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

Optimal. Leaf size=138 \[ -\frac{2 (80 A-3 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(55 A-6 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{A x}{a^4}-\frac{(10 A-3 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

(A*x)/a^4 - ((55*A - 6*B)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (2*(80*A - 3*B)*Tan[c + d*x])/(105*
a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - ((10*A - 3*B)*Tan[c + d*x])/
(35*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.267809, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3922, 3919, 3794} \[ -\frac{2 (80 A-3 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(55 A-6 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{A x}{a^4}-\frac{(10 A-3 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a^4 - ((55*A - 6*B)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (2*(80*A - 3*B)*Tan[c + d*x])/(105*
a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - ((10*A - 3*B)*Tan[c + d*x])/
(35*a*d*(a + a*Sec[c + d*x])^3)

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))^4} \, dx &=-\frac{(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{-7 a A+3 a (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{(10 A-3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{35 a^2 A-2 a^2 (10 A-3 B) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(55 A-6 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(10 A-3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{-105 a^3 A+a^3 (55 A-6 B) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac{A x}{a^4}-\frac{(55 A-6 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(10 A-3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(2 (80 A-3 B)) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac{A x}{a^4}-\frac{(55 A-6 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{(10 A-3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{2 (80 A-3 B) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.755103, size = 329, normalized size = 2.38 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (8260 A \sin \left (c+\frac{d x}{2}\right )-7140 A \sin \left (c+\frac{3 d x}{2}\right )+3780 A \sin \left (2 c+\frac{3 d x}{2}\right )-2800 A \sin \left (2 c+\frac{5 d x}{2}\right )+840 A \sin \left (3 c+\frac{5 d x}{2}\right )-520 A \sin \left (3 c+\frac{7 d x}{2}\right )+3675 A d x \cos \left (c+\frac{d x}{2}\right )+2205 A d x \cos \left (c+\frac{3 d x}{2}\right )+2205 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+735 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+735 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+105 A d x \cos \left (3 c+\frac{7 d x}{2}\right )+105 A d x \cos \left (4 c+\frac{7 d x}{2}\right )-9940 A \sin \left (\frac{d x}{2}\right )+3675 A d x \cos \left (\frac{d x}{2}\right )-1260 B \sin \left (c+\frac{d x}{2}\right )+882 B \sin \left (c+\frac{3 d x}{2}\right )-630 B \sin \left (2 c+\frac{3 d x}{2}\right )+294 B \sin \left (2 c+\frac{5 d x}{2}\right )-210 B \sin \left (3 c+\frac{5 d x}{2}\right )+72 B \sin \left (3 c+\frac{7 d x}{2}\right )+1260 B \sin \left (\frac{d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*A*d*x*Cos[(d*x)/2] + 3675*A*d*x*Cos[c + (d*x)/2] + 2205*A*d*x*Cos[c + (3*d*
x)/2] + 2205*A*d*x*Cos[2*c + (3*d*x)/2] + 735*A*d*x*Cos[2*c + (5*d*x)/2] + 735*A*d*x*Cos[3*c + (5*d*x)/2] + 10
5*A*d*x*Cos[3*c + (7*d*x)/2] + 105*A*d*x*Cos[4*c + (7*d*x)/2] - 9940*A*Sin[(d*x)/2] + 1260*B*Sin[(d*x)/2] + 82
60*A*Sin[c + (d*x)/2] - 1260*B*Sin[c + (d*x)/2] - 7140*A*Sin[c + (3*d*x)/2] + 882*B*Sin[c + (3*d*x)/2] + 3780*
A*Sin[2*c + (3*d*x)/2] - 630*B*Sin[2*c + (3*d*x)/2] - 2800*A*Sin[2*c + (5*d*x)/2] + 294*B*Sin[2*c + (5*d*x)/2]
+ 840*A*Sin[3*c + (5*d*x)/2] - 210*B*Sin[3*c + (5*d*x)/2] - 520*A*Sin[3*c + (7*d*x)/2] + 72*B*Sin[3*c + (7*d*
x)/2]))/(13440*a^4*d)

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Maple [A]  time = 0.068, size = 177, normalized size = 1.3 \begin{align*}{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{3\,B}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{11\,A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{15\,A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{8\,d{a}^{4}}\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}^{4}}} \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))^4,x)

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*B-1/8/d/a^4*tan(1/2*d*x+1/2*c)^5*A+3/40/d/a^
4*tan(1/2*d*x+1/2*c)^5*B+11/24/d/a^4*A*tan(1/2*d*x+1/2*c)^3-1/8/d/a^4*B*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*A*tan(
1/2*d*x+1/2*c)+1/8/d/a^4*B*tan(1/2*d*x+1/2*c)+2/d/a^4*A*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.47748, size = 271, normalized size = 1.96 \begin{align*} -\frac{5 \, A{\left (\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{77 \, \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{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{336 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac{3 \, B{\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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^4) - 3*B*(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 = 0.470773, size = 475, normalized size = 3.44 \begin{align*} \frac{105 \, A d x \cos \left (d x + c\right )^{4} + 420 \, A d x \cos \left (d x + c\right )^{3} + 630 \, A d x \cos \left (d x + c\right )^{2} + 420 \, A d x \cos \left (d x + c\right ) + 105 \, A d x -{\left (4 \,{\left (65 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (620 \, A - 39 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (535 \, A - 24 \, B\right )} \cos \left (d x + c\right ) + 160 \, A - 6 \, 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 )}} \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))^4,x, algorithm="fricas")

[Out]

1/105*(105*A*d*x*cos(d*x + c)^4 + 420*A*d*x*cos(d*x + c)^3 + 630*A*d*x*cos(d*x + c)^2 + 420*A*d*x*cos(d*x + c)
+ 105*A*d*x - (4*(65*A - 9*B)*cos(d*x + c)^3 + (620*A - 39*B)*cos(d*x + c)^2 + (535*A - 24*B)*cos(d*x + c) +
160*A - 6*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*c
os(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 ^{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{\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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A/(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*se
c(c + d*x)/(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 = 1.20594, size = 208, normalized size = 1.51 \begin{align*} \frac{\frac{840 \,{\left (d x + c\right )} A}{a^{4}} + \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 63 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1575 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, 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))^4,x, algorithm="giac")

[Out]

1/840*(840*(d*x + c)*A/a^4 + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 - 105*A*a^24
*tan(1/2*d*x + 1/2*c)^5 + 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*B*a^24*ta
n(1/2*d*x + 1/2*c)^3 - 1575*A*a^24*tan(1/2*d*x + 1/2*c) + 105*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d