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

Optimal. Leaf size=148 \[ \frac {(8 A+13 B+36 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

1/105*(23*A-2*B-54*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+1/105*(8*A+13*B+36*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-
1/7*(A-B+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-1/35*(6*A+B-8*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.41, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4084, 4008, 4000, 3794} \[ \frac {(8 A+13 B+36 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

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])^4,x]

[Out]

((23*A - 2*B - 54*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((8*A + 13*B + 36*C)*Tan[c + d*x])/(105*
a^4*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - ((6*A + B
- 8*C)*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

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]

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 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 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)]

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))^4} \, dx &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^2(c+d x) (a (5 A+2 B-2 C)-a (A-B-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (-3 a^2 (6 A+B-8 C)+5 a^2 (A-B-6 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B+36 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B+36 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.70, size = 200, normalized size = 1.35 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (-35 (5 A+4 B) \sin \left (c+\frac {d x}{2}\right )+70 (4 A+2 B+3 C) \sin \left (\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+56 B \sin \left (2 c+\frac {5 d x}{2}\right )+8 B \sin \left (3 c+\frac {7 d x}{2}\right )+126 C \sin \left (c+\frac {3 d x}{2}\right )+42 C \sin \left (2 c+\frac {5 d x}{2}\right )+6 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(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*(4*A + 2*B + 3*C)*Sin[(d*x)/2] - 35*(5*A + 4*B)*Sin[c + (d*x)/2] + 168*A*Sin[
c + (3*d*x)/2] + 168*B*Sin[c + (3*d*x)/2] + 126*C*Sin[c + (3*d*x)/2] - 105*A*Sin[2*c + (3*d*x)/2] + 91*A*Sin[2
*c + (5*d*x)/2] + 56*B*Sin[2*c + (5*d*x)/2] + 42*C*Sin[2*c + (5*d*x)/2] + 13*A*Sin[3*c + (7*d*x)/2] + 8*B*Sin[
3*c + (7*d*x)/2] + 6*C*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)

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fricas [A]  time = 0.41, size = 135, normalized size = 0.91 \[ \frac {{\left ({\left (13 \, A + 8 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 8 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A + 52 \, B + 39 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 13 \, B + 36 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

1/105*((13*A + 8*B + 6*C)*cos(d*x + c)^3 + 4*(13*A + 8*B + 6*C)*cos(d*x + c)^2 + (32*A + 52*B + 39*C)*cos(d*x
+ c) + 8*A + 13*B + 36*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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giac [A]  time = 0.28, size = 171, normalized size = 1.16 \[ \frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

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

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maple [A]  time = 0.92, size = 106, normalized size = 0.72 \[ \frac {\frac {\left (A -B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 C -A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A +B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}} \]

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))^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*C-A-B)*tan(1/2*d*x+1/2*c)^5+1/3*(-A+B+3*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 = 0.37, size = 259, normalized size = 1.75 \[ \frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

1/840*(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/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + A*(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*C*(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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mupad [B]  time = 3.30, size = 99, normalized size = 0.67 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B-3\,C\right )}{40\,a^4\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{8\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-A+3\,C\right )}{24\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) - (tan(c/2 + (d*x)/2)^5*(A + B - 3*C))/(40*a^4*d) + (tan(c/2 + (
d*x)/2)*(A + B + C))/(8*a^4*d) + (tan(c/2 + (d*x)/2)^3*(B - A + 3*C))/(24*a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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))**4,x)

[Out]

(Integral(A*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(B*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) + Integral(C*sec(c + d*x)**4/(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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