3.61 \(\int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=162 \[ -\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {76 \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {13 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {11 \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

13/2*arctanh(sin(d*x+c))/a^3/d-152/15*tan(d*x+c)/a^3/d+13/2*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*sec(d*x+c)^4*tan(d
*x+c)/d/(a+a*sec(d*x+c))^3-11/15*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-76/15*sec(d*x+c)^2*tan(d*x+c)/
d/(a^3+a^3*sec(d*x+c))

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Rubi [A]  time = 0.29, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {76 \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {13 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {11 \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(13*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (152*Tan[c + d*x])/(15*a^3*d) + (13*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d
) - (Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (11*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a +
a*Sec[c + d*x])^2) - (76*Sec[c + d*x]^2*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 8

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

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 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]

Rubi steps

\begin {align*} \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^4(c+d x) (4 a-7 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^3(c+d x) \left (33 a^2-43 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec ^2(c+d x) \left (152 a^3-195 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {152 \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {13 \int \sec ^3(c+d x) \, dx}{a^3}\\ &=\frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {13 \int \sec (c+d x) \, dx}{2 a^3}+\frac {152 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.05, size = 351, normalized size = 2.17 \[ -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \left (-4329 \sin \left (c-\frac {d x}{2}\right )+1989 \sin \left (c+\frac {d x}{2}\right )-3575 \sin \left (2 c+\frac {d x}{2}\right )-475 \sin \left (c+\frac {3 d x}{2}\right )+2005 \sin \left (2 c+\frac {3 d x}{2}\right )-2275 \sin \left (3 c+\frac {3 d x}{2}\right )+2673 \sin \left (c+\frac {5 d x}{2}\right )+105 \sin \left (2 c+\frac {5 d x}{2}\right )+1593 \sin \left (3 c+\frac {5 d x}{2}\right )-975 \sin \left (4 c+\frac {5 d x}{2}\right )+1325 \sin \left (2 c+\frac {7 d x}{2}\right )+255 \sin \left (3 c+\frac {7 d x}{2}\right )+875 \sin \left (4 c+\frac {7 d x}{2}\right )-195 \sin \left (5 c+\frac {7 d x}{2}\right )+304 \sin \left (3 c+\frac {9 d x}{2}\right )+90 \sin \left (4 c+\frac {9 d x}{2}\right )+214 \sin \left (5 c+\frac {9 d x}{2}\right )-1235 \sin \left (\frac {d x}{2}\right )+3805 \sin \left (\frac {3 d x}{2}\right )\right ) \sec ^2(c+d x)+24960 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{480 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/480*(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(24960*Cos[(c + d*x)/2]^5*(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*(-1235*Sin[(d*x)/2] + 3805*Sin[(3*d
*x)/2] - 4329*Sin[c - (d*x)/2] + 1989*Sin[c + (d*x)/2] - 3575*Sin[2*c + (d*x)/2] - 475*Sin[c + (3*d*x)/2] + 20
05*Sin[2*c + (3*d*x)/2] - 2275*Sin[3*c + (3*d*x)/2] + 2673*Sin[c + (5*d*x)/2] + 105*Sin[2*c + (5*d*x)/2] + 159
3*Sin[3*c + (5*d*x)/2] - 975*Sin[4*c + (5*d*x)/2] + 1325*Sin[2*c + (7*d*x)/2] + 255*Sin[3*c + (7*d*x)/2] + 875
*Sin[4*c + (7*d*x)/2] - 195*Sin[5*c + (7*d*x)/2] + 304*Sin[3*c + (9*d*x)/2] + 90*Sin[4*c + (9*d*x)/2] + 214*Si
n[5*c + (9*d*x)/2])))/(a^3*d*(1 + Sec[c + d*x])^3)

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fricas [A]  time = 0.81, size = 206, normalized size = 1.27 \[ \frac {195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (304 \, \cos \left (d x + c\right )^{4} + 717 \, \cos \left (d x + c\right )^{3} + 479 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 15\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(195*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 195*
(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(304*cos(d*
x + c)^4 + 717*cos(d*x + c)^3 + 479*cos(d*x + c)^2 + 45*cos(d*x + c) - 15)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5
+ 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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giac [A]  time = 3.53, size = 139, normalized size = 0.86 \[ \frac {\frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 390*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*tan(1/2*
d*x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5
+ 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 0.42, size = 181, normalized size = 1.12 \[ -\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d \,a^{3}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {1}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{3}}-\frac {1}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5-2/3/d/a^3*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*tan(1/2*d*x+1/2*c)+1/2/d/a^3/(tan(1
/2*d*x+1/2*c)-1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)-13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^3/(tan(1/2*d*x
+1/2*c)+1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)+13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [A]  time = 0.62, size = 211, normalized size = 1.30 \[ -\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 4
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos
(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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mupad [B]  time = 0.68, size = 141, normalized size = 0.87 \[ \frac {13\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^3\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + a/cos(c + d*x))^3),x)

[Out]

(13*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^5/(20*a^3*d) - (2*tan(c/2 + (d*x)/2)^3)/(3*a^3*d)
- (5*tan(c/2 + (d*x)/2) - 7*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 +
a^3)) - (31*tan(c/2 + (d*x)/2))/(4*a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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