3.1 \(\int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^5 \, dx\)

Optimal. Leaf size=196 \[ \frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}+\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {a^2 c^5 \tan (e+f x)}{f}-\frac {19 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^2 c^5 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}+\frac {a^2 c^5 \tan (e+f x) \sec ^3(e+f x)}{8 f}-\frac {3 a^2 c^5 \tan ^3(e+f x) \sec (e+f x)}{4 f}+\frac {17 a^2 c^5 \tan (e+f x) \sec (e+f x)}{16 f}+a^2 c^5 x \]

[Out]

a^2*c^5*x-19/16*a^2*c^5*arctanh(sin(f*x+e))/f-a^2*c^5*tan(f*x+e)/f+17/16*a^2*c^5*sec(f*x+e)*tan(f*x+e)/f+1/8*a
^2*c^5*sec(f*x+e)^3*tan(f*x+e)/f+1/3*a^2*c^5*tan(f*x+e)^3/f-3/4*a^2*c^5*sec(f*x+e)*tan(f*x+e)^3/f-1/6*a^2*c^5*
sec(f*x+e)^3*tan(f*x+e)^3/f+3/5*a^2*c^5*tan(f*x+e)^5/f

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3904, 3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}+\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {a^2 c^5 \tan (e+f x)}{f}-\frac {19 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^2 c^5 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}+\frac {a^2 c^5 \tan (e+f x) \sec ^3(e+f x)}{8 f}-\frac {3 a^2 c^5 \tan ^3(e+f x) \sec (e+f x)}{4 f}+\frac {17 a^2 c^5 \tan (e+f x) \sec (e+f x)}{16 f}+a^2 c^5 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*c^5*x - (19*a^2*c^5*ArcTanh[Sin[e + f*x]])/(16*f) - (a^2*c^5*Tan[e + f*x])/f + (17*a^2*c^5*Sec[e + f*x]*Ta
n[e + f*x])/(16*f) + (a^2*c^5*Sec[e + f*x]^3*Tan[e + f*x])/(8*f) + (a^2*c^5*Tan[e + f*x]^3)/(3*f) - (3*a^2*c^5
*Sec[e + f*x]*Tan[e + f*x]^3)/(4*f) - (a^2*c^5*Sec[e + f*x]^3*Tan[e + f*x]^3)/(6*f) + (3*a^2*c^5*Tan[e + f*x]^
5)/(5*f)

Rule 8

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

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 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 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 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 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])

