3.43 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=150 \[ -\frac {70 c^4 \tan (e+f x)}{3 a^2 f}+\frac {35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac {35 c^4 \tan (e+f x) \sec (e+f x)}{6 a^2 f}-\frac {14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^3}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

35/2*c^4*arctanh(sin(f*x+e))/a^2/f-70/3*c^4*tan(f*x+e)/a^2/f+35/6*c^4*sec(f*x+e)*tan(f*x+e)/a^2/f+2/3*c*(c-c*s
ec(f*x+e))^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^2-14/3*(c^2-c^2*sec(f*x+e))^2*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3957, 3788, 3767, 8, 4046, 3770} \[ -\frac {70 c^4 \tan (e+f x)}{3 a^2 f}+\frac {35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac {35 c^4 \tan (e+f x) \sec (e+f x)}{6 a^2 f}-\frac {14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^3}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*c^4*ArcTanh[Sin[e + f*x]])/(2*a^2*f) - (70*c^4*Tan[e + f*x])/(3*a^2*f) + (35*c^4*Sec[e + f*x]*Tan[e + f*x]
)/(6*a^2*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (14*(c^2 - c^2*Sec[e +
f*x])^2*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

Rule 8

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

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 3770

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx &=\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {(7 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (35 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^2 \, dx}{3 a^2}\\ &=\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (35 c^2\right ) \int \sec (e+f x) \left (c^2+c^2 \sec ^2(e+f x)\right ) \, dx}{3 a^2}-\frac {\left (70 c^4\right ) \int \sec ^2(e+f x) \, dx}{3 a^2}\\ &=\frac {35 c^4 \sec (e+f x) \tan (e+f x)}{6 a^2 f}+\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (35 c^4\right ) \int \sec (e+f x) \, dx}{2 a^2}+\frac {\left (70 c^4\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{3 a^2 f}\\ &=\frac {35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac {70 c^4 \tan (e+f x)}{3 a^2 f}+\frac {35 c^4 \sec (e+f x) \tan (e+f x)}{6 a^2 f}+\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.97, size = 349, normalized size = 2.33 \[ \frac {c^4 \sin ^3\left (\frac {1}{2} (e+f x)\right ) \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (-32 \tan \left (\frac {e}{2}\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )-32 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \csc ^3\left (\frac {1}{2} (e+f x)\right )+3 \cot ^3\left (\frac {1}{2} (e+f x)\right ) \left (-\frac {24 \sin (f x)}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {1}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {1}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}-70 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+70 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-256 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \cot ^2\left (\frac {1}{2} (e+f x)\right ) \csc \left (\frac {1}{2} (e+f x)\right )\right )}{3 a^2 f (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*Cos[(e + f*x)/2]*Sec[e + f*x]^2*Sin[(e + f*x)/2]^3*(-256*Cot[(e + f*x)/2]^2*Csc[(e + f*x)/2]*Sec[e/2]*Sin
[(f*x)/2] - 32*Csc[(e + f*x)/2]^3*Sec[e/2]*Sin[(f*x)/2] + 3*Cot[(e + f*x)/2]^3*(-70*Log[Cos[(e + f*x)/2] - Sin
[(e + f*x)/2]] + 70*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2) - (C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^(-2) - (24*Sin[f*x])/((Cos[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]))) - 32*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2
*Tan[e/2]))/(3*a^2*f*(1 + Sec[e + f*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 197, normalized size = 1.31 \[ \frac {105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + 2 \, c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + 2 \, c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (164 \, c^{4} \cos \left (f x + e\right )^{3} + 229 \, c^{4} \cos \left (f x + e\right )^{2} + 30 \, c^{4} \cos \left (f x + e\right ) - 3 \, c^{4}\right )} \sin \left (f x + e\right )}{12 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(105*(c^4*cos(f*x + e)^4 + 2*c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*log(sin(f*x + e) + 1) - 105*(c^4*co
s(f*x + e)^4 + 2*c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*log(-sin(f*x + e) + 1) - 2*(164*c^4*cos(f*x + e)^3 +
229*c^4*cos(f*x + e)^2 + 30*c^4*cos(f*x + e) - 3*c^4)*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)

________________________________________________________________________________________

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)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,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*((-4/3*tan((f*x+exp(1))/2)^3*c^4*a^4-12*tan((f*x+exp(1))/2
)*c^4*a^4)/a^6-(-13*tan((f*x+exp(1))/2)^3*c^4+11*tan((f*x+exp(1))/2)*c^4)*1/2/a^2/(tan((f*x+exp(1))/2)^2-1)^2-
35*c^4*1/4/a^2*ln(abs(tan((f*x+exp(1))/2)-1))+35*c^4*1/4/a^2*ln(abs(tan((f*x+exp(1))/2)+1)))

________________________________________________________________________________________

maple [A]  time = 0.74, size = 186, normalized size = 1.24 \[ -\frac {8 c^{4} \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 f \,a^{2}}-\frac {24 c^{4} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{2}}+\frac {c^{4}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )^{2}}+\frac {13 c^{4}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {35 c^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{2 f \,a^{2}}-\frac {c^{4}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )^{2}}+\frac {13 c^{4}}{2 f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {35 c^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{2 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3/f*c^4/a^2*tan(1/2*e+1/2*f*x)^3-24/f*c^4/a^2*tan(1/2*e+1/2*f*x)+1/2/f*c^4/a^2/(tan(1/2*e+1/2*f*x)-1)^2+13/
2/f*c^4/a^2/(tan(1/2*e+1/2*f*x)-1)-35/2/f*c^4/a^2*ln(tan(1/2*e+1/2*f*x)-1)-1/2/f*c^4/a^2/(tan(1/2*e+1/2*f*x)+1
)^2+13/2/f*c^4/a^2/(tan(1/2*e+1/2*f*x)+1)+35/2/f*c^4/a^2*ln(tan(1/2*e+1/2*f*x)+1)

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 531, normalized size = 3.54 \[ -\frac {c^{4} {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + 4 \, c^{4} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 6 \, c^{4} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {4 \, c^{4} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {c^{4} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c^4*(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) + 4*c^4*((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))) + 6*c^4*((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) + 4*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x +
e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)
/a^2)/f

________________________________________________________________________________________

mupad [B]  time = 1.68, size = 136, normalized size = 0.91 \[ \frac {13\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-11\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a^2\right )}-\frac {24\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}-\frac {8\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}+\frac {35\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*c^4*tan(e/2 + (f*x)/2)^3 - 11*c^4*tan(e/2 + (f*x)/2))/(f*(a^2*tan(e/2 + (f*x)/2)^4 - 2*a^2*tan(e/2 + (f*x)
/2)^2 + a^2)) - (24*c^4*tan(e/2 + (f*x)/2))/(a^2*f) - (8*c^4*tan(e/2 + (f*x)/2)^3)/(3*a^2*f) + (35*c^4*atanh(t
an(e/2 + (f*x)/2)))/(a^2*f)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**4*(Integral(sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-4*sec(e + f*x)**2/(sec(e +
f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(6*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + In
tegral(-4*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))/a**2

________________________________________________________________________________________