\(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [141]

Optimal result

Integrand size = 31, antiderivative size = 104 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac {(2 A+7 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/5*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/3*(A-C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/15*(2*A+7*
C)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4164, 4085, 3879} \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A+7 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(A-C) \tan (c+d x)}{3 a d (a \sec (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[In]

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

[Out]

-1/5*((A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((A - C)*Tan[c + d*x])/(3*a*d*(a + a*Sec
[c + d*x])^2) + ((2*A + 7*C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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 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 4164

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

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

Mathematica [A] (verified)

Time = 1.76 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (20 (2 A+C) \sin \left (\frac {d x}{2}\right )-30 A \sin \left (c+\frac {d x}{2}\right )+20 A \sin \left (c+\frac {3 d x}{2}\right )+10 C \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+7 A \sin \left (2 c+\frac {5 d x}{2}\right )+2 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{240 a^3 d} \]

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(20*(2*A + C)*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] + 20*A*Sin[c + (3*d*x)/2] + 10
*C*Sin[c + (3*d*x)/2] - 15*A*Sin[2*c + (3*d*x)/2] + 7*A*Sin[2*c + (5*d*x)/2] + 2*C*Sin[2*c + (5*d*x)/2]))/(240
*a^3*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {10 \left (-A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+5 A +5 C \right )}{20 a^{3} d}\)	\(58\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{4 d \,a^{3}}\)	\(88\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{4 d \,a^{3}}\)	\(88\)
risch	\(\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 A \,{\mathrm e}^{i \left (d x +c \right )}+10 C \,{\mathrm e}^{i \left (d x +c \right )}+7 A +2 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(102\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}-\frac {\left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (4 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 a d}+\frac {\left (19 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{2}}\)	\(139\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {{\left ({\left (7 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A + C\right )} \cos \left (d x + c\right ) + 2 \, A + 7 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/15*((7*A + 2*C)*cos(d*x + c)^2 + 6*(A + C)*cos(d*x + c) + 2*A + 7*C)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*
a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F]

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

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d
*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

[In]

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

[Out]

1/60*(C*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 + A*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d
*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]

[In]

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

[Out]

1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 - 10*A*tan(1/2*d*x + 1/2*c)^3 + 10*C*tan(1/2*d*x
+ 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) + 15*C*tan(1/2*d*x + 1/2*c))/(a^3*d)

Mupad [B] (verification not implemented)

Time = 14.81 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.66 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{4\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-2\,C\right )}{12\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + a/cos(c + d*x))^3),x)

[Out]

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