3.32 \(\int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=148 \[ \frac {c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac {c^4 \tan (e+f x) \sec ^2(e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}-\frac {23 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}+\frac {14 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac {3 c^4 \tan (e+f x)}{a^3 f (\sec (e+f x)+1)^3}+\frac {c^4 x}{a^3} \]

[Out]

c^4*x/a^3+c^4*arctanh(sin(f*x+e))/a^3/f-3*c^4*tan(f*x+e)/a^3/f/(1+sec(f*x+e))^3-1/5*c^4*sec(f*x+e)^2*tan(f*x+e
)/a^3/f/(1+sec(f*x+e))^3+14/5*c^4*tan(f*x+e)/a^3/f/(1+sec(f*x+e))^2-23/5*c^4*tan(f*x+e)/a^3/f/(1+sec(f*x+e))

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Rubi [A]  time = 0.61, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797, 3799, 4000, 3816, 4008, 3998, 3770} \[ \frac {c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac {c^4 \tan (e+f x) \sec ^2(e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}-\frac {23 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}+\frac {14 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac {3 c^4 \tan (e+f x)}{a^3 f (\sec (e+f x)+1)^3}+\frac {c^4 x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*x)/a^3 + (c^4*ArcTanh[Sin[e + f*x]])/(a^3*f) - (3*c^4*Tan[e + f*x])/(a^3*f*(1 + Sec[e + f*x])^3) - (c^4*S
ec[e + f*x]^2*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])^3) + (14*c^4*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])
^2) - (23*c^4*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x]))

Rule 3770

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

Rule 3777

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3796

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

Rule 3797

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(2*m + 1)), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d*csc[e + f*x])/c)^n, (a + b*csc[e + f*x])^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] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3922

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4000

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

Rule 4008

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

Rubi steps

\begin {align*} \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx &=\frac {\int \left (\frac {c^4}{(1+\sec (e+f x))^3}-\frac {4 c^4 \sec (e+f x)}{(1+\sec (e+f x))^3}+\frac {6 c^4 \sec ^2(e+f x)}{(1+\sec (e+f x))^3}-\frac {4 c^4 \sec ^3(e+f x)}{(1+\sec (e+f x))^3}+\frac {c^4 \sec ^4(e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3}\\ &=\frac {c^4 \int \frac {1}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac {c^4 \int \frac {\sec ^4(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac {\left (4 c^4\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac {\left (4 c^4\right ) \int \frac {\sec ^3(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac {\left (6 c^4\right ) \int \frac {\sec ^2(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}\\ &=-\frac {3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac {c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {c^4 \int \frac {(2-5 \sec (e+f x)) \sec ^2(e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac {c^4 \int \frac {-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac {\left (4 c^4\right ) \int \frac {\sec (e+f x) (-3+5 \sec (e+f x))}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac {\left (8 c^4\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}+\frac {\left (18 c^4\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}\\ &=-\frac {3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac {c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac {14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}+\frac {c^4 \int \frac {15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac {c^4 \int \frac {\sec (e+f x) (-14+15 \sec (e+f x))}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac {\left (8 c^4\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac {\left (6 c^4\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{5 a^3}-\frac {\left (28 c^4\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac {c^4 x}{a^3}-\frac {3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac {c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac {14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac {6 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))}+\frac {c^4 \int \sec (e+f x) \, dx}{a^3}-\frac {\left (22 c^4\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac {\left (29 c^4\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac {c^4 x}{a^3}+\frac {c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac {3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac {c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac {14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac {23 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 1.20, size = 231, normalized size = 1.56 \[ \frac {c^4 (\cos (e+f x)-1)^4 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \left (8 \tan \left (\frac {e}{2}\right ) \cot ^3\left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )-4 \tan \left (\frac {e}{2}\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right )+5 \cot ^5\left (\frac {1}{2} (e+f x)\right ) \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+f x\right )-\sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) (8 \cos (e+f x)+3 \cos (2 (e+f x))+9) \csc ^5\left (\frac {1}{2} (e+f x)\right )\right )}{10 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*(-1 + Cos[e + f*x])^4*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*(5*Cot[(e + f*x)/2]^5*(f*x - Log[Cos[(e + f*x)/
2] - Sin[(e + f*x)/2]] + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - (9 + 8*Cos[e + f*x] + 3*Cos[2*(e + f*x)])
*Csc[(e + f*x)/2]^5*Sec[e/2]*Sin[(f*x)/2] + 8*Cot[(e + f*x)/2]^3*Csc[(e + f*x)/2]^2*Tan[e/2] - 4*Cot[(e + f*x)
/2]*Csc[(e + f*x)/2]^4*Tan[e/2]))/(10*a^3*f*(1 + Cos[e + f*x])^3)

