3.148 \(\int \frac{\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=161 \[ \frac{(16 A-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(8 A-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{2 (2 A-5 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((8*A - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*A - 21
5*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c +
d*x])^4) + (2*(2*A - 5*C)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.484186, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4085, 4019, 4008, 3998, 3770, 3794} \[ \frac{(16 A-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(8 A-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{2 (2 A-5 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((8*A - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*A - 21
5*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c +
d*x])^4) + (2*(2*A - 5*C)*Sec[c + d*x]^2*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 4019

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

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 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{\sec ^3(c+d x) (-a (4 A-3 C)-7 a C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^2(c+d x) \left (-4 a^2 (2 A-5 C)-35 a^2 C \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) \left (2 a^3 (8 A-55 C)+105 a^3 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(16 A-215 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}+\frac{C \int \sec (c+d x) \, dx}{a^4}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(16 A-215 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.19314, size = 283, normalized size = 1.76 \[ -\frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (6720 C \cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec \left (\frac{c}{2}\right ) \left (-70 (2 A-31 C) \sin \left (c+\frac{d x}{2}\right )+168 A \sin \left (c+\frac{3 d x}{2}\right )+56 A \sin \left (2 c+\frac{5 d x}{2}\right )+8 A \sin \left (3 c+\frac{7 d x}{2}\right )+70 (2 A-49 C) \sin \left (\frac{d x}{2}\right )-2625 C \sin \left (c+\frac{3 d x}{2}\right )+735 C \sin \left (2 c+\frac{3 d x}{2}\right )-1015 C \sin \left (2 c+\frac{5 d x}{2}\right )+105 C \sin \left (3 c+\frac{5 d x}{2}\right )-160 C \sin \left (3 c+\frac{7 d x}{2}\right )\right )\right )}{210 a^4 d (\sec (c+d x)+1)^4 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(6720*C*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c/2]*(70*(2*A - 49*C)*Sin[(d*x)/2] - 70*(2*A
- 31*C)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] - 2625*C*Sin[c + (3*d*x)/2] + 735*C*Sin[2*c + (3*d*x)/2]
+ 56*A*Sin[2*c + (5*d*x)/2] - 1015*C*Sin[2*c + (5*d*x)/2] + 105*C*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (7*d*x)
/2] - 160*C*Sin[3*c + (7*d*x)/2])))/(210*a^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^4)

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Maple [A]  time = 0.068, size = 199, normalized size = 1.2 \begin{align*}{\frac{A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{15\,C}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{11\,C}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{C}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 0.961218, size = 308, normalized size = 1.91 \begin{align*} -\frac{5 \, C{\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{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \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}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*C*((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 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) +
1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - 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/(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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Fricas [A]  time = 0.51241, size = 624, normalized size = 3.88 \begin{align*} \frac{105 \,{\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (A - 20 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (32 \, A - 535 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (13 \, A - 155 \, C\right )} \cos \left (d x + c\right ) + 13 \, A - 260 \, C\right )} \sin \left (d x + c\right )}{210 \,{\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)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/210*(105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(sin(d*x + c
) + 1) - 105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(-sin(d*x
+ c) + 1) + 2*(8*(A - 20*C)*cos(d*x + c)^3 + (32*A - 535*C)*cos(d*x + c)^2 + 4*(13*A - 155*C)*cos(d*x + c) + 1
3*A - 260*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*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 ^{3}{\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 ^{5}{\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)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**3/(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)**5/(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 = 1.24557, size = 246, normalized size = 1.53 \begin{align*} \frac{\frac{840 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{840 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{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} + 21 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 105 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1575 \, C 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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (15*A*a^2
4*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 21*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*C*a^24*ta
n(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*A*a^24*tan(1
/2*d*x + 1/2*c) + 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d