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

Optimal. Leaf size=172 \[ \frac {(5 A+12 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(5 A+12 C) \tan (c+d x)}{a^2 d}-\frac {(2 A+5 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 (2 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(2 A+5 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-(2*A+5*C)*arctanh(sin(d*x+c))/a^2/d+(5*A+12*C)*tan(d*x+c)/a^2/d-(2*A+5*C)*sec(d*x+c)*tan(d*x+c)/a^2/d-2/3*(2*
A+5*C)*sec(d*x+c)^3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^2+1/3
*(5*A+12*C)*tan(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.33, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4085, 4019, 3787, 3768, 3770, 3767} \[ \frac {(5 A+12 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(5 A+12 C) \tan (c+d x)}{a^2 d}-\frac {(2 A+5 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 (2 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(2 A+5 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*A + 5*C)*ArcTanh[Sin[c + d*x]])/(a^2*d)) + ((5*A + 12*C)*Tan[c + d*x])/(a^2*d) - ((2*A + 5*C)*Sec[c + d*
x]*Tan[c + d*x])/(a^2*d) - (2*(2*A + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)
*Sec[c + d*x]^4*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((5*A + 12*C)*Tan[c + d*x]^3)/(3*a^2*d)

Rule 3767

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 3768

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 3770

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

Rule 3787

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 4019

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 4085

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*} \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^4(c+d x) (a (A+4 C)-3 a (A+2 C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 (2 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \sec ^3(c+d x) \left (6 a^2 (2 A+5 C)-3 a^2 (5 A+12 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 (2 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 (2 A+5 C)) \int \sec ^3(c+d x) \, dx}{a^2}+\frac {(5 A+12 C) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(2 A+5 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {2 (2 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 A+5 C) \int \sec (c+d x) \, dx}{a^2}-\frac {(5 A+12 C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {(2 A+5 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {(5 A+12 C) \tan (c+d x)}{a^2 d}-\frac {(2 A+5 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {2 (2 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(5 A+12 C) \tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 3.04, size = 623, normalized size = 3.62 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (-60 A \sin \left (c-\frac {d x}{2}\right )+24 A \sin \left (c+\frac {d x}{2}\right )-60 A \sin \left (2 c+\frac {d x}{2}\right )-4 A \sin \left (c+\frac {3 d x}{2}\right )+36 A \sin \left (2 c+\frac {3 d x}{2}\right )-34 A \sin \left (3 c+\frac {3 d x}{2}\right )+42 A \sin \left (c+\frac {5 d x}{2}\right )+24 A \sin \left (3 c+\frac {5 d x}{2}\right )-18 A \sin \left (4 c+\frac {5 d x}{2}\right )+24 A \sin \left (2 c+\frac {7 d x}{2}\right )+3 A \sin \left (3 c+\frac {7 d x}{2}\right )+15 A \sin \left (4 c+\frac {7 d x}{2}\right )-6 A \sin \left (5 c+\frac {7 d x}{2}\right )+10 A \sin \left (3 c+\frac {9 d x}{2}\right )+3 A \sin \left (4 c+\frac {9 d x}{2}\right )+7 A \sin \left (5 c+\frac {9 d x}{2}\right )-3 (8 A+C) \sin \left (\frac {d x}{2}\right )+(66 A+155 C) \sin \left (\frac {3 d x}{2}\right )-153 C \sin \left (c-\frac {d x}{2}\right )+21 C \sin \left (c+\frac {d x}{2}\right )-135 C \sin \left (2 c+\frac {d x}{2}\right )+25 C \sin \left (c+\frac {3 d x}{2}\right )+45 C \sin \left (2 c+\frac {3 d x}{2}\right )-85 C \sin \left (3 c+\frac {3 d x}{2}\right )+99 C \sin \left (c+\frac {5 d x}{2}\right )+21 C \sin \left (2 c+\frac {5 d x}{2}\right )+33 C \sin \left (3 c+\frac {5 d x}{2}\right )-45 C \sin \left (4 c+\frac {5 d x}{2}\right )+57 C \sin \left (2 c+\frac {7 d x}{2}\right )+18 C \sin \left (3 c+\frac {7 d x}{2}\right )+24 C \sin \left (4 c+\frac {7 d x}{2}\right )-15 C \sin \left (5 c+\frac {7 d x}{2}\right )+24 C \sin \left (3 c+\frac {9 d x}{2}\right )+11 C \sin \left (4 c+\frac {9 d x}{2}\right )+13 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )+192 (2 A+5 C) \cos ^3\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 (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{24 a^2 d (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(A + C*Sec[c + d*x]^2)*(192*(2*A + 5*C)*Cos[(c + d*x)/2]^3*(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]^3*(-3*(8*A + C)*Sin[(d*x)/
2] + (66*A + 155*C)*Sin[(3*d*x)/2] - 60*A*Sin[c - (d*x)/2] - 153*C*Sin[c - (d*x)/2] + 24*A*Sin[c + (d*x)/2] +
21*C*Sin[c + (d*x)/2] - 60*A*Sin[2*c + (d*x)/2] - 135*C*Sin[2*c + (d*x)/2] - 4*A*Sin[c + (3*d*x)/2] + 25*C*Sin
[c + (3*d*x)/2] + 36*A*Sin[2*c + (3*d*x)/2] + 45*C*Sin[2*c + (3*d*x)/2] - 34*A*Sin[3*c + (3*d*x)/2] - 85*C*Sin
[3*c + (3*d*x)/2] + 42*A*Sin[c + (5*d*x)/2] + 99*C*Sin[c + (5*d*x)/2] + 21*C*Sin[2*c + (5*d*x)/2] + 24*A*Sin[3
*c + (5*d*x)/2] + 33*C*Sin[3*c + (5*d*x)/2] - 18*A*Sin[4*c + (5*d*x)/2] - 45*C*Sin[4*c + (5*d*x)/2] + 24*A*Sin
[2*c + (7*d*x)/2] + 57*C*Sin[2*c + (7*d*x)/2] + 3*A*Sin[3*c + (7*d*x)/2] + 18*C*Sin[3*c + (7*d*x)/2] + 15*A*Si
n[4*c + (7*d*x)/2] + 24*C*Sin[4*c + (7*d*x)/2] - 6*A*Sin[5*c + (7*d*x)/2] - 15*C*Sin[5*c + (7*d*x)/2] + 10*A*S
in[3*c + (9*d*x)/2] + 24*C*Sin[3*c + (9*d*x)/2] + 3*A*Sin[4*c + (9*d*x)/2] + 11*C*Sin[4*c + (9*d*x)/2] + 7*A*S
in[5*c + (9*d*x)/2] + 13*C*Sin[5*c + (9*d*x)/2])))/(24*a^2*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])
^2)

