (\left (a^3 c^3\right ) \int \left (c \sec (e+f x) \tan ^6(e+f x)-c \sec ^2(e+f x) \tan ^6(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c^4\right ) \int \sec (e+f x) \tan ^6(e+f x) \, dx\right )+\left (a^3 c^4\right ) \int \sec ^2(e+f x) \tan ^6(e+f x) \, dx\\ &=-\frac {a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac {1}{6} \left (5 a^3 c^4\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\frac {\left (a^3 c^4\right ) \operatorname {Subst}\left (\int x^6 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac {a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac {a^3 c^4 \tan ^7(e+f x)}{7 f}-\frac {1}{8} \left (5 a^3 c^4\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {5 a^3 c^4 \sec (e+f x) \tan (e+f x)}{16 f}+\frac {5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac {a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac {a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac {1}{16} \left (5 a^3 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac {5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {5 a^3 c^4 \sec (e+f x) \tan (e+f x)}{16 f}+\frac {5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac {a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac {a^3 c^4 \tan ^7(e+f x)}{7 f}\\ \end {align*}

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Mathematica [A]  time = 1.64, size = 102, normalized size = 0.84 \[ \frac {a^3 c^4 \left (3360 \tanh ^{-1}(\sin (e+f x))-(-840 \sin (e+f x)+595 \sin (2 (e+f x))+504 \sin (3 (e+f x))+196 \sin (4 (e+f x))-168 \sin (5 (e+f x))+231 \sin (6 (e+f x))+24 \sin (7 (e+f x))) \sec ^7(e+f x)\right )}{10752 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^4*(3360*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^7*(-840*Sin[e + f*x] + 595*Sin[2*(e + f*x)] + 504*Sin[3*(e
+ f*x)] + 196*Sin[4*(e + f*x)] - 168*Sin[5*(e + f*x)] + 231*Sin[6*(e + f*x)] + 24*Sin[7*(e + f*x)])))/(10752*
f)

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fricas [A]  time = 0.51, size = 177, normalized size = 1.46 \[ \frac {105 \, a^{3} c^{4} \cos \left (f x + e\right )^{7} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a^{3} c^{4} \cos \left (f x + e\right )^{7} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (48 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 231 \, a^{3} c^{4} \cos \left (f x + e\right )^{5} - 144 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} - 182 \, a^{3} c^{4} \cos \left (f x + e\right )^{3} + 144 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 56 \, a^{3} c^{4} \cos \left (f x + e\right ) - 48 \, a^{3} c^{4}\right )} \sin \left (f x + e\right )}{672 \, f \cos \left (f x + e\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/672*(105*a^3*c^4*cos(f*x + e)^7*log(sin(f*x + e) + 1) - 105*a^3*c^4*cos(f*x + e)^7*log(-sin(f*x + e) + 1) -
2*(48*a^3*c^4*cos(f*x + e)^6 + 231*a^3*c^4*cos(f*x + e)^5 - 144*a^3*c^4*cos(f*x + e)^4 - 182*a^3*c^4*cos(f*x +
e)^3 + 144*a^3*c^4*cos(f*x + e)^2 + 56*a^3*c^4*cos(f*x + e) - 48*a^3*c^4)*sin(f*x + e))/(f*cos(f*x + e)^7)

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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))^3*(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*(-5*a^3*c^4/32*ln(abs(tan((f*x+exp(1))/2)-1))+5*a^3*c^4/32
*ln(abs(tan((f*x+exp(1))/2)+1))+(-105*tan((f*x+exp(1))/2)^13*a^3*c^4+700*tan((f*x+exp(1))/2)^11*a^3*c^4-1981*t
an((f*x+exp(1))/2)^9*a^3*c^4-3072*tan((f*x+exp(1))/2)^7*a^3*c^4+1981*tan((f*x+exp(1))/2)^5*a^3*c^4-700*tan((f*
x+exp(1))/2)^3*a^3*c^4+105*tan((f*x+exp(1))/2)*a^3*c^4)*1/336/(tan((f*x+exp(1))/2)^2-1)^7)

