3.1.9 \(\int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^4} \, dx\) [9]

Optimal. Leaf size=133 \[ \frac {a^2 x}{c^4}-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))} \]

[Out]

a^2*x/c^4-4/7*a^2*tan(f*x+e)/c^4/f/(1-sec(f*x+e))^4-12/35*a^2*tan(f*x+e)/c^4/f/(1-sec(f*x+e))^3-59/105*a^2*tan
(f*x+e)/c^4/f/(1-sec(f*x+e))^2-164/105*a^2*tan(f*x+e)/c^4/f/(1-sec(f*x+e))

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Rubi [A]
time = 0.33, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3988, 3862, 4007, 4004, 3879, 3881, 3882} \begin {gather*} -\frac {164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}-\frac {59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}+\frac {a^2 x}{c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/c^4 - (4*a^2*Tan[e + f*x])/(7*c^4*f*(1 - Sec[e + f*x])^4) - (12*a^2*Tan[e + f*x])/(35*c^4*f*(1 - Sec[e
+ f*x])^3) - (59*a^2*Tan[e + f*x])/(105*c^4*f*(1 - Sec[e + f*x])^2) - (164*a^2*Tan[e + f*x])/(105*c^4*f*(1 -
Sec[e + f*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 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]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^4} \, dx &=\frac {\int \left (\frac {a^2}{(1-\sec (e+f x))^4}+\frac {2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^4}+\frac {a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^4}\right ) \, dx}{c^4}\\ &=\frac {a^2 \int \frac {1}{(1-\sec (e+f x))^4} \, dx}{c^4}+\frac {a^2 \int \frac {\sec ^2(e+f x)}{(1-\sec (e+f x))^4} \, dx}{c^4}+\frac {\left (2 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^4} \, dx}{c^4}\\ &=-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {a^2 \int \frac {-7-3 \sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}-\frac {\left (4 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}+\frac {\left (6 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{7 c^4}\\ &=-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}+\frac {a^2 \int \frac {35+20 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}-\frac {\left (8 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}+\frac {\left (12 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{35 c^4}\\ &=-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {a^2 \int \frac {-105-55 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{105 c^4}-\frac {\left (8 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{105 c^4}+\frac {\left (4 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{35 c^4}\\ &=\frac {a^2 x}{c^4}-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {4 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}+\frac {\left (32 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{21 c^4}\\ &=\frac {a^2 x}{c^4}-\frac {4 a^2 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}-\frac {12 a^2 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {59 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {164 a^2 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 227, normalized size = 1.71 \begin {gather*} \frac {a^2 \csc \left (\frac {e}{2}\right ) \csc ^7\left (\frac {1}{2} (e+f x)\right ) \left (3675 f x \cos \left (\frac {f x}{2}\right )-3675 f x \cos \left (e+\frac {f x}{2}\right )-2205 f x \cos \left (e+\frac {3 f x}{2}\right )+2205 f x \cos \left (2 e+\frac {3 f x}{2}\right )+735 f x \cos \left (2 e+\frac {5 f x}{2}\right )-735 f x \cos \left (3 e+\frac {5 f x}{2}\right )-105 f x \cos \left (3 e+\frac {7 f x}{2}\right )+105 f x \cos \left (4 e+\frac {7 f x}{2}\right )-11900 \sin \left (\frac {f x}{2}\right )-10430 \sin \left (e+\frac {f x}{2}\right )+8568 \sin \left (e+\frac {3 f x}{2}\right )+4830 \sin \left (2 e+\frac {3 f x}{2}\right )-3206 \sin \left (2 e+\frac {5 f x}{2}\right )-1260 \sin \left (3 e+\frac {5 f x}{2}\right )+638 \sin \left (3 e+\frac {7 f x}{2}\right )\right )}{13440 c^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Csc[e/2]*Csc[(e + f*x)/2]^7*(3675*f*x*Cos[(f*x)/2] - 3675*f*x*Cos[e + (f*x)/2] - 2205*f*x*Cos[e + (3*f*x)
/2] + 2205*f*x*Cos[2*e + (3*f*x)/2] + 735*f*x*Cos[2*e + (5*f*x)/2] - 735*f*x*Cos[3*e + (5*f*x)/2] - 105*f*x*Co
s[3*e + (7*f*x)/2] + 105*f*x*Cos[4*e + (7*f*x)/2] - 11900*Sin[(f*x)/2] - 10430*Sin[e + (f*x)/2] + 8568*Sin[e +
(3*f*x)/2] + 4830*Sin[2*e + (3*f*x)/2] - 3206*Sin[2*e + (5*f*x)/2] - 1260*Sin[3*e + (5*f*x)/2] + 638*Sin[3*e
+ (7*f*x)/2]))/(13440*c^4*f)

