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

Optimal. Leaf size=108 \[ -\frac{2 (11 A-B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{A x}{a^3}-\frac{(7 A-2 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

(A*x)/a^3 - ((A - B)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((7*A - 2*B)*Tan[c + d*x])/(15*a*d*(a + a*Se
c[c + d*x])^2) - (2*(11*A - B)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.186089, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3922, 3919, 3794} \[ -\frac{2 (11 A-B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{A x}{a^3}-\frac{(7 A-2 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a^3 - ((A - B)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((7*A - 2*B)*Tan[c + d*x])/(15*a*d*(a + a*Se
c[c + d*x])^2) - (2*(11*A - B)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 3922

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

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{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{\int \frac{-5 a A+2 a (A-B) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-2 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{15 a^2 A-a^2 (7 A-2 B) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{A x}{a^3}-\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-2 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(2 (11 A-B)) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{A x}{a^3}-\frac{(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-2 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{2 (11 A-B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.566556, size = 241, normalized size = 2.23 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (270 A \sin \left (c+\frac{d x}{2}\right )-230 A \sin \left (c+\frac{3 d x}{2}\right )+90 A \sin \left (2 c+\frac{3 d x}{2}\right )-64 A \sin \left (2 c+\frac{5 d x}{2}\right )+150 A d x \cos \left (c+\frac{d x}{2}\right )+75 A d x \cos \left (c+\frac{3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac{5 d x}{2}\right )-370 A \sin \left (\frac{d x}{2}\right )+150 A d x \cos \left (\frac{d x}{2}\right )-60 B \sin \left (c+\frac{d x}{2}\right )+40 B \sin \left (c+\frac{3 d x}{2}\right )-30 B \sin \left (2 c+\frac{3 d x}{2}\right )+14 B \sin \left (2 c+\frac{5 d x}{2}\right )+80 B \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*A*d*x*Cos[(d*x)/2] + 150*A*d*x*Cos[c + (d*x)/2] + 75*A*d*x*Cos[c + (3*d*x)/2
] + 75*A*d*x*Cos[2*c + (3*d*x)/2] + 15*A*d*x*Cos[2*c + (5*d*x)/2] + 15*A*d*x*Cos[3*c + (5*d*x)/2] - 370*A*Sin[
(d*x)/2] + 80*B*Sin[(d*x)/2] + 270*A*Sin[c + (d*x)/2] - 60*B*Sin[c + (d*x)/2] - 230*A*Sin[c + (3*d*x)/2] + 40*
B*Sin[c + (3*d*x)/2] + 90*A*Sin[2*c + (3*d*x)/2] - 30*B*Sin[2*c + (3*d*x)/2] - 64*A*Sin[2*c + (5*d*x)/2] + 14*
B*Sin[2*c + (5*d*x)/2]))/(480*a^3*d)

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Maple [A]  time = 0.065, size = 137, normalized size = 1.3 \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*A+1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*B+1/3/d/a^3*A*tan(1/2*d*x+1/2*c)^3-1/6/d/a^
3*B*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*A*tan(1/2*d*x+1/2*c)+1/4/d/a^3*B*tan(1/2*d*x+1/2*c)+2/d/a^3*A*arctan(tan(1/
2*d*x+1/2*c))

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Maxima [A]  time = 1.46819, size = 216, normalized size = 2. \begin{align*} -\frac{A{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \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}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac{B{\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}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*(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)/d

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Fricas [A]  time = 0.460595, size = 351, normalized size = 3.25 \begin{align*} \frac{15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x -{\left ({\left (32 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (17 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, 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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*A*d*x*cos(d*x + c)^3 + 45*A*d*x*cos(d*x + c)^2 + 45*A*d*x*cos(d*x + c) + 15*A*d*x - ((32*A - 7*B)*cos
(d*x + c)^2 + 3*(17*A - 2*B)*cos(d*x + c) + 22*A - 2*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 ^{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{\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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(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 = 1.29854, size = 163, normalized size = 1.51 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )} A}{a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*A/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(
1/2*d*x + 1/2*c)^3 + 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 15*B*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d