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

Optimal. Leaf size=173 \[ \frac{(16 A+12 B-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(4 A+3 B-10 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((8*A + 6*B - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*
A + 12*B - 215*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*
d*(a + a*Sec[c + d*x])^4) + ((4*A + 3*B - 10*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 = 0.496624, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4084, 4019, 4008, 3998, 3770, 3794} \[ \frac{(16 A+12 B-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac{(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(4 A+3 B-10 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + 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 + 6*B - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*
A + 12*B - 215*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*
d*(a + a*Sec[c + d*x])^4) + ((4*A + 3*B - 10*C)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, 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+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B+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 B-3 C)+7 a C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B-10 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 (2 a^2 (4 A+3 B-10 C)+35 a^2 C \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B-10 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+6 B-55 C)-105 a^3 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(16 A+12 B-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+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(16 A+12 B-215 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.81439, size = 335, normalized size = 1.94 \[ -\frac{\left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (6720 C \cos ^8\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (70 (2 A+3 B-49 C) \sin \left (\frac{d x}{2}\right )-70 (2 A-31 C) \sin \left (c+\frac{d x}{2}\right )+168 A \sin \left (c+\frac{3 d x}{2}\right )+56 A \sin \left (2 c+\frac{5 d x}{2}\right )+8 A \sin \left (3 c+\frac{7 d x}{2}\right )+126 B \sin \left (c+\frac{3 d x}{2}\right )+42 B \sin \left (2 c+\frac{5 d x}{2}\right )+6 B \sin \left (3 c+\frac{7 d x}{2}\right )-2625 C \sin \left (c+\frac{3 d x}{2}\right )+735 C \sin \left (2 c+\frac{3 d x}{2}\right )-1015 C \sin \left (2 c+\frac{5 d x}{2}\right )+105 C \sin \left (3 c+\frac{5 d x}{2}\right )-160 C \sin \left (3 c+\frac{7 d x}{2}\right )\right )\right )}{210 a^4 d (\cos (c+d x)+1)^4 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(6720*C*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
- Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Cos[(c + d*x)/2]*Sec[c/2]*(70*(2*A + 3*B - 49*C)*Sin[(d*x)/2] -
70*(2*A - 31*C)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] + 126*B*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] + 42*B*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] + 6*B*Sin[3*c + (7*d*x)/2] - 160*C*Sin[3*c
+ (7*d*x)/2])))/(210*a^4*d*(1 + Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.077, size = 277, normalized size = 1.6 \begin{align*}{\frac{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{C}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,B}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\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{B}{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 ) }+{\frac{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 ) }-{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 0.991262, size = 423, normalized size = 2.45 \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}} - \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(sec(d*x+c)^3*(A+B*sec(d*x+c)+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 - 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 = 0.537018, size = 660, normalized size = 3.82 \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 (2 \,{\left (4 \, A + 3 \, B - 80 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (32 \, A + 24 \, B - 535 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (52 \, A + 39 \, B - 620 \, C\right )} \cos \left (d x + c\right ) + 13 \, A + 36 \, B - 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+B*sec(d*x+c)+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*(2*(4*A + 3*B - 80*C)*cos(d*x + c)^3 + (32*A + 24*B - 535*C)*cos(d*x + c)^2 + (52*A + 39*B - 620
*C)*cos(d*x + c) + 13*A + 36*B - 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*cos(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{B \sec ^{4}{\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+B*sec(d*x+c)+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(B*sec(c + d*x)**4/(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 = 1.23529, size = 335, normalized size = 1.94 \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 \, B 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} - 63 \, B 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} - 105 \, B 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 ) - 105 \, B 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+B*sec(d*x+c)+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*B*a^24*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 - 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2
*d*x + 1/2*c)^3 - 105*B*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) - 105*B*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