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fricas [A]  time = 0.48, size = 242, normalized size = 1.64 \[ \frac {10 \, c^{4} f x \cos \left (f x + e\right )^{3} + 30 \, c^{4} f x \cos \left (f x + e\right )^{2} + 30 \, c^{4} f x \cos \left (f x + e\right ) + 10 \, c^{4} f x + 5 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 5 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 16 \, {\left (3 \, c^{4} \cos \left (f x + e\right )^{2} + 4 \, c^{4} \cos \left (f x + e\right ) + 3 \, c^{4}\right )} \sin \left (f x + e\right )}{10 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(10*c^4*f*x*cos(f*x + e)^3 + 30*c^4*f*x*cos(f*x + e)^2 + 30*c^4*f*x*cos(f*x + e) + 10*c^4*f*x + 5*(c^4*co
s(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + 3*c^4*cos(f*x + e) + c^4)*log(sin(f*x + e) + 1) - 5*(c^4*cos(f*x + e)^3
+ 3*c^4*cos(f*x + e)^2 + 3*c^4*cos(f*x + e) + c^4)*log(-sin(f*x + e) + 1) - 16*(3*c^4*cos(f*x + e)^2 + 4*c^4*c
os(f*x + e) + 3*c^4)*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3
*f)

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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((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^3,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*(-c^4*1/2/a^3*ln(abs(tan((f*x+exp(1))/2)-1))+c^4*1/2/a^3*l
n(abs(tan((f*x+exp(1))/2)+1))+2*c^4*1/2/a^3*(f*x+exp(1))/2+(-2/5*tan((f*x+exp(1))/2)^5*c^4*a^12-2*tan((f*x+exp
(1))/2)*c^4*a^12)/a^15)

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maple [A]  time = 0.77, size = 110, normalized size = 0.74 \[ -\frac {4 c^{4} \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 f \,a^{3}}-\frac {4 c^{4} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{3}}-\frac {c^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,a^{3}}+\frac {c^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,a^{3}}+\frac {2 c^{4} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5/f*c^4/a^3*tan(1/2*e+1/2*f*x)^5-4/f*c^4/a^3*tan(1/2*e+1/2*f*x)-1/f*c^4/a^3*ln(tan(1/2*e+1/2*f*x)-1)+1/f*c^
4/a^3*ln(tan(1/2*e+1/2*f*x)+1)+2/f*c^4/a^3*arctan(tan(1/2*e+1/2*f*x))

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maxima [B]  time = 0.44, size = 396, normalized size = 2.68 \[ -\frac {c^{4} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + c^{4} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}}\right )} + \frac {4 \, c^{4} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {4 \, c^{4} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {18 \, c^{4} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(c^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(
cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e)
+ 1) - 1)/a^3) + c^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) - 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f
*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 4*c^4*(15*sin(f*x + e
)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + 4
*c^4*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x
+ e) + 1)^5)/a^3 - 18*c^4*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

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mupad [B]  time = 1.42, size = 50, normalized size = 0.34 \[ \frac {c^4\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5}+f\,x\right )}{a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^4*(2*atanh(tan(e/2 + (f*x)/2)) - 4*tan(e/2 + (f*x)/2) - (4*tan(e/2 + (f*x)/2)^5)/5 + f*x))/(a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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