Rubi steps

\begin {align*} \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^5 \, dx &=\left (a^2 c^2\right ) \int (c-c \sec (e+f x))^3 \tan ^4(e+f x) \, dx\\ &=\left (a^2 c^2\right ) \int \left (c^3 \tan ^4(e+f x)-3 c^3 \sec (e+f x) \tan ^4(e+f x)+3 c^3 \sec ^2(e+f x) \tan ^4(e+f x)-c^3 \sec ^3(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^5\right ) \int \tan ^4(e+f x) \, dx-\left (a^2 c^5\right ) \int \sec ^3(e+f x) \tan ^4(e+f x) \, dx-\left (3 a^2 c^5\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\left (3 a^2 c^5\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}+\frac {1}{2} \left (a^2 c^5\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\left (a^2 c^5\right ) \int \tan ^2(e+f x) \, dx+\frac {1}{4} \left (9 a^2 c^5\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx+\frac {\left (3 a^2 c^5\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a^2 c^5 \tan (e+f x)}{f}+\frac {9 a^2 c^5 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}+\frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}-\frac {1}{8} \left (a^2 c^5\right ) \int \sec ^3(e+f x) \, dx+\left (a^2 c^5\right ) \int 1 \, dx-\frac {1}{8} \left (9 a^2 c^5\right ) \int \sec (e+f x) \, dx\\ &=a^2 c^5 x-\frac {9 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^2 c^5 \tan (e+f x)}{f}+\frac {17 a^2 c^5 \sec (e+f x) \tan (e+f x)}{16 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}+\frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}-\frac {1}{16} \left (a^2 c^5\right ) \int \sec (e+f x) \, dx\\ &=a^2 c^5 x-\frac {19 a^2 c^5 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^2 c^5 \tan (e+f x)}{f}+\frac {17 a^2 c^5 \sec (e+f x) \tan (e+f x)}{16 f}+\frac {a^2 c^5 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^5 \tan ^3(e+f x)}{3 f}-\frac {3 a^2 c^5 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^5 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}+\frac {3 a^2 c^5 \tan ^5(e+f x)}{5 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.13, size = 165, normalized size = 0.84 \[ \frac {a^2 c^5 \sec ^6(e+f x) \left (-210 \sin (e+f x)-120 \sin (2 (e+f x))+865 \sin (3 (e+f x))-768 \sin (4 (e+f x))+435 \sin (5 (e+f x))-88 \sin (6 (e+f x))+1800 (e+f x) \cos (2 (e+f x))+720 e \cos (4 (e+f x))+720 f x \cos (4 (e+f x))+120 e \cos (6 (e+f x))+120 f x \cos (6 (e+f x))-4560 \cos ^6(e+f x) \tanh ^{-1}(\sin (e+f x))+1200 e+1200 f x\right )}{3840 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^5*Sec[e + f*x]^6*(1200*e + 1200*f*x - 4560*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^6 + 1800*(e + f*x)*Cos[2*
(e + f*x)] + 720*e*Cos[4*(e + f*x)] + 720*f*x*Cos[4*(e + f*x)] + 120*e*Cos[6*(e + f*x)] + 120*f*x*Cos[6*(e + f
*x)] - 210*Sin[e + f*x] - 120*Sin[2*(e + f*x)] + 865*Sin[3*(e + f*x)] - 768*Sin[4*(e + f*x)] + 435*Sin[5*(e +
f*x)] - 88*Sin[6*(e + f*x)]))/(3840*f)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 179, normalized size = 0.91 \[ \frac {480 \, a^{2} c^{5} f x \cos \left (f x + e\right )^{6} - 285 \, a^{2} c^{5} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) + 285 \, a^{2} c^{5} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (176 \, a^{2} c^{5} \cos \left (f x + e\right )^{5} - 435 \, a^{2} c^{5} \cos \left (f x + e\right )^{4} + 208 \, a^{2} c^{5} \cos \left (f x + e\right )^{3} + 110 \, a^{2} c^{5} \cos \left (f x + e\right )^{2} - 144 \, a^{2} c^{5} \cos \left (f x + e\right ) + 40 \, a^{2} c^{5}\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(480*a^2*c^5*f*x*cos(f*x + e)^6 - 285*a^2*c^5*cos(f*x + e)^6*log(sin(f*x + e) + 1) + 285*a^2*c^5*cos(f*x
+ e)^6*log(-sin(f*x + e) + 1) - 2*(176*a^2*c^5*cos(f*x + e)^5 - 435*a^2*c^5*cos(f*x + e)^4 + 208*a^2*c^5*cos(
f*x + e)^3 + 110*a^2*c^5*cos(f*x + e)^2 - 144*a^2*c^5*cos(f*x + e) + 40*a^2*c^5)*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((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,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*(-2*a^2*c^5/2*(f*x+exp(1))/2-19*a^2*c^5/32*ln(abs(tan((f*
x+exp(1))/2)-1))+19*a^2*c^5/32*ln(abs(tan((f*x+exp(1))/2)+1))-(525*tan((f*x+exp(1))/2)^11*a^2*c^5-3135*tan((f*
x+exp(1))/2)^9*a^2*c^5+1746*tan((f*x+exp(1))/2)^7*a^2*c^5-366*tan((f*x+exp(1))/2)^5*a^2*c^5-95*tan((f*x+exp(1)
)/2)^3*a^2*c^5+45*tan((f*x+exp(1))/2)*a^2*c^5)*1/240/(tan((f*x+exp(1))/2)^2-1)^6)