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maple [A]  time = 1.95, size = 192, normalized size = 1.59 \[ \frac {13 a^{3} c^{4} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{24 f}-\frac {11 a^{3} c^{4} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}+\frac {5 a^{3} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f}-\frac {a^{3} c^{4} \tan \left (f x +e \right )}{7 f}+\frac {3 a^{3} c^{4} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{7 f}-\frac {3 a^{3} c^{4} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{7 f}-\frac {a^{3} c^{4} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{6 f}+\frac {a^{3} c^{4} \tan \left (f x +e \right ) \left (\sec ^{6}\left (f x +e \right )\right )}{7 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/24/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^3-11/16*a^3*c^4*sec(f*x+e)*tan(f*x+e)/f+5/16/f*a^3*c^4*ln(sec(f*x+e)+tan
(f*x+e))-1/7*a^3*c^4*tan(f*x+e)/f+3/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^2-3/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^4-
1/6/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^5+1/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^6

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maxima [B]  time = 0.36, size = 368, normalized size = 3.04 \[ \frac {96 \, {\left (5 \, \tan \left (f x + e\right )^{7} + 21 \, \tan \left (f x + e\right )^{5} + 35 \, \tan \left (f x + e\right )^{3} + 35 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} - 672 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} + 3360 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} + 35 \, a^{3} c^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 630 \, a^{3} c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 2520 \, a^{3} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 3360 \, a^{3} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 3360 \, a^{3} c^{4} \tan \left (f x + e\right )}{3360 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/3360*(96*(5*tan(f*x + e)^7 + 21*tan(f*x + e)^5 + 35*tan(f*x + e)^3 + 35*tan(f*x + e))*a^3*c^4 - 672*(3*tan(f
*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^4 + 3360*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^3*c^4 + 35
*a^3*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*s
in(f*x + e)^2 - 1) - 15*log(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 630*a^3*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))
+ 2520*a^3*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)) + 3360*a^
3*c^4*log(sec(f*x + e) + tan(f*x + e)) - 3360*a^3*c^4*tan(f*x + e))/f

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mupad [B]  time = 5.71, size = 252, normalized size = 2.08 \[ \frac {5\,a^3\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f}-\frac {\frac {5\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{13}}{8}-\frac {25\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{6}+\frac {283\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}+\frac {128\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{7}-\frac {283\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{24}+\frac {25\,a^3\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{6}-\frac {5\,a^3\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^3*c^4*atanh(tan(e/2 + (f*x)/2)))/(8*f) - ((25*a^3*c^4*tan(e/2 + (f*x)/2)^3)/6 - (283*a^3*c^4*tan(e/2 + (f
*x)/2)^5)/24 + (128*a^3*c^4*tan(e/2 + (f*x)/2)^7)/7 + (283*a^3*c^4*tan(e/2 + (f*x)/2)^9)/24 - (25*a^3*c^4*tan(
e/2 + (f*x)/2)^11)/6 + (5*a^3*c^4*tan(e/2 + (f*x)/2)^13)/8 - (5*a^3*c^4*tan(e/2 + (f*x)/2))/8)/(f*(7*tan(e/2 +
(f*x)/2)^2 - 21*tan(e/2 + (f*x)/2)^4 + 35*tan(e/2 + (f*x)/2)^6 - 35*tan(e/2 + (f*x)/2)^8 + 21*tan(e/2 + (f*x)
/2)^10 - 7*tan(e/2 + (f*x)/2)^12 + tan(e/2 + (f*x)/2)^14 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} c^{4} \left (\int \sec {\left (e + f x \right )}\, dx + \int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 3 \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 3 \sec ^{4}{\left (e + f x \right )}\, dx + \int 3 \sec ^{5}{\left (e + f x \right )}\, dx + \int \left (- 3 \sec ^{6}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{7}{\left (e + f x \right )}\right )\, dx + \int \sec ^{8}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3*(c-c*sec(f*x+e))**4,x)

[Out]

a**3*c**4*(Integral(sec(e + f*x), x) + Integral(-sec(e + f*x)**2, x) + Integral(-3*sec(e + f*x)**3, x) + Integ
ral(3*sec(e + f*x)**4, x) + Integral(3*sec(e + f*x)**5, x) + Integral(-3*sec(e + f*x)**6, x) + Integral(-sec(e
+ f*x)**7, x) + Integral(sec(e + f*x)**8, x))

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