3.1.70 \(\int \frac {\sec ^7(c+d x)}{(a+a \sec (c+d x))^4} \, dx\) [70]

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

[Out]

21/2*arctanh(sin(d*x+c))/a^4/d-576/35*tan(d*x+c)/a^4/d+21/2*sec(d*x+c)*tan(d*x+c)/a^4/d-43/35*sec(d*x+c)^3*tan
(d*x+c)/a^4/d/(1+sec(d*x+c))^2-288/35*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*sec(d*x+c)^5*tan(d*x+c)
/d/(a+a*sec(d*x+c))^4-2/5*sec(d*x+c)^4*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]
time = 0.27, antiderivative size = 193, normalized size of antiderivative = 1.00, 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 = {3901, 4104, 3872, 3852, 8, 3853, 3855} \begin {gather*} -\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} \end {gather*}

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 8

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

Rule 3852

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 3853

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 3855

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

Rule 3872

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 3901

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)),
Int[(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] /;
FreeQ[{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 4104

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]

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 \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(403\) vs. \(2(193)=386\).
time = 1.50, size = 403, normalized size = 2.09 \begin {gather*} -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-24402 \sin \left (\frac {d x}{2}\right )+55556 \sin \left (\frac {3 d x}{2}\right )-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 )\right )\right )}{2240 a^4 d (1+\sec (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2240*(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[Cos[(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])))/(a^4*d*(1 + Sec[c + d*x])^4)

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Maple [A]
time = 0.07, size = 148, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\)	\(148\)
default	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\)	\(148\)
risch	\(-\frac {i \left (735 \,{\mathrm e}^{10 i \left (d x +c \right )}+5145 \,{\mathrm e}^{9 i \left (d x +c \right )}+16660 \,{\mathrm e}^{8 i \left (d x +c \right )}+34300 \,{\mathrm e}^{7 i \left (d x +c \right )}+51842 \,{\mathrm e}^{6 i \left (d x +c \right )}+61054 \,{\mathrm e}^{5 i \left (d x +c \right )}+55556 \,{\mathrm e}^{4 i \left (d x +c \right )}+39788 \,{\mathrm e}^{3 i \left (d x +c \right )}+21351 \,{\mathrm e}^{2 i \left (d x +c \right )}+7329 \,{\mathrm e}^{i \left (d x +c \right )}+1152\right )}{35 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{4} d}\)	\(191\)

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,method=_RETURNVERBOSE)

[Out]

1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7-9/5*tan(1/2*d*x+1/2*c)^5-13*tan(1/2*d*x+1/2*c)^3-111*tan(1/2*d*x+1/2*c)+4
/(tan(1/2*d*x+1/2*c)-1)^2+36/(tan(1/2*d*x+1/2*c)-1)-84*ln(tan(1/2*d*x+1/2*c)-1)-4/(tan(1/2*d*x+1/2*c)+1)^2+36/
(tan(1/2*d*x+1/2*c)+1)+84*ln(tan(1/2*d*x+1/2*c)+1))

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Maxima [A]
time = 0.28, size = 231, normalized size = 1.20 \begin {gather*} -\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 {gather*}

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 = 2.99, size = 250, normalized size = 1.30 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.49, size = 155, normalized size = 0.80 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.75, size = 160, normalized size = 0.83 \begin {gather*} \frac {21\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4\,d}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^4\,d}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {111\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a + a/cos(c + d*x))^4),x)

[Out]

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

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