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

Optimal. Leaf size=102 \[ \frac{(2 A+3 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 A+3 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

((A - B)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((2*A + 3*B)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^
2) + ((2*A + 3*B)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.114332, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4000, 3796, 3794} \[ \frac{(2 A+3 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 A+3 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

((A - B)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((2*A + 3*B)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^
2) + ((2*A + 3*B)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a}\\ &=\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(2 A+3 B) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(2 A+3 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.324102, size = 135, normalized size = 1.32 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-15 (2 A+B) \sin \left (c+\frac{d x}{2}\right )+5 (8 A+3 B) \sin \left (\frac{d x}{2}\right )+20 A \sin \left (c+\frac{3 d x}{2}\right )-15 A \sin \left (2 c+\frac{3 d x}{2}\right )+7 A \sin \left (2 c+\frac{5 d x}{2}\right )+15 B \sin \left (c+\frac{3 d x}{2}\right )+3 B \sin \left (2 c+\frac{5 d x}{2}\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(8*A + 3*B)*Sin[(d*x)/2] - 15*(2*A + B)*Sin[c + (d*x)/2] + 20*A*Sin[c + (3*d*x)/
2] + 15*B*Sin[c + (3*d*x)/2] - 15*A*Sin[2*c + (3*d*x)/2] + 7*A*Sin[2*c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2]
))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.061, size = 64, normalized size = 0.6 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{A-B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,A}{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 ) \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))/(a+a*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.01165, size = 155, normalized size = 1.52 \begin{align*} \frac{\frac{A{\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 \, B{\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)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(A*(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*B*(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 = 0.444781, size = 227, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (7 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A + 3 \, B\right )} \sin \left (d x + c\right )}{15 \,{\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)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((7*A + 3*B)*cos(d*x + c)^2 + 3*(2*A + 3*B)*cos(d*x + c) + 2*A + 3*B)*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{\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 ^{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}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d
*x)**2/(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 = 1.32072, size = 101, normalized size = 0.99 \begin{align*} \frac{3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} 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))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 - 10*A*tan(1/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x
+ 1/2*c) + 15*B*tan(1/2*d*x + 1/2*c))/(a^3*d)