\(\int \frac {\sec ^4(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [138]

Optimal result

Integrand size = 33, antiderivative size = 198 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

1/2*(2*A+13*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(11*A+76*C)*tan(d*x+c)/a^3/d+1/2*(2*A+13*C)*sec(d*x+c)*tan(d*x+c
)/a^3/d-1/5*(A+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(A+11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*
sec(d*x+c))^2-1/15*(11*A+76*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4170, 4104, 3872, 3852, 8, 3853, 3855} \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A+76 C) \tan (c+d x)}{15 a^3 d}-\frac {(11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(2 A+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(A+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

((2*A + 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(11*A + 76*C)*Tan[c + d*x])/(15*a^3*d) + ((2*A + 13*C)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A + C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((A +
11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((11*A + 76*C)*Sec[c + d*x]^2*Tan[c + d*
x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4104

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

Rule 4170

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^4(c+d x) (-a (A-4 C)-a (2 A+7 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec ^2(c+d x) \left (2 a^3 (11 A+76 C)-15 a^3 (2 A+13 C) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A+13 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac {(2 (11 A+76 C)) \int \sec ^2(c+d x) \, dx}{15 a^3} \\ & = \frac {(2 A+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A+13 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(2 (11 A+76 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = \frac {(2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(632\) vs. \(2(198)=396\).

Time = 3.95 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.19 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (1920 (2 A+13 C) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-5 (98 A+247 C) \sin \left (\frac {d x}{2}\right )+5 (106 A+761 C) \sin \left (\frac {3 d x}{2}\right )-654 A \sin \left (c-\frac {d x}{2}\right )-4329 C \sin \left (c-\frac {d x}{2}\right )+654 A \sin \left (c+\frac {d x}{2}\right )+1989 C \sin \left (c+\frac {d x}{2}\right )-490 A \sin \left (2 c+\frac {d x}{2}\right )-3575 C \sin \left (2 c+\frac {d x}{2}\right )-350 A \sin \left (c+\frac {3 d x}{2}\right )-475 C \sin \left (c+\frac {3 d x}{2}\right )+530 A \sin \left (2 c+\frac {3 d x}{2}\right )+2005 C \sin \left (2 c+\frac {3 d x}{2}\right )-350 A \sin \left (3 c+\frac {3 d x}{2}\right )-2275 C \sin \left (3 c+\frac {3 d x}{2}\right )+378 A \sin \left (c+\frac {5 d x}{2}\right )+2673 C \sin \left (c+\frac {5 d x}{2}\right )-150 A \sin \left (2 c+\frac {5 d x}{2}\right )+105 C \sin \left (2 c+\frac {5 d x}{2}\right )+378 A \sin \left (3 c+\frac {5 d x}{2}\right )+1593 C \sin \left (3 c+\frac {5 d x}{2}\right )-150 A \sin \left (4 c+\frac {5 d x}{2}\right )-975 C \sin \left (4 c+\frac {5 d x}{2}\right )+190 A \sin \left (2 c+\frac {7 d x}{2}\right )+1325 C \sin \left (2 c+\frac {7 d x}{2}\right )-30 A \sin \left (3 c+\frac {7 d x}{2}\right )+255 C \sin \left (3 c+\frac {7 d x}{2}\right )+190 A \sin \left (4 c+\frac {7 d x}{2}\right )+875 C \sin \left (4 c+\frac {7 d x}{2}\right )-30 A \sin \left (5 c+\frac {7 d x}{2}\right )-195 C \sin \left (5 c+\frac {7 d x}{2}\right )+44 A \sin \left (3 c+\frac {9 d x}{2}\right )+304 C \sin \left (3 c+\frac {9 d x}{2}\right )+90 C \sin \left (4 c+\frac {9 d x}{2}\right )+44 A \sin \left (5 c+\frac {9 d x}{2}\right )+214 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )\right )}{240 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \]

[In]

