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

Optimal. Leaf size=138 \[ \frac{4 (2 A+9 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac{(23 A-54 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 (3 A-4 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

((23*A - 54*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + (4*(2*A + 9*C)*Tan[c + d*x])/(105*a^4*d*(1 + S
ec[c + d*x])) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*(3*A - 4*C)*Tan[c + d*
x])/(35*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.374343, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4085, 4008, 4000, 3794} \[ \frac{4 (2 A+9 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac{(23 A-54 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 (3 A-4 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((23*A - 54*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + (4*(2*A + 9*C)*Tan[c + d*x])/(105*a^4*d*(1 + S
ec[c + d*x])) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*(3*A - 4*C)*Tan[c + d*
x])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b*(
2*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, 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 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 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+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A+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 C)+a (A-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (3 A-4 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) \left (6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(23 A-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (3 A-4 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(4 (2 A+9 C)) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac{(23 A-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (3 A-4 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{4 (2 A+9 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.641402, size = 151, normalized size = 1.09 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (-175 A \sin \left (c+\frac{d x}{2}\right )+168 A \sin \left (c+\frac{3 d x}{2}\right )-105 A \sin \left (2 c+\frac{3 d x}{2}\right )+91 A \sin \left (2 c+\frac{5 d x}{2}\right )+13 A \sin \left (3 c+\frac{7 d x}{2}\right )+70 (4 A+3 C) \sin \left (\frac{d x}{2}\right )+126 C \sin \left (c+\frac{3 d x}{2}\right )+42 C \sin \left (2 c+\frac{5 d x}{2}\right )+6 C \sin \left (3 c+\frac{7 d x}{2}\right )\right )}{6720 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(A + 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 + 3*C)*Sin[(d*x)/2] - 175*A*Sin[c + (d*x)/2] + 168*A*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] + 42*C*Sin[2*c + (5*d*x)/2
] + 13*A*Sin[3*c + (7*d*x)/2] + 6*C*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)

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Maple [A]  time = 0.067, size = 88, normalized size = 0.6 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{A+C}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{-A+3\,C}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{-A+3\,C}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\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)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 0.974169, size = 236, normalized size = 1.71 \begin{align*} \frac{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(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/(co
s(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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Fricas [A]  time = 0.462843, size = 308, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (32 \, A + 39 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((13*A + 6*C)*cos(d*x + c)^3 + 4*(13*A + 6*C)*cos(d*x + c)^2 + (32*A + 39*C)*cos(d*x + c) + 8*A + 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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+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(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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Giac [A]  time = 1.23181, size = 158, normalized size = 1.14 \begin{align*} \frac{15 \, A \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} + 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} + 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 \, 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)^2*(A+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*C*tan(1/2*d*x + 1/2*c)^7 - 21*A*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 + 105*C*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105
*C*tan(1/2*d*x + 1/2*c))/(a^4*d)