\(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx\) [229]

Optimal result

Integrand size = 31, antiderivative size = 115 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]

[Out]

1/5*(c-d)^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+2/15*(c-d)*(c+4*d)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2+1/15*(2*c^2+6
*c*d+7*d^2)*tan(f*x+e)/f/(a^3+a^3*sec(f*x+e))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4072, 91, 79, 37} \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {(c-d)^2 \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}+\frac {2 (c+4 d) (c-d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2} \]

[In]

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

[Out]

((c - d)^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (2*(c - d)*(c + 4*d)*Tan[e + f*x])/(15*a*f*(a + a*Sec[
e + f*x])^2) + ((2*c^2 + 6*c*d + 7*d^2)*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

Rule 37

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

Rule 79

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 4072

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[a^2*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]
])), Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*((c + d*x)^n/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^2}{\sqrt {a-a x} (a+a x)^{7/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\tan (e+f x) \text {Subst}\left (\int \frac {a^3 \left (2 c^2+6 c d-3 d^2\right )+5 a^3 d^2 x}{\sqrt {a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {\left (\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.73 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\left (2 c^2+6 c d+7 d^2+6 \left (c^2+3 c d+d^2\right ) \cos (e+f x)+\left (7 c^2+6 c d+2 d^2\right ) \cos ^2(e+f x)\right ) \sin (e+f x)}{15 a^3 f (1+\cos (e+f x))^3} \]

[In]

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

[Out]

((2*c^2 + 6*c*d + 7*d^2 + 6*(c^2 + 3*c*d + d^2)*Cos[e + f*x] + (7*c^2 + 6*c*d + 2*d^2)*Cos[e + f*x]^2)*Sin[e +
f*x])/(15*a^3*f*(1 + Cos[e + f*x])^3)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {\left (\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {10 \left (-c^{2}+d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+5 \left (c +d \right )^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{20 a^{3} f}\)	\(67\)
derivativedivides	\(\frac {\frac {\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {2 \left (-c -d \right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-c -d \right )^{2}}{4 f \,a^{3}}\)	\(74\)
default	\(\frac {\frac {\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {2 \left (-c -d \right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-c -d \right )^{2}}{4 f \,a^{3}}\)	\(74\)
risch	\(\frac {2 i \left (15 c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+30 c^{2} {\mathrm e}^{3 i \left (f x +e \right )}+30 c d \,{\mathrm e}^{3 i \left (f x +e \right )}+40 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+30 c d \,{\mathrm e}^{2 i \left (f x +e \right )}+20 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+20 c^{2} {\mathrm e}^{i \left (f x +e \right )}+30 d \,{\mathrm e}^{i \left (f x +e \right )} c +10 d^{2} {\mathrm e}^{i \left (f x +e \right )}+7 c^{2}+6 c d +2 d^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\)	\(161\)
norman	\(\frac {\frac {\left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{20 a f}+\frac {\left (c^{2}+2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a f}-\frac {\left (2 c^{2}+3 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {\left (4 c^{2}-3 c d -d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 a f}+\frac {\left (19 c^{2}+12 c d -d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{30 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} a^{2}}\)	\(179\)

[In]

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

[Out]

1/20*((c-d)^2*tan(1/2*f*x+1/2*e)^4+10/3*(-c^2+d^2)*tan(1/2*f*x+1/2*e)^2+5*(c+d)^2)*tan(1/2*f*x+1/2*e)/a^3/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left ({\left (7 \, c^{2} + 6 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, c^{2} + 6 \, c d + 7 \, d^{2} + 6 \, {\left (c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

[In]

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

[Out]

1/15*((7*c^2 + 6*c*d + 2*d^2)*cos(f*x + e)^2 + 2*c^2 + 6*c*d + 7*d^2 + 6*(c^2 + 3*c*d + d^2)*cos(f*x + e))*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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.60 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {d^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {6 \, c d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

[In]

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

[Out]

1/60*(d^2*(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 + c^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*s
in(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + 6*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x +
e) + 1)^5)/a^3)/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 6 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 10 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{60 \, a^{3} f} \]

[In]

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

[Out]

1/60*(3*c^2*tan(1/2*f*x + 1/2*e)^5 - 6*c*d*tan(1/2*f*x + 1/2*e)^5 + 3*d^2*tan(1/2*f*x + 1/2*e)^5 - 10*c^2*tan(
1/2*f*x + 1/2*e)^3 + 10*d^2*tan(1/2*f*x + 1/2*e)^3 + 15*c^2*tan(1/2*f*x + 1/2*e) + 30*c*d*tan(1/2*f*x + 1/2*e)
+ 15*d^2*tan(1/2*f*x + 1/2*e))/(a^3*f)

Mupad [B] (verification not implemented)

Time = 13.75 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (c+d\right )}^2}{4\,a^3\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^2-2\,d^2\right )}{12\,a^3\,f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\left (c-d\right )}^2}{20\,a^3\,f} \]

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(c + d)^2)/(4*a^3*f) - (tan(e/2 + (f*x)/2)^3*(2*c^2 - 2*d^2))/(12*a^3*f) + (tan(e/2 + (f*x
)/2)^5*(c - d)^2)/(20*a^3*f)