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

Optimal. Leaf size=132 \[ \frac{(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (
(3*A + 2*B - 7*C)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + ((6*A + 4*B - 29*C)*Tan[c + d*x])/(15*d*(a^3
+ a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.350824, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4084, 4008, 3998, 3770, 3794} \[ \frac{(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

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])^3,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (
(3*A + 2*B - 7*C)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + ((6*A + 4*B - 29*C)*Tan[c + d*x])/(15*d*(a^3
+ a^3*Sec[c + d*x]))

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 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

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))^3} \, dx &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^2(c+d x) (a (3 A+2 B-2 C)+5 a C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (-2 a^2 (3 A+2 B-7 C)-15 a^2 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(6 A+4 B-29 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac{C \int \sec (c+d x) \, dx}{a^3}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.61372, size = 277, normalized size = 2.1 \[ -\frac{\left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (240 C \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (5 (3 A+4 B-29 C) \sin \left (\frac{d x}{2}\right )-15 (A-5 C) \sin \left (c+\frac{d x}{2}\right )+15 A \sin \left (c+\frac{3 d x}{2}\right )+3 A \sin \left (2 c+\frac{5 d x}{2}\right )+10 B \sin \left (c+\frac{3 d x}{2}\right )+2 B \sin \left (2 c+\frac{5 d x}{2}\right )-95 C \sin \left (c+\frac{3 d x}{2}\right )+15 C \sin \left (2 c+\frac{3 d x}{2}\right )-22 C \sin \left (2 c+\frac{5 d x}{2}\right )\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

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])^3,x]

[Out]

-((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(240*C*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
- Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*A + 4*B - 29*C)*Sin[(d*x)/2] - 1
5*(A - 5*C)*Sin[c + (d*x)/2] + 15*A*Sin[c + (3*d*x)/2] + 10*B*Sin[c + (3*d*x)/2] - 95*C*Sin[c + (3*d*x)/2] + 1
5*C*Sin[2*c + (3*d*x)/2] + 3*A*Sin[2*c + (5*d*x)/2] + 2*B*Sin[2*c + (5*d*x)/2] - 22*C*Sin[2*c + (5*d*x)/2])))/
(15*a^3*d*(1 + Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.069, size = 197, normalized size = 1.5 \begin{align*}{\frac{B}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{7\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}} \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))^3,x)

[Out]

1/6/d/a^3*tan(1/2*d*x+1/2*c)^3*B+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C+1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*B-1/20/d/a
^3*tan(1/2*d*x+1/2*c)^5*A-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)-1/d/a^3*ln(tan(1/2*
d*x+1/2*c)-1)*C+1/4/d/a^3*A*tan(1/2*d*x+1/2*c)+1/4/d/a^3*B*tan(1/2*d*x+1/2*c)-1/3/d/a^3*C*tan(1/2*d*x+1/2*c)^3

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Maxima [A]  time = 0.967131, size = 313, normalized size = 2.37 \begin{align*} -\frac{C{\left (\frac{\frac{105 \, \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{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac{B{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{3 \, A{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c
)^5/(cos(d*x + c) + 1)^5)/a^3 - 3*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/
a^3)/d

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Fricas [A]  time = 0.511033, size = 505, normalized size = 3.83 \begin{align*} \frac{15 \,{\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (3 \, A + 2 \, B - 22 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A + 2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 7 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\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))^3,x, algorithm="fricas")

[Out]

1/30*(15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - 15*(C*cos(d*x
+ c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*log(-sin(d*x + c) + 1) + 2*((3*A + 2*B - 22*C)*cos(d*x + c
)^2 + 3*(3*A + 2*B - 17*C)*cos(d*x + c) + 3*A + 7*B - 32*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(
d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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

[Out]

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

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Giac [A]  time = 1.31097, size = 243, normalized size = 1.84 \begin{align*} \frac{\frac{60 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(60*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - (3*A*a^12*ta
n(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*B*a^12*tan(1/2*d
*x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^12*tan(1/2*d*x + 1/2*c) - 15*B*a^12*tan(1/2*d*x + 1/
2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d