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

Optimal. Leaf size=136 \[ -\frac {8 (20 A-C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(55 A-8 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {A x}{a^4}-\frac {2 (5 A-2 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

A*x/a^4-1/105*(55*A-8*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-8/105*(20*A-C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*
(A+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-2/35*(5*A-2*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.26, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4053, 3922, 3919, 3794} \[ -\frac {8 (20 A-C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(55 A-8 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {A x}{a^4}-\frac {2 (5 A-2 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a^4 - ((55*A - 8*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (8*(20*A - C)*Tan[c + d*x])/(105*a^
4*d*(1 + Sec[c + d*x])) - ((A + C)*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*(5*A - 2*C)*Tan[c + d*x])/(
35*a*d*(a + a*Sec[c + d*x])^3)

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 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 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 4053

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

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {-7 a A+a (3 A-4 C) \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (5 A-2 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {35 a^2 A-4 a^2 (5 A-2 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(55 A-8 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (5 A-2 C) \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-8 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac {A x}{a^4}-\frac {(55 A-8 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (5 A-2 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(8 (20 A-C)) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac {A x}{a^4}-\frac {(55 A-8 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (5 A-2 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {8 (20 A-C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.15, size = 315, normalized size = 2.32 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (8260 A \sin \left (c+\frac {d x}{2}\right )-7140 A \sin \left (c+\frac {3 d x}{2}\right )+3780 A \sin \left (2 c+\frac {3 d x}{2}\right )-2800 A \sin \left (2 c+\frac {5 d x}{2}\right )+840 A \sin \left (3 c+\frac {5 d x}{2}\right )-520 A \sin \left (3 c+\frac {7 d x}{2}\right )+3675 A d x \cos \left (c+\frac {d x}{2}\right )+2205 A d x \cos \left (c+\frac {3 d x}{2}\right )+2205 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+735 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+735 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+105 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+105 A d x \cos \left (4 c+\frac {7 d x}{2}\right )-9940 A \sin \left (\frac {d x}{2}\right )+3675 A d x \cos \left (\frac {d x}{2}\right )-350 C \sin \left (c+\frac {d x}{2}\right )+336 C \sin \left (c+\frac {3 d x}{2}\right )-210 C \sin \left (2 c+\frac {3 d x}{2}\right )+182 C \sin \left (2 c+\frac {5 d x}{2}\right )+26 C \sin \left (3 c+\frac {7 d x}{2}\right )+560 C \sin \left (\frac {d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(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*(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] + 560*C*Sin[(d*x)/2] + 826
0*A*Sin[c + (d*x)/2] - 350*C*Sin[c + (d*x)/2] - 7140*A*Sin[c + (3*d*x)/2] + 336*C*Sin[c + (3*d*x)/2] + 3780*A*
Sin[2*c + (3*d*x)/2] - 210*C*Sin[2*c + (3*d*x)/2] - 2800*A*Sin[2*c + (5*d*x)/2] + 182*C*Sin[2*c + (5*d*x)/2] +
840*A*Sin[3*c + (5*d*x)/2] - 520*A*Sin[3*c + (7*d*x)/2] + 26*C*Sin[3*c + (7*d*x)/2]))/(13440*a^4*d)

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fricas [A]  time = 0.43, size = 182, normalized size = 1.34 \[ \frac {105 \, A d x \cos \left (d x + c\right )^{4} + 420 \, A d x \cos \left (d x + c\right )^{3} + 630 \, A d x \cos \left (d x + c\right )^{2} + 420 \, A d x \cos \left (d x + c\right ) + 105 \, A d x - {\left (13 \, {\left (20 \, A - C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (155 \, A - 13 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (535 \, A - 32 \, C\right )} \cos \left (d x + c\right ) + 160 \, A - 8 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(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 - (13*(20*A - C)*cos(d*x + c)^3 + 4*(155*A - 13*C)*cos(d*x + c)^2 + (535*A - 32*C)*cos(d*x + c) +
160*A - 8*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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giac [A]  time = 0.25, size = 154, normalized size = 1.13 \[ \frac {\frac {840 \, {\left (d x + c\right )} A}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(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*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 105*A*a^24
*tan(1/2*d*x + 1/2*c)^5 - 21*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 35*C*a^24*tan
(1/2*d*x + 1/2*c)^3 - 1575*A*a^24*tan(1/2*d*x + 1/2*c) + 105*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.81, size = 177, normalized size = 1.30 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(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*C*tan(1/2*d*x+1/2*c)^7-1/8/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/40/d/a^
4*C*tan(1/2*d*x+1/2*c)^5+11/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A-1/24/d/a^4*C*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*C*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 = 0.42, size = 201, normalized size = 1.48 \[ -\frac {5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(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) - C*(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 = 2.64, size = 163, normalized size = 1.20 \[ \frac {A\,x}{a^4}+\frac {\left (\frac {13\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-\frac {52\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {16\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}+\frac {13\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{210}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {5\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{28}-\frac {11\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{140}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*x)/a^4 + ((A*sin(c/2 + (d*x)/2))/56 + (C*sin(c/2 + (d*x)/2))/56 - cos(c/2 + (d*x)/2)^2*((5*A*sin(c/2 + (d*x
)/2))/28 + (11*C*sin(c/2 + (d*x)/2))/140) - cos(c/2 + (d*x)/2)^6*((52*A*sin(c/2 + (d*x)/2))/21 - (13*C*sin(c/2
+ (d*x)/2))/105) + cos(c/2 + (d*x)/2)^4*((16*A*sin(c/2 + (d*x)/2))/21 + (13*C*sin(c/2 + (d*x)/2))/210))/(a^4*
d*cos(c/2 + (d*x)/2)^7)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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