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Maple [A]
time = 0.16, size = 77, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {a^{2} \left (4 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {3}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {4}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 f \,c^{4}}\)	\(77\)
default	\(\frac {a^{2} \left (4 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {3}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {4}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 f \,c^{4}}\)	\(77\)
risch	\(\frac {a^{2} x}{c^{4}}+\frac {2 i a^{2} \left (630 \,{\mathrm e}^{6 i \left (f x +e \right )}-2415 \,{\mathrm e}^{5 i \left (f x +e \right )}+5215 \,{\mathrm e}^{4 i \left (f x +e \right )}-5950 \,{\mathrm e}^{3 i \left (f x +e \right )}+4284 \,{\mathrm e}^{2 i \left (f x +e \right )}-1603 \,{\mathrm e}^{i \left (f x +e \right )}+319\right )}{105 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7}}\)	\(103\)
norman	\(\frac {\frac {a^{2} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {a^{2}}{14 c f}-\frac {13 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{35 c f}+\frac {29 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{30 c f}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {2 a^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {a^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\)	\(170\)

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))^4,x,method=_RETURNVERBOSE)

[Out]

1/2/f*a^2/c^4*(4*arctan(tan(1/2*f*x+1/2*e))-1/7/tan(1/2*f*x+1/2*e)^7+3/5/tan(1/2*f*x+1/2*e)^5-4/3/tan(1/2*f*x+
1/2*e)^3+4/tan(1/2*f*x+1/2*e))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (125) = 250\).
time = 0.50, size = 320, normalized size = 2.41 \begin {gather*} \frac {5 \, a^{2} {\left (\frac {336 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{4}} + \frac {{\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {77 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {315 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}\right )} + \frac {a^{2} {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac {6 \, a^{2} {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, f} \end {gather*}

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))^4,x, algorithm="maxima")

[Out]

1/840*(5*a^2*(336*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^4 + (21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 77*s
in(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3)*(cos(f*x + e) + 1)^7/(c^4*si
n(f*x + e)^7)) + a^2*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 105*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 - 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 6*a^2*(21*sin(f*x + e)^2/(
cos(f*x + e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(co
s(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7))/f

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Fricas [A]
time = 3.96, size = 184, normalized size = 1.38 \begin {gather*} \frac {319 \, a^{2} \cos \left (f x + e\right )^{4} - 327 \, a^{2} \cos \left (f x + e\right )^{3} - 95 \, a^{2} \cos \left (f x + e\right )^{2} + 387 \, a^{2} \cos \left (f x + e\right ) - 164 \, a^{2} + 105 \, {\left (a^{2} f x \cos \left (f x + e\right )^{3} - 3 \, a^{2} f x \cos \left (f x + e\right )^{2} + 3 \, a^{2} f x \cos \left (f x + e\right ) - a^{2} f x\right )} \sin \left (f x + e\right )}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )} \end {gather*}

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))^4,x, algorithm="fricas")

[Out]

1/105*(319*a^2*cos(f*x + e)^4 - 327*a^2*cos(f*x + e)^3 - 95*a^2*cos(f*x + e)^2 + 387*a^2*cos(f*x + e) - 164*a^
2 + 105*(a^2*f*x*cos(f*x + e)^3 - 3*a^2*f*x*cos(f*x + e)^2 + 3*a^2*f*x*cos(f*x + e) - a^2*f*x)*sin(f*x + e))/(
(c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e))

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

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))**4,x)

[Out]

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

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Giac [A]
time = 0.52, size = 88, normalized size = 0.66 \begin {gather*} \frac {\frac {210 \, {\left (f x + e\right )} a^{2}}{c^{4}} + \frac {420 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 140 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 63 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, a^{2}}{c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}}}{210 \, f} \end {gather*}

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))^4,x, algorithm="giac")

[Out]

1/210*(210*(f*x + e)*a^2/c^4 + (420*a^2*tan(1/2*f*x + 1/2*e)^6 - 140*a^2*tan(1/2*f*x + 1/2*e)^4 + 63*a^2*tan(1
/2*f*x + 1/2*e)^2 - 15*a^2)/(c^4*tan(1/2*f*x + 1/2*e)^7))/f

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Mupad [B]
time = 1.50, size = 124, normalized size = 0.93 \begin {gather*} \frac {a^2\,x}{c^4}-\frac {\frac {a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{14}-\frac {3\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{10}+\frac {2\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{3}-2\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{c^4\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7} \end {gather*}

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))^4,x)

[Out]

(a^2*x)/c^4 - ((a^2*cos(e/2 + (f*x)/2)^7)/14 - 2*a^2*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^6 + (2*a^2*cos(e/2
+ (f*x)/2)^3*sin(e/2 + (f*x)/2)^4)/3 - (3*a^2*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^2)/10)/(c^4*f*sin(e/2 +
(f*x)/2)^7)

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