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fricas [A]  time = 0.47, size = 237, normalized size = 1.38 \[ -\frac {3 \, {\left ({\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (5 \, A + 12 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (14 \, A + 33 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*((2*A + 5*C)*cos(d*x + c)^5 + 2*(2*A + 5*C)*cos(d*x + c)^4 + (2*A + 5*C)*cos(d*x + c)^3)*log(sin(d*x +
c) + 1) - 3*((2*A + 5*C)*cos(d*x + c)^5 + 2*(2*A + 5*C)*cos(d*x + c)^4 + (2*A + 5*C)*cos(d*x + c)^3)*log(-sin
(d*x + c) + 1) - 2*(2*(5*A + 12*C)*cos(d*x + c)^4 + (14*A + 33*C)*cos(d*x + c)^3 + 3*(A + 2*C)*cos(d*x + c)^2
- C*cos(d*x + c) + C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

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giac [A]  time = 1.52, size = 225, normalized size = 1.31 \[ -\frac {\frac {6 \, {\left (2 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (2 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {4 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(2*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(2*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 4*(3*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^5 - 6*A*tan(1/2*d*x + 1/2*c)^3 - 20*C*tan(1/2*
d*x + 1/2*c)^3 + 3*A*tan(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (
A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(1/2
*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.73, size = 338, normalized size = 1.97 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {9 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}-\frac {C}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3 C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}}-\frac {5 C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3 C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*A*tan(1/2*d*x+1/2*c)+9/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-5/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)+2/d/a^2*A*ln(tan(1/2*d*x+1
/2*c)-1)+5/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-1/3/d/a^2*C/(tan(1/2*d*x+1/2*c)-1)^3-3/2/d/a^2*C/(tan(1/2*d*x+1/2*
c)-1)^2-2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-5/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-5/d/a^2/(tan(1/2*d*x+1/2*c)+1)*C
-1/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)-1/3/d/a^2*C/(tan(1/2*d*x+1/2*c)+1)^3+3/2/d/a^2*C/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [B]  time = 0.38, size = 379, normalized size = 2.20 \[ \frac {C {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(C*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) + A*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c
)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*
sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 2.74, size = 197, normalized size = 1.15 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A+C\right )}{a^2}+\frac {A+5\,C}{2\,a^2}\right )}{d}-\frac {\left (2\,A+10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A-\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+5\,C\right )}{a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((2*(A + C))/a^2 + (A + 5*C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^5*(2*A + 10*C) - tan(c/2 +
(d*x)/2)^3*(4*A + (40*C)/3) + tan(c/2 + (d*x)/2)*(2*A + 6*C))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 - 3*a^2*tan(c/2 +
(d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2)) - (2*atanh(tan(c/2 + (d*x)/2))*(2*A + 5*C))/(a^2*d) + (tan(c/2
+ (d*x)/2)^3*(A + C))/(6*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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