3.70 \(\int \frac{\sec ^7(c+d x)}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=193 \[ -\frac{576 \tan (c+d x)}{35 a^4 d}+\frac{21 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{43 \tan (c+d x) \sec ^3(c+d x)}{35 a^4 d (\sec (c+d x)+1)^2}-\frac{288 \tan (c+d x) \sec ^2(c+d x)}{35 a^4 d (\sec (c+d x)+1)}+\frac{21 \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{\tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

(21*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (576*Tan[c + d*x])/(35*a^4*d) + (21*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d
) - (43*Sec[c + d*x]^3*Tan[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])^2) - (288*Sec[c + d*x]^2*Tan[c + d*x])/(35*a
^4*d*(1 + Sec[c + d*x])) - (Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Sec[c + d*x]^4*Tan[
c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.393318, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3816, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac{576 \tan (c+d x)}{35 a^4 d}+\frac{21 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{43 \tan (c+d x) \sec ^3(c+d x)}{35 a^4 d (\sec (c+d x)+1)^2}-\frac{288 \tan (c+d x) \sec ^2(c+d x)}{35 a^4 d (\sec (c+d x)+1)}+\frac{21 \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{\tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + a*Sec[c + d*x])^4,x]

[Out]

(21*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (576*Tan[c + d*x])/(35*a^4*d) + (21*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d
) - (43*Sec[c + d*x]^3*Tan[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])^2) - (288*Sec[c + d*x]^2*Tan[c + d*x])/(35*a
^4*d*(1 + Sec[c + d*x])) - (Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Sec[c + d*x]^4*Tan[
c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

Rule 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec ^7(c+d x)}{(a+a \sec (c+d x))^4} \, dx &=-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{\sec ^5(c+d x) (5 a-9 a \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^4(c+d x) \left (56 a^2-73 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{43 \sec ^3(c+d x) \tan (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^3(c+d x) \left (387 a^3-477 a^3 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{43 \sec ^3(c+d x) \tan (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \sec ^2(c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{\int \sec ^2(c+d x) \left (1728 a^4-2205 a^4 \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac{43 \sec ^3(c+d x) \tan (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \sec ^2(c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{576 \int \sec ^2(c+d x) \, dx}{35 a^4}+\frac{21 \int \sec ^3(c+d x) \, dx}{a^4}\\ &=\frac{21 \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{43 \sec ^3(c+d x) \tan (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \sec ^2(c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{21 \int \sec (c+d x) \, dx}{2 a^4}+\frac{576 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{35 a^4 d}\\ &=\frac{21 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{576 \tan (c+d x)}{35 a^4 d}+\frac{21 \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{43 \sec ^3(c+d x) \tan (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \sec ^2(c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.5704, size = 403, normalized size = 2.09 \[ -\frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (\sec \left (\frac{c}{2}\right ) \sec (c) \left (-61054 \sin \left (c-\frac{d x}{2}\right )+33614 \sin \left (c+\frac{d x}{2}\right )-51842 \sin \left (2 c+\frac{d x}{2}\right )-12460 \sin \left (c+\frac{3 d x}{2}\right )+33716 \sin \left (2 c+\frac{3 d x}{2}\right )-34300 \sin \left (3 c+\frac{3 d x}{2}\right )+39788 \sin \left (c+\frac{5 d x}{2}\right )-2940 \sin \left (2 c+\frac{5 d x}{2}\right )+26068 \sin \left (3 c+\frac{5 d x}{2}\right )-16660 \sin \left (4 c+\frac{5 d x}{2}\right )+21351 \sin \left (2 c+\frac{7 d x}{2}\right )+1295 \sin \left (3 c+\frac{7 d x}{2}\right )+14911 \sin \left (4 c+\frac{7 d x}{2}\right )-5145 \sin \left (5 c+\frac{7 d x}{2}\right )+7329 \sin \left (3 c+\frac{9 d x}{2}\right )+1225 \sin \left (4 c+\frac{9 d x}{2}\right )+5369 \sin \left (5 c+\frac{9 d x}{2}\right )-735 \sin \left (6 c+\frac{9 d x}{2}\right )+1152 \sin \left (4 c+\frac{11 d x}{2}\right )+280 \sin \left (5 c+\frac{11 d x}{2}\right )+872 \sin \left (6 c+\frac{11 d x}{2}\right )-24402 \sin \left (\frac{d x}{2}\right )+55556 \sin \left (\frac{3 d x}{2}\right )\right ) \sec ^2(c+d x)+376320 \cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{2240 a^4 d (\sec (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + a*Sec[c + d*x])^4,x]