________________________________________________________________________________________

maple [A]  time = 1.58, size = 186, normalized size = 0.95 \[ -\frac {11 a^{2} c^{5} \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{24 f}+\frac {29 a^{2} c^{5} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}-\frac {19 c^{5} a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f}-\frac {11 a^{2} c^{5} \tan \left (f x +e \right )}{15 f}-\frac {13 c^{5} a^{2} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{15 f}+a^{2} c^{5} x +\frac {a^{2} c^{5} e}{f}+\frac {3 c^{5} a^{2} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f}-\frac {c^{5} a^{2} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{6 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^5,x)

[Out]

-11/24*a^2*c^5*sec(f*x+e)^3*tan(f*x+e)/f+29/16*a^2*c^5*sec(f*x+e)*tan(f*x+e)/f-19/16/f*c^5*a^2*ln(sec(f*x+e)+t
an(f*x+e))-11/15*a^2*c^5*tan(f*x+e)/f-13/15/f*c^5*a^2*tan(f*x+e)*sec(f*x+e)^2+a^2*c^5*x+1/f*a^2*c^5*e+3/5/f*c^
5*a^2*tan(f*x+e)*sec(f*x+e)^4-1/6/f*c^5*a^2*tan(f*x+e)*sec(f*x+e)^5

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 334, normalized size = 1.70 \[ \frac {96 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c^{5} - 800 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{5} + 480 \, {\left (f x + e\right )} a^{2} c^{5} + 5 \, a^{2} c^{5} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 30 \, a^{2} c^{5} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 600 \, a^{2} c^{5} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 1440 \, a^{2} c^{5} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 480 \, a^{2} c^{5} \tan \left (f x + e\right )}{480 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(96*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^5 - 800*(tan(f*x + e)^3 + 3*tan(f*x +
e))*a^2*c^5 + 480*(f*x + e)*a^2*c^5 + 5*a^2*c^5*(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^5*(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)) - 600*a^2*c^5*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) +
1) + log(sin(f*x + e) - 1)) - 1440*a^2*c^5*log(sec(f*x + e) + tan(f*x + e)) + 480*a^2*c^5*tan(f*x + e))/f

________________________________________________________________________________________

mupad [B]  time = 2.53, size = 228, normalized size = 1.16 \[ a^2\,c^5\,x-\frac {-\frac {35\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}+\frac {209\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{8}-\frac {291\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{20}+\frac {61\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20}+\frac {19\,a^2\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24}-\frac {3\,a^2\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {19\,a^2\,c^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{8\,f} \]

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))^5,x)

[Out]

a^2*c^5*x - ((19*a^2*c^5*tan(e/2 + (f*x)/2)^3)/24 + (61*a^2*c^5*tan(e/2 + (f*x)/2)^5)/20 - (291*a^2*c^5*tan(e/
2 + (f*x)/2)^7)/20 + (209*a^2*c^5*tan(e/2 + (f*x)/2)^9)/8 - (35*a^2*c^5*tan(e/2 + (f*x)/2)^11)/8 - (3*a^2*c^5*
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)) - (19*a^2*c^5*atanh(tan(e/2 + (f*x)/
2)))/(8*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} c^{5} \left (\int \left (-1\right )\, dx + \int 3 \sec {\left (e + f x \right )}\, dx + \int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 5 \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 5 \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx + \int \left (- 3 \sec ^{6}{\left (e + f x \right )}\right )\, dx + \int \sec ^{7}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**5,x)

[Out]

-a**2*c**5*(Integral(-1, x) + Integral(3*sec(e + f*x), x) + Integral(-sec(e + f*x)**2, x) + Integral(-5*sec(e
+ f*x)**3, x) + Integral(5*sec(e + f*x)**4, x) + Integral(sec(e + f*x)**5, x) + Integral(-3*sec(e + f*x)**6, x
) + Integral(sec(e + f*x)**7, x))

________________________________________________________________________________________