Integrate[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/240*(Cos[(c + d*x)/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(1920*(2*A + 13*C)*Cos[(c + d*x)/2]^5*(Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-
5*(98*A + 247*C)*Sin[(d*x)/2] + 5*(106*A + 761*C)*Sin[(3*d*x)/2] - 654*A*Sin[c - (d*x)/2] - 4329*C*Sin[c - (d*
x)/2] + 654*A*Sin[c + (d*x)/2] + 1989*C*Sin[c + (d*x)/2] - 490*A*Sin[2*c + (d*x)/2] - 3575*C*Sin[2*c + (d*x)/2
] - 350*A*Sin[c + (3*d*x)/2] - 475*C*Sin[c + (3*d*x)/2] + 530*A*Sin[2*c + (3*d*x)/2] + 2005*C*Sin[2*c + (3*d*x
)/2] - 350*A*Sin[3*c + (3*d*x)/2] - 2275*C*Sin[3*c + (3*d*x)/2] + 378*A*Sin[c + (5*d*x)/2] + 2673*C*Sin[c + (5
*d*x)/2] - 150*A*Sin[2*c + (5*d*x)/2] + 105*C*Sin[2*c + (5*d*x)/2] + 378*A*Sin[3*c + (5*d*x)/2] + 1593*C*Sin[3
*c + (5*d*x)/2] - 150*A*Sin[4*c + (5*d*x)/2] - 975*C*Sin[4*c + (5*d*x)/2] + 190*A*Sin[2*c + (7*d*x)/2] + 1325*
C*Sin[2*c + (7*d*x)/2] - 30*A*Sin[3*c + (7*d*x)/2] + 255*C*Sin[3*c + (7*d*x)/2] + 190*A*Sin[4*c + (7*d*x)/2] +
875*C*Sin[4*c + (7*d*x)/2] - 30*A*Sin[5*c + (7*d*x)/2] - 195*C*Sin[5*c + (7*d*x)/2] + 44*A*Sin[3*c + (9*d*x)/
2] + 304*C*Sin[3*c + (9*d*x)/2] + 90*C*Sin[4*c + (9*d*x)/2] + 44*A*Sin[5*c + (9*d*x)/2] + 214*C*Sin[5*c + (9*d
*x)/2])))/(a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {-240 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {13 C}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+240 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {13 C}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-22 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\left (\frac {108 A}{11}+\frac {783 C}{11}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {51 A}{11}+\frac {717 C}{22}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {76 C}{11}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {153 A}{11}+\frac {2331 C}{22}\right ) \cos \left (d x +c \right )+\frac {97 A}{11}+\frac {677 C}{11}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(175\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (26 C +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(194\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (26 C +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(194\)
norman	\(\frac {-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}-\frac {\left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {\left (7 A +59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (51 C +7 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (475 C +71 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (721 C +101 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (1165 C +173 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}+\frac {\left (6613 C +953 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a^{2}}-\frac {\left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(276\)
risch	\(-\frac {i \left (30 A \,{\mathrm e}^{8 i \left (d x +c \right )}+195 C \,{\mathrm e}^{8 i \left (d x +c \right )}+150 A \,{\mathrm e}^{7 i \left (d x +c \right )}+975 C \,{\mathrm e}^{7 i \left (d x +c \right )}+350 A \,{\mathrm e}^{6 i \left (d x +c \right )}+2275 C \,{\mathrm e}^{6 i \left (d x +c \right )}+490 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3575 C \,{\mathrm e}^{5 i \left (d x +c \right )}+654 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4329 C \,{\mathrm e}^{4 i \left (d x +c \right )}+530 A \,{\mathrm e}^{3 i \left (d x +c \right )}+3805 C \,{\mathrm e}^{3 i \left (d x +c \right )}+378 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2673 C \,{\mathrm e}^{2 i \left (d x +c \right )}+190 A \,{\mathrm e}^{i \left (d x +c \right )}+1325 C \,{\mathrm e}^{i \left (d x +c \right )}+44 A +304 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{3} d}\)	\(323\)

[In]

int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/240*(-240*(1+cos(2*d*x+2*c))*(13/2*C+A)*ln(tan(1/2*d*x+1/2*c)-1)+240*(1+cos(2*d*x+2*c))*(13/2*C+A)*ln(tan(1/
2*d*x+1/2*c)+1)-22*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4*((108/11*A+783/11*C)*cos(2*d*x+2*c)+(51/11*A+717/22
*C)*cos(3*d*x+3*c)+(A+76/11*C)*cos(4*d*x+4*c)+(153/11*A+2331/22*C)*cos(d*x+c)+97/11*A+677/11*C))/d/a^3/(1+cos(
2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (11 \, A + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A + 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (64 \, A + 479 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((2*A + 13*C)*cos(d*x + c)^5 + 3*(2*A + 13*C)*cos(d*x + c)^4 + 3*(2*A + 13*C)*cos(d*x + c)^3 + (2*A +
13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((2*A + 13*C)*cos(d*x + c)^5 + 3*(2*A + 13*C)*cos(d*x + c)^4
+ 3*(2*A + 13*C)*cos(d*x + c)^3 + (2*A + 13*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(11*A + 76*C)*co
s(d*x + c)^4 + 3*(34*A + 239*C)*cos(d*x + c)^3 + (64*A + 479*C)*cos(d*x + c)^2 + 45*C*cos(d*x + c) - 15*C)*sin
(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**4*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c
+ d*x)**6/(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 = 0.20 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.67 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*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 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + A*((105*sin(d*x + c)/(cos(d*x
+ c) + 1) + 20*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 - 60*log(sin(
d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (2 \, A + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (2 \, A + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(2*A + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(2*A + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(7*C*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3
*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 40*C*a^1
2*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 15.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.98 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {13\,C}{2}\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (A+5\,C\right )}{4\,a^3}-\frac {2\,A-10\,C}{4\,a^3}\right )}{d}-\frac {5\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {A+5\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^3),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A + (13*C)/2))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^3) + (3*(A + 5*C)
)/(4*a^3) - (2*A - 10*C)/(4*a^3)))/d - (5*C*tan(c/2 + (d*x)/2) - 7*C*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (
d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (A + 5*C)/(12*a^3)))
/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)