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

Optimal. Leaf size=154 \[ \frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{(8 A-B-6 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((8*A - B - 6*C)*Tan[c + d*x])/(35*a*d
*(a + a*Sec[c + d*x])^3) + ((6*A + 8*B + 13*C)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((6*A + 8*B
+ 13*C)*Tan[c + d*x])/(105*d*(a^4 + a^4*Sec[c + d*x]))

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Rubi [A]  time = 0.260498, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4078, 4000, 3796, 3794} \[ \frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{(8 A-B-6 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((8*A - B - 6*C)*Tan[c + d*x])/(35*a*d
*(a + a*Sec[c + d*x])^3) + ((6*A + 8*B + 13*C)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((6*A + 8*B
+ 13*C)*Tan[c + d*x])/(105*d*(a^4 + a^4*Sec[c + d*x]))

Rule 4078

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

Rule 4000

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

Rule 3796

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

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 (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{\int \frac{\sec (c+d x) (a (6 A+B-C)-a (2 A-2 B-5 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(6 A+8 B+13 C) \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{(6 A+8 B+13 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.762154, size = 231, normalized size = 1.5 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (-35 (18 A+5 B+4 C) \sin \left (c+\frac{d x}{2}\right )+70 (9 A+4 B+2 C) \sin \left (\frac{d x}{2}\right )+441 A \sin \left (c+\frac{3 d x}{2}\right )-315 A \sin \left (2 c+\frac{3 d x}{2}\right )+147 A \sin \left (2 c+\frac{5 d x}{2}\right )-105 A \sin \left (3 c+\frac{5 d x}{2}\right )+36 A \sin \left (3 c+\frac{7 d x}{2}\right )+168 B \sin \left (c+\frac{3 d x}{2}\right )-105 B \sin \left (2 c+\frac{3 d x}{2}\right )+91 B \sin \left (2 c+\frac{5 d x}{2}\right )+13 B \sin \left (3 c+\frac{7 d x}{2}\right )+168 C \sin \left (c+\frac{3 d x}{2}\right )+56 C \sin \left (2 c+\frac{5 d x}{2}\right )+8 C \sin \left (3 c+\frac{7 d x}{2}\right )\right )}{6720 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(70*(9*A + 4*B + 2*C)*Sin[(d*x)/2] - 35*(18*A + 5*B + 4*C)*Sin[c + (d*x)/2] + 441
*A*Sin[c + (3*d*x)/2] + 168*B*Sin[c + (3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] - 315*A*Sin[2*c + (3*d*x)/2] - 105
*B*Sin[2*c + (3*d*x)/2] + 147*A*Sin[2*c + (5*d*x)/2] + 91*B*Sin[2*c + (5*d*x)/2] + 56*C*Sin[2*c + (5*d*x)/2] -
105*A*Sin[3*c + (5*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] + 13*B*Sin[3*c + (7*d*x)/2] + 8*C*Sin[3*c + (7*d*x)/2]
))/(6720*a^4*d)

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Maple [A]  time = 0.078, size = 108, normalized size = 0.7 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{-A+B-C}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{3\,A-C-B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{-3\,A-B+C}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +C\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)

[Out]

1/8/d/a^4*(1/7*(-A+B-C)*tan(1/2*d*x+1/2*c)^7+1/5*(3*A-C-B)*tan(1/2*d*x+1/2*c)^5+1/3*(-3*A-B+C)*tan(1/2*d*x+1/2
*c)^3+A*tan(1/2*d*x+1/2*c)+B*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 0.989747, size = 350, normalized size = 2.27 \begin{align*} \frac{\frac{C{\left (\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac{B{\left (\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac{3 \, A{\left (\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(C*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + B*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

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Fricas [A]  time = 0.473608, size = 343, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (36 \, A + 13 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (39 \, A + 52 \, B + 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (6 \, A + 8 \, B + 13 \, C\right )} \cos \left (d x + c\right ) + 6 \, A + 8 \, B + 13 \, C\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((36*A + 13*B + 8*C)*cos(d*x + c)^3 + (39*A + 52*B + 32*C)*cos(d*x + c)^2 + 4*(6*A + 8*B + 13*C)*cos(d*x
+ c) + 6*A + 8*B + 13*C)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2
+ 4*a^4*d*cos(d*x + c) + a^4*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) +
Integral(B*sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(C*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x
))/a**4

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Giac [A]  time = 1.2063, size = 231, normalized size = 1.5 \begin{align*} -\frac{15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 63 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{840 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 63*A*tan(1/2
*d*x + 1/2*c)^5 + 21*B*tan(1/2*d*x + 1/2*c)^5 + 21*C*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1/2*c)^3 + 3
5*B*tan(1/2*d*x + 1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 105*B*tan(1/2*d*x + 1/
2*c) - 105*C*tan(1/2*d*x + 1/2*c))/(a^4*d)