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^4*(376320*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[C
os[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-24402*Sin[(d*x)/2] + 55556*Sin[(3*d*x)
/2] - 61054*Sin[c - (d*x)/2] + 33614*Sin[c + (d*x)/2] - 51842*Sin[2*c + (d*x)/2] - 12460*Sin[c + (3*d*x)/2] +
33716*Sin[2*c + (3*d*x)/2] - 34300*Sin[3*c + (3*d*x)/2] + 39788*Sin[c + (5*d*x)/2] - 2940*Sin[2*c + (5*d*x)/2]
+ 26068*Sin[3*c + (5*d*x)/2] - 16660*Sin[4*c + (5*d*x)/2] + 21351*Sin[2*c + (7*d*x)/2] + 1295*Sin[3*c + (7*d*
x)/2] + 14911*Sin[4*c + (7*d*x)/2] - 5145*Sin[5*c + (7*d*x)/2] + 7329*Sin[3*c + (9*d*x)/2] + 1225*Sin[4*c + (9
*d*x)/2] + 5369*Sin[5*c + (9*d*x)/2] - 735*Sin[6*c + (9*d*x)/2] + 1152*Sin[4*c + (11*d*x)/2] + 280*Sin[5*c + (
11*d*x)/2] + 872*Sin[6*c + (11*d*x)/2])))/(2240*a^4*d*(1 + Sec[c + d*x])^4)

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Maple [A]  time = 0.047, size = 200, normalized size = 1. \begin{align*} -{\frac{1}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{9}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{13}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{111}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{9}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{21}{2\,d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{9}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{21}{2\,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)^7/(a+a*sec(d*x+c))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7-9/40/d/a^4*tan(1/2*d*x+1/2*c)^5-13/8/d/a^4*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*t
an(1/2*d*x+1/2*c)-1/2/d/a^4/(tan(1/2*d*x+1/2*c)+1)^2+9/2/d/a^4/(tan(1/2*d*x+1/2*c)+1)+21/2/d/a^4*ln(tan(1/2*d*
x+1/2*c)+1)+1/2/d/a^4/(tan(1/2*d*x+1/2*c)-1)^2+9/2/d/a^4/(tan(1/2*d*x+1/2*c)-1)-21/2/d/a^4*ln(tan(1/2*d*x+1/2*
c)-1)

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Maxima [A]  time = 1.12619, size = 312, normalized size = 1.62 \begin{align*} -\frac{\frac{280 \,{\left (\frac{7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac{2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{63 \, \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}}}{a^{4}} - \frac{2940 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{2940 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/280*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 1)
+ 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*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 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^4)/d

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Fricas [A]  time = 2.08836, size = 670, normalized size = 3.47 \begin{align*} \frac{735 \,{\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 735 \,{\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (1152 \, \cos \left (d x + c\right )^{5} + 3873 \, \cos \left (d x + c\right )^{4} + 4548 \, \cos \left (d x + c\right )^{3} + 2012 \, \cos \left (d x + c\right )^{2} + 140 \, \cos \left (d x + c\right ) - 35\right )} \sin \left (d x + c\right )}{140 \,{\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/140*(735*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(
d*x + c) + 1) - 735*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)
*log(-sin(d*x + c) + 1) - 2*(1152*cos(d*x + c)^5 + 3873*cos(d*x + c)^4 + 4548*cos(d*x + c)^3 + 2012*cos(d*x +
c)^2 + 140*cos(d*x + c) - 35)*sin(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x +
c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{7}{\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)**7/(a+a*sec(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**7/(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 = 1.4268, size = 209, normalized size = 1.08 \begin{align*} \frac{\frac{2940 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{2940 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{280 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac{5 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 63 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 455 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3885 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/280*(2940*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 2940*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 280*(9*tan(
1/2*d*x + 1/2*c)^3 - 7*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - (5*a^24*tan(1/2*d*x + 1/2*
c)^7 + 63*a^24*tan(1/2*d*x + 1/2*c)^5 + 455*a^24*tan(1/2*d*x + 1/2*c)^3 + 3885*a^24*tan(1/2*d*x + 1/2*c))/a^28
)/d