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

Optimal. Leaf size=112 \[ \frac{(A-B+4 C) \tan (c+d x)}{3 a^2 d}+\frac{(B-2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(B-2 C) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.282613, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4084, 4008, 3787, 3770, 3767, 8} \[ \frac{(A-B+4 C) \tan (c+d x)}{3 a^2 d}+\frac{(B-2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(B-2 C) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

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)]

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 8

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

Rubi steps

\begin{align*} \int \frac{\sec ^2(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 ^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) (a (A+2 B-2 C)+a (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{(A-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 (A-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{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(B-2 C) \int \sec (c+d x) \, dx}{a^2}+\frac{(A-B+4 C) \int \sec ^2(c+d x) \, dx}{3 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{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(A-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{(A-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{(A-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 = 2.07581, size = 312, normalized size = 2.79 \[ \frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\tan \left (\frac{c}{2}\right ) (A-B+C) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) (A-B+C) \sin \left (\frac{d x}{2}\right )+2 \sec \left (\frac{c}{2}\right ) (A-4 B+7 C) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )+\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{6 C \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-6 (B-2 C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((A - B + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(A - 4*B +
7*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]^3*(-6*(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]]) + (6*C*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + S
in[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (A - B + C)*Cos[(c +
d*x)/2]*Tan[c/2]))/(3*a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.063, size = 243, normalized size = 2.2 \begin{align*}{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{2}}}-{\frac{C}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{2}}}-{\frac{C}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(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*tan(1/2*d*x+1/2*c)^3*B+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+1/2/d/a^2*A
*tan(1/2*d*x+1/2*c)-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 = 0.96502, size = 387, normalized size = 3.46 \begin{align*} \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 )} + \frac{A{\left (\frac{3 \, \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}}\right )}}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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*((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) + A*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 0.522094, size = 513, normalized size = 4.58 \begin{align*} \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 ({\left (A - 4 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, A - 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+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*((A - 4*B + 10*C)*cos(d*x + c)^2 + (2*A - 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))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{2}{\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 ^{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + 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

________________________________________________________________________________________

Giac [A]  time = 1.2245, size = 244, normalized size = 2.18 \begin{align*} \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{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} + 3 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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*(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) + (A*a^4*tan(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 + 3*A*a^4*tan(1/2*d*x + 1/2*c) - 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