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

Optimal. Leaf size=150 \[ -\frac {4 (A+4 C) \tan (c+d x)}{3 a^2 d}+\frac {(2 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {2 (A+4 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {(2 A+7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

1/2*(2*A+7*C)*arctanh(sin(d*x+c))/a^2/d-4/3*(A+4*C)*tan(d*x+c)/a^2/d+1/2*(2*A+7*C)*sec(d*x+c)*tan(d*x+c)/a^2/d
-2/3*(A+4*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))
^2

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4085, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac {4 (A+4 C) \tan (c+d x)}{3 a^2 d}+\frac {(2 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {2 (A+4 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {(2 A+7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^3(c+d x) (3 a C-a (2 A+5 C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 (A+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \sec ^2(c+d x) \left (4 a^2 (A+4 C)-3 a^2 (2 A+7 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 (A+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 (A+4 C)) \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac {(2 A+7 C) \int \sec ^3(c+d x) \, dx}{a^2}\\ &=\frac {(2 A+7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {2 (A+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 A+7 C) \int \sec (c+d x) \, dx}{2 a^2}+\frac {(4 (A+4 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=\frac {(2 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {4 (A+4 C) \tan (c+d x)}{3 a^2 d}+\frac {(2 A+7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {2 (A+4 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 2.14, size = 513, normalized size = 3.42 \[ -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-36 A \sin \left (c-\frac {d x}{2}\right )+36 A \sin \left (c+\frac {d x}{2}\right )-20 A \sin \left (2 c+\frac {d x}{2}\right )-18 A \sin \left (c+\frac {3 d x}{2}\right )+22 A \sin \left (2 c+\frac {3 d x}{2}\right )-18 A \sin \left (3 c+\frac {3 d x}{2}\right )+18 A \sin \left (c+\frac {5 d x}{2}\right )-6 A \sin \left (2 c+\frac {5 d x}{2}\right )+18 A \sin \left (3 c+\frac {5 d x}{2}\right )-6 A \sin \left (4 c+\frac {5 d x}{2}\right )+8 A \sin \left (2 c+\frac {7 d x}{2}\right )+8 A \sin \left (4 c+\frac {7 d x}{2}\right )-2 (10 A+7 C) \sin \left (\frac {d x}{2}\right )+(22 A+97 C) \sin \left (\frac {3 d x}{2}\right )-126 C \sin \left (c-\frac {d x}{2}\right )+42 C \sin \left (c+\frac {d x}{2}\right )-98 C \sin \left (2 c+\frac {d x}{2}\right )-3 C \sin \left (c+\frac {3 d x}{2}\right )+37 C \sin \left (2 c+\frac {3 d x}{2}\right )-63 C \sin \left (3 c+\frac {3 d x}{2}\right )+75 C \sin \left (c+\frac {5 d x}{2}\right )+15 C \sin \left (2 c+\frac {5 d x}{2}\right )+39 C \sin \left (3 c+\frac {5 d x}{2}\right )-21 C \sin \left (4 c+\frac {5 d x}{2}\right )+32 C \sin \left (2 c+\frac {7 d x}{2}\right )+12 C \sin \left (3 c+\frac {7 d x}{2}\right )+20 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )+96 (2 A+7 C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{24 a^2 d (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Cos[(c + d*x)/2]*(A + C*Sec[c + d*x]^2)*(96*(2*A + 7*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]^2*(-2*(10*A + 7*C)*Si
n[(d*x)/2] + (22*A + 97*C)*Sin[(3*d*x)/2] - 36*A*Sin[c - (d*x)/2] - 126*C*Sin[c - (d*x)/2] + 36*A*Sin[c + (d*x
)/2] + 42*C*Sin[c + (d*x)/2] - 20*A*Sin[2*c + (d*x)/2] - 98*C*Sin[2*c + (d*x)/2] - 18*A*Sin[c + (3*d*x)/2] - 3
*C*Sin[c + (3*d*x)/2] + 22*A*Sin[2*c + (3*d*x)/2] + 37*C*Sin[2*c + (3*d*x)/2] - 18*A*Sin[3*c + (3*d*x)/2] - 63
*C*Sin[3*c + (3*d*x)/2] + 18*A*Sin[c + (5*d*x)/2] + 75*C*Sin[c + (5*d*x)/2] - 6*A*Sin[2*c + (5*d*x)/2] + 15*C*
Sin[2*c + (5*d*x)/2] + 18*A*Sin[3*c + (5*d*x)/2] + 39*C*Sin[3*c + (5*d*x)/2] - 6*A*Sin[4*c + (5*d*x)/2] - 21*C
*Sin[4*c + (5*d*x)/2] + 8*A*Sin[2*c + (7*d*x)/2] + 32*C*Sin[2*c + (7*d*x)/2] + 12*C*Sin[3*c + (7*d*x)/2] + 8*A
*Sin[4*c + (7*d*x)/2] + 20*C*Sin[4*c + (7*d*x)/2])))/(a^2*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^
2)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 222, normalized size = 1.48 \[ \frac {3 \, {\left ({\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, {\left (A + 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (10 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, C \cos \left (d x + c\right ) - 3 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*((2*A + 7*C)*cos(d*x + c)^4 + 2*(2*A + 7*C)*cos(d*x + c)^3 + (2*A + 7*C)*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) - 3*((2*A + 7*C)*cos(d*x + c)^4 + 2*(2*A + 7*C)*cos(d*x + c)^3 + (2*A + 7*C)*cos(d*x + c)^2)*log(-sin
(d*x + c) + 1) - 2*(8*(A + 4*C)*cos(d*x + c)^3 + (10*A + 43*C)*cos(d*x + c)^2 + 6*C*cos(d*x + c) - 3*C)*sin(d*
x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [A]  time = 0.33, size = 171, normalized size = 1.14 \[ \frac {\frac {3 \, {\left (2 \, A + 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (2 \, A + 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*A + 7*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(2*A + 7*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^2 + 6*(5*C*tan(1/2*d*x + 1/2*c)^3 - 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (A*a^4*t
an(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) + 21*C*a^4*tan(1/2*d*x + 1
/2*c))/a^6)/d

________________________________________________________________________________________

maple [A]  time = 0.66, size = 249, normalized size = 1.66 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 d \,a^{2}}+\frac {C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5 C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 d \,a^{2}}-\frac {C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5 C}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 288, normalized size = 1.92 \[ -\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*(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) + s
in(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*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/
a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2))/d

________________________________________________________________________________________

mupad [B]  time = 2.69, size = 144, normalized size = 0.96 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {7\,C}{2}\right )}{a^2\,d}-\frac {3\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^2}+\frac {2\,C}{a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A + (7*C)/2))/(a^2*d) - (3*C*tan(c/2 + (d*x)/2) - 5*C*tan(c/2 + (d*x)/2)^3)/(d*(
a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^2) + (2*
C)/a^2))/d - (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

________________________________________________________________________________________