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

Optimal. Leaf size=194 \[ \frac {(5 A-8 B+12 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(5 A-8 B+12 C) \tan (c+d x)}{a^2 d}-\frac {(4 A-7 B+10 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 A-7 B+10 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-1/2*(4*A-7*B+10*C)*arctanh(sin(d*x+c))/a^2/d+(5*A-8*B+12*C)*tan(d*x+c)/a^2/d-1/2*(4*A-7*B+10*C)*sec(d*x+c)*ta
n(d*x+c)/a^2/d-1/3*(4*A-7*B+10*C)*sec(d*x+c)^3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*sec(d*x+c)^4*tan(d*
x+c)/d/(a+a*sec(d*x+c))^2+1/3*(5*A-8*B+12*C)*tan(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.36, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4084, 4019, 3787, 3768, 3770, 3767} \[ \frac {(5 A-8 B+12 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(5 A-8 B+12 C) \tan (c+d x)}{a^2 d}-\frac {(4 A-7 B+10 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 A-7 B+10 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*A - 7*B + 10*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + ((5*A - 8*B + 12*C)*Tan[c + d*x])/(a^2*d) - ((4*A - 7*
B + 10*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) - ((4*A - 7*B + 10*C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*a^2*d*(1
+ Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((5*A - 8*B + 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 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, 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+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B+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 B+4 C)+3 a (A-B+2 C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(4 A-7 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+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 (-3 a^2 (4 A-7 B+10 C)+3 a^2 (5 A-8 B+12 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(4 A-7 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-7 B+10 C) \int \sec ^3(c+d x) \, dx}{a^2}+\frac {(5 A-8 B+12 C) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(4 A-7 B+10 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(4 A-7 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-7 B+10 C) \int \sec (c+d x) \, dx}{2 a^2}-\frac {(5 A-8 B+12 C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {(4 A-7 B+10 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(5 A-8 B+12 C) \tan (c+d x)}{a^2 d}-\frac {(4 A-7 B+10 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(4 A-7 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(5 A-8 B+12 C) \tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 6.45, size = 1069, normalized size = 5.51 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(4*A - 7*B + 10*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + B*Sec[c + d*x] +
C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (4*(4*A - 7*
B + 10*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + B*Sec[c + d*x] + C*Sec[c + d*
x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]*Sec[
c/2]*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-48*A*Sin[(d*x)/2] + 45*B*Sin[(d*x)/2] - 6
*C*Sin[(d*x)/2] + 132*A*Sin[(3*d*x)/2] - 201*B*Sin[(3*d*x)/2] + 310*C*Sin[(3*d*x)/2] - 120*A*Sin[c - (d*x)/2]
+ 195*B*Sin[c - (d*x)/2] - 306*C*Sin[c - (d*x)/2] + 48*A*Sin[c + (d*x)/2] - 51*B*Sin[c + (d*x)/2] + 42*C*Sin[c
+ (d*x)/2] - 120*A*Sin[2*c + (d*x)/2] + 189*B*Sin[2*c + (d*x)/2] - 270*C*Sin[2*c + (d*x)/2] - 8*A*Sin[c + (3*
d*x)/2] - B*Sin[c + (3*d*x)/2] + 50*C*Sin[c + (3*d*x)/2] + 72*A*Sin[2*c + (3*d*x)/2] - 81*B*Sin[2*c + (3*d*x)/
2] + 90*C*Sin[2*c + (3*d*x)/2] - 68*A*Sin[3*c + (3*d*x)/2] + 119*B*Sin[3*c + (3*d*x)/2] - 170*C*Sin[3*c + (3*d
*x)/2] + 84*A*Sin[c + (5*d*x)/2] - 129*B*Sin[c + (5*d*x)/2] + 198*C*Sin[c + (5*d*x)/2] - 9*B*Sin[2*c + (5*d*x)
/2] + 42*C*Sin[2*c + (5*d*x)/2] + 48*A*Sin[3*c + (5*d*x)/2] - 57*B*Sin[3*c + (5*d*x)/2] + 66*C*Sin[3*c + (5*d*
x)/2] - 36*A*Sin[4*c + (5*d*x)/2] + 63*B*Sin[4*c + (5*d*x)/2] - 90*C*Sin[4*c + (5*d*x)/2] + 48*A*Sin[2*c + (7*
d*x)/2] - 75*B*Sin[2*c + (7*d*x)/2] + 114*C*Sin[2*c + (7*d*x)/2] + 6*A*Sin[3*c + (7*d*x)/2] - 15*B*Sin[3*c + (
7*d*x)/2] + 36*C*Sin[3*c + (7*d*x)/2] + 30*A*Sin[4*c + (7*d*x)/2] - 39*B*Sin[4*c + (7*d*x)/2] + 48*C*Sin[4*c +
(7*d*x)/2] - 12*A*Sin[5*c + (7*d*x)/2] + 21*B*Sin[5*c + (7*d*x)/2] - 30*C*Sin[5*c + (7*d*x)/2] + 20*A*Sin[3*c
+ (9*d*x)/2] - 32*B*Sin[3*c + (9*d*x)/2] + 48*C*Sin[3*c + (9*d*x)/2] + 6*A*Sin[4*c + (9*d*x)/2] - 12*B*Sin[4*
c + (9*d*x)/2] + 22*C*Sin[4*c + (9*d*x)/2] + 14*A*Sin[5*c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2] + 26*C*Sin[
5*c + (9*d*x)/2]))/(48*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

