3.1.32 \(\int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx\) [32]

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

[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 = 0.44, 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 = {3988, 3862, 4007, 4004, 3879, 3881, 3882, 3884, 4085, 3901, 4093, 4083, 3855} \begin {gather*} \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} \end {gather*}

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 3855

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

Rule 3862

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 3879

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 3881

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 3882

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 3884

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 3901

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)),
Int[(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] /;
FreeQ[{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 3988

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/c)*csc[e + f*x])^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 4004

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 4007

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 4083

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 4085

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 4093

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 = 1.32, size = 231, normalized size = 1.56 \begin {gather*} \frac {c^4 (-1+\cos (e+f x))^4 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \left (5 \cot ^5\left (\frac {1}{2} (e+f x)\right ) \left (f x-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )-(9+8 \cos (e+f x)+3 \cos (2 (e+f x))) \csc ^5\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+8 \cot ^3\left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {e}{2}\right )-4 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {e}{2}\right )\right )}{10 a^3 f (1+\cos (e+f x))^3} \end {gather*}

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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Maple [A]
time = 0.16, size = 77, normalized size = 0.52

method	result	size
derivativedivides	\(\frac {4 c^{4} \left (-\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f \,a^{3}}\)	\(77\)
default	\(\frac {4 c^{4} \left (-\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f \,a^{3}}\)	\(77\)
risch	\(\frac {c^{4} x}{a^{3}}-\frac {16 i c^{4} \left (5 \,{\mathrm e}^{4 i \left (f x +e \right )}+10 \,{\mathrm e}^{3 i \left (f x +e \right )}+20 \,{\mathrm e}^{2 i \left (f x +e \right )}+10 \,{\mathrm e}^{i \left (f x +e \right )}+3\right )}{5 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}+\frac {c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{3} f}-\frac {c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{3} f}\)	\(128\)
norman	\(\frac {\frac {c^{4} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {c^{4} x}{a}+\frac {4 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {12 c^{4} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}+\frac {64 c^{4} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}-\frac {32 c^{4} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}+\frac {12 c^{4} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}-\frac {4 c^{4} \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}+\frac {3 c^{4} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {3 c^{4} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} a^{2}}+\frac {c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{3} f}-\frac {c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{3} f}\)	\(265\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (152) = 304\).
time = 0.50, size = 430, normalized size = 2.91 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 3.19, size = 259, normalized size = 1.75 \begin {gather*} \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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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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Giac [A]
time = 0.56, size = 102, normalized size = 0.69 \begin {gather*} \frac {\frac {5 \, {\left (f x + e\right )} c^{4}}{a^{3}} + \frac {5 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac {5 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (a^{12} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5 \, a^{12} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15}}}{5 \, f} \end {gather*}

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]

1/5*(5*(f*x + e)*c^4/a^3 + 5*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 5*c^4*log(abs(tan(1/2*f*x + 1/2*e) -
1))/a^3 - 4*(a^12*c^4*tan(1/2*f*x + 1/2*e)^5 + 5*a^12*c^4*tan(1/2*f*x + 1/2*e))/a^15)/f

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Mupad [B]
time = 1.42, size = 50, normalized size = 0.34 \begin {gather*} \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} \end {gather*}

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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