\(\int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx\) [8]

Optimal result

Integrand size = 26, antiderivative size = 102 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {a^2 x}{c^3}-\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac {8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac {23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))} \]

[Out]

a^2*x/c^3-4/5*a^2*tan(f*x+e)/c^3/f/(1-sec(f*x+e))^3-8/15*a^2*tan(f*x+e)/c^3/f/(1-sec(f*x+e))^2-23/15*a^2*tan(f
*x+e)/c^3/f/(1-sec(f*x+e))

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3988, 3862, 4007, 4004, 3879, 3881, 3882} \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=-\frac {23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}-\frac {8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac {a^2 x}{c^3} \]

[In]

Int[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^3,x]

[Out]

(a^2*x)/c^3 - (4*a^2*Tan[e + f*x])/(5*c^3*f*(1 - Sec[e + f*x])^3) - (8*a^2*Tan[e + f*x])/(15*c^3*f*(1 - Sec[e
+ f*x])^2) - (23*a^2*Tan[e + f*x])/(15*c^3*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*} \text {integral}& = \frac {\int \left (\frac {a^2}{(1-\sec (e+f x))^3}+\frac {2 a^2 \sec (e+f x)}{(1-\sec (e+f x))^3}+\frac {a^2 \sec ^2(e+f x)}{(1-\sec (e+f x))^3}\right ) \, dx}{c^3} \\ & = \frac {a^2 \int \frac {1}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac {a^2 \int \frac {\sec ^2(e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3}+\frac {\left (2 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^3} \, dx}{c^3} \\ & = -\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac {a^2 \int \frac {-5-2 \sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}-\frac {\left (3 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3}+\frac {\left (4 a^2\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{5 c^3} \\ & = -\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac {8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}+\frac {a^2 \int \frac {15+7 \sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3}-\frac {a^2 \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{5 c^3}+\frac {\left (4 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3} \\ & = \frac {a^2 x}{c^3}-\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac {8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac {a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))}+\frac {\left (22 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{15 c^3} \\ & = \frac {a^2 x}{c^3}-\frac {4 a^2 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}-\frac {8 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac {23 a^2 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.48 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {a^{3/2} \tan (e+f x) \left (\sqrt {a} \sqrt {c} \left (43-11 \sec (e+f x)-31 \sec ^2(e+f x)+23 \sec ^3(e+f x)\right )-60 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sec ^2(e+f x) \sin ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {-a c \tan ^2(e+f x)}\right )}{15 c^{7/2} f (-1+\sec (e+f x))^3 (1+\sec (e+f x))} \]

[In]

Integrate[(a + a*Sec[e + f*x])^2/(c - c*Sec[e + f*x])^3,x]

[Out]

(a^(3/2)*Tan[e + f*x]*(Sqrt[a]*Sqrt[c]*(43 - 11*Sec[e + f*x] - 31*Sec[e + f*x]^2 + 23*Sec[e + f*x]^3) - 60*Arc
Tanh[Sqrt[-(a*c*Tan[e + f*x]^2)]/(Sqrt[a]*Sqrt[c])]*Sec[e + f*x]^2*Sin[(e + f*x)/2]^4*Sqrt[-(a*c*Tan[e + f*x]^
2)]))/(15*c^(7/2)*f*(-1 + Sec[e + f*x])^3*(1 + Sec[e + f*x]))

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.53

method	result	size
parallelrisch	\(\frac {a^{2} \left (3 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-10 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 f x +30 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c^{3} f}\)	\(54\)
derivativedivides	\(\frac {a^{2} \left (\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\)	\(63\)
default	\(\frac {a^{2} \left (\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\)	\(63\)
risch	\(\frac {a^{2} x}{c^{3}}+\frac {2 i a^{2} \left (75 \,{\mathrm e}^{4 i \left (f x +e \right )}-180 \,{\mathrm e}^{3 i \left (f x +e \right )}+250 \,{\mathrm e}^{2 i \left (f x +e \right )}-140 \,{\mathrm e}^{i \left (f x +e \right )}+43\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}\)	\(81\)
norman	\(\frac {\frac {a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}-\frac {a^{2}}{5 c f}+\frac {13 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 c f}-\frac {8 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 c f}+\frac {2 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c f}-\frac {a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\)	\(148\)

[In]

int((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/15*a^2*(3*cot(1/2*f*x+1/2*e)^5-10*cot(1/2*f*x+1/2*e)^3+15*f*x+30*cot(1/2*f*x+1/2*e))/c^3/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {43 \, a^{2} \cos \left (f x + e\right )^{3} - 11 \, a^{2} \cos \left (f x + e\right )^{2} - 31 \, a^{2} \cos \left (f x + e\right ) + 23 \, a^{2} + 15 \, {\left (a^{2} f x \cos \left (f x + e\right )^{2} - 2 \, a^{2} f x \cos \left (f x + e\right ) + a^{2} f x\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \]

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(43*a^2*cos(f*x + e)^3 - 11*a^2*cos(f*x + e)^2 - 31*a^2*cos(f*x + e) + 23*a^2 + 15*(a^2*f*x*cos(f*x + e)^
2 - 2*a^2*f*x*cos(f*x + e) + a^2*f*x)*sin(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin
(f*x + e))

Sympy [F]

\[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=- \frac {a^{2} \left (\int \frac {2 \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \]

[In]

integrate((a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**3,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (90) = 180\).

Time = 0.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.11 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {a^{2} {\left (\frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} - \frac {2 \, a^{2} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac {3 \, a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \]

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(a^2*(120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^3 - (20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 105*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5)) - 2*a^2*(10*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) - 3
*a^2*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} a^{2}}{c^{3}} + \frac {30 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{2}}{c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{15 \, f} \]

[In]

integrate((a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*a^2/c^3 + (30*a^2*tan(1/2*f*x + 1/2*e)^4 - 10*a^2*tan(1/2*f*x + 1/2*e)^2 + 3*a^2)/(c^3*tan(
1/2*f*x + 1/2*e)^5))/f

Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94 \[ \int \frac {(a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx=\frac {a^2\,x}{c^3}+\frac {\frac {a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5}-\frac {2\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+2\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{c^3\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

[In]

int((a + a/cos(e + f*x))^2/(c - c/cos(e + f*x))^3,x)

[Out]

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