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

Optimal. Leaf size=108 \[ -\frac {(B-4 C) \tan (c+d x)}{3 a^2 d}+\frac {(B-2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(B-2 C) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4072, 4019, 4008, 3787, 3770, 3767, 8} \[ -\frac {(B-4 C) \tan (c+d x)}{3 a^2 d}+\frac {(B-2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(B-2 C) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B - 2*C)*ArcTanh[Sin[c + d*x]])/(a^2*d) - ((B - 4*C)*Tan[c + d*x])/(3*a^2*d) - ((B - 2*C)*Tan[c + d*x])/(a^2
*d*(1 + Sec[c + d*x])) + ((B - C)*Sec[c + d*x]^2*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 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 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 4072

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 1.14, size = 245, normalized size = 2.27 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left ((4 B-7 C) \tan ^3\left (\frac {1}{2} (c+d x)\right )+(13 C-4 B) \tan \left (\frac {1}{2} (c+d x)\right )+16 (B-C) \sin ^8\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)-4 (B-C) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)-3 (B-2 C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 (B-2 C) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(-3*(B - 2*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]]) - 4*(B - C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 16*(B - C)*Csc[c + d*x]^5*Sin[(c + d*x)
/2]^8 + (-4*B + 13*C)*Tan[(c + d*x)/2] + 3*(B - 2*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]])*Tan[(c + d*x)/2]^2 + (4*B - 7*C)*Tan[(c + d*x)/2]^3))/(3*a^2*d)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 195, normalized size = 1.81 \[ \frac {3 \, {\left ({\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 2 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 2 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, B - 14 \, C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.27, size = 151, normalized size = 1.40 \[ \frac {\frac {6 \, {\left (B - 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (B - 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(B - 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(B - 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 -
12*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^2) - (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*
d*x + 1/2*c)^3 + 9*B*a^4*tan(1/2*d*x + 1/2*c) - 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

________________________________________________________________________________________

maple [A]  time = 0.67, size = 205, normalized size = 1.90 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{2}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{2}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(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*B*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*B*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*
tan(1/2*d*x+1/2*c)-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^2/(tan(1/2*d*x+
1/2*c)-1)*C+1/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-1/d/a^2/(tan(1/2*d*x+1/2*c)+
1)*C

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 244, normalized size = 2.26 \[ \frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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*((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))) - B*((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 = 2.86, size = 120, normalized size = 1.11 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-2\,C\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{a^2}+\frac {B-3\,C}{2\,a^2}\right )}{d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(B - 2*C))/(a^2*d) - (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d) - (tan(c/2 + (d*x)/
2)*((B - C)/a^2 + (B - 3*C)/(2*a^2)))/d - (2*C*tan(c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x)/2)^2 - a^2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(B*sec(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

________________________________________________________________________________________