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

Optimal. Leaf size=138 \[ \frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-1/7*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(4*A-11*B)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3+1/105*(8*A+13*B)*
tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+1/105*(8*A+13*B)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))

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Rubi [A]  time = 0.26, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4008, 4000, 3796, 3794} \[ \frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A - B)*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((4*A - 11*B)*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x]
)^3) + ((8*A + 13*B)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((8*A + 13*B)*Tan[c + d*x])/(105*d*(a^
4 + a^4*Sec[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]

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 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)]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec (c+d x) (-4 a (A-B)-7 a B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(8 A+13 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(8 A+13 B) \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 = 0.42, size = 163, normalized size = 1.18 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-35 (5 A+4 B) \sin \left (c+\frac {d x}{2}\right )+140 (2 A+B) \sin \left (\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+56 B \sin \left (2 c+\frac {5 d x}{2}\right )+8 B \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(140*(2*A + B)*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] - 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] + 13*A*Sin[3*c + (7*d*x)/2] + 8*B*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)

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fricas [A]  time = 0.43, size = 124, normalized size = 0.90 \[ \frac {{\left ({\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (8 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 13 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((13*A + 8*B)*cos(d*x + c)^3 + 4*(13*A + 8*B)*cos(d*x + c)^2 + 4*(8*A + 13*B)*cos(d*x + c) + 8*A + 13*B)
*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 = 0.30, size = 117, normalized size = 0.85 \[ \frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(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 - 21*A*tan(1/2*d*x + 1/2*c)^5 - 21*B*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*A*tan(1/2*d*x + 1/2*c) + 105*
B*tan(1/2*d*x + 1/2*c))/(a^4*d)

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maple [A]  time = 0.72, size = 88, normalized size = 0.64 \[ \frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 174, normalized size = 1.26 \[ \frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(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)/d

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mupad [B]  time = 1.98, size = 84, normalized size = 0.61 \[ -\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B\right )}{40\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{24\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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