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fricas [A]  time = 0.48, size = 271, normalized size = 1.40 \[ -\frac {3 \, {\left ({\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (5 \, A - 8 \, B + 12 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (28 \, A - 43 \, B + 66 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (A - B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*((4*A - 7*B + 10*C)*cos(d*x + c)^5 + 2*(4*A - 7*B + 10*C)*cos(d*x + c)^4 + (4*A - 7*B + 10*C)*cos(d*x
+ c)^3)*log(sin(d*x + c) + 1) - 3*((4*A - 7*B + 10*C)*cos(d*x + c)^5 + 2*(4*A - 7*B + 10*C)*cos(d*x + c)^4 +
(4*A - 7*B + 10*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(5*A - 8*B + 12*C)*cos(d*x + c)^4 + (28*A - 4
3*B + 66*C)*cos(d*x + c)^3 + 6*(A - B + 2*C)*cos(d*x + c)^2 + (3*B - 2*C)*cos(d*x + c) + 2*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 = 0.31, size = 303, normalized size = 1.56 \[ -\frac {\frac {3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(4*A - 7*B + 10*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*A - 7*B + 10*C)*log(abs(tan(1/2*d*x +
1/2*c) - 1))/a^2 + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 30*C*tan(1/2*d*x + 1/2*c)^5
- 12*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(1/2*d*x + 1/2*c)^3 - 40*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x +
1/2*c) - 9*B*tan(1/2*d*x + 1/2*c) + 18*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*t
an(1/2*d*x + 1/2*c)^3 - B*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) - 21*B*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 = 0.78, size = 506, normalized size = 2.61 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{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 {7 B \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 {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {5 C}{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 {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 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 {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {5 C}{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 {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+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*B*tan(1/2*d*x+1/2*c)^3+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)-7/2/d/a^2*B*tan(1/2*d*x+1/2*c)+9/2/d/a^2*C*tan(1/2*d*x+1/2*c)+1/2/d/a^2/(tan(1/2*d*x+1/2*c
)-1)^2*B-3/2/d/a^2*C/(tan(1/2*d*x+1/2*c)-1)^2-1/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)+5/2/d/a^2/(tan(1/2*d*x+1/2*c)-1
)*B-5/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C+2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)-7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B+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)^2-1/
2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*B-1/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)+5/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B-5/d/a^2
/(tan(1/2*d*x+1/2*c)+1)*C-2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)+7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B-5/d/a^2*ln(t
an(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

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maxima [B]  time = 0.37, size = 567, normalized size = 2.92 \[ \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 )} - B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \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 {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \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+B*sec(d*x+c)+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) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) +
sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*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 + 1
2*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 = 3.43, size = 218, normalized size = 1.12 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-3\,B+5\,C}{2\,a^2}+\frac {2\,\left (A-B+C\right )}{a^2}\right )}{d}-\frac {\left (2\,A-5\,B+10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,B-4\,A-\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-3\,B+6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-7\,B+10\,C\right )}{a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + 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)*((A - 3*B + 5*C)/(2*a^2) + (2*(A - B + C))/a^2))/d - (tan(c/2 + (d*x)/2)*(2*A - 3*B + 6*C)
+ tan(c/2 + (d*x)/2)^5*(2*A - 5*B + 10*C) - tan(c/2 + (d*x)/2)^3*(4*A - 8*B + (40*C)/3))/(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)) - (atanh(tan(c/2 + (d*x)/2))*(4*A -
7*B + 10*C))/(a^2*d) + (tan(c/2 + (d*x)/2)^3*(A - B + C))/(6*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+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(B*sec(c + d*x)**5/(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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