3.883 \(\int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=134 \[ -\frac {\tan ^8(c+d x)}{8 a d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 a d}-\frac {5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a d}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{64 a d}-\frac {5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

[Out]

-5/128*arctanh(sin(d*x+c))/a/d-5/128*sec(d*x+c)*tan(d*x+c)/a/d+5/64*sec(d*x+c)^3*tan(d*x+c)/a/d-5/48*sec(d*x+c
)^3*tan(d*x+c)^3/a/d+1/8*sec(d*x+c)^3*tan(d*x+c)^5/a/d-1/8*tan(d*x+c)^8/a/d

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Rubi [A]  time = 0.24, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2835, 2611, 3768, 3770, 2607, 30} \[ -\frac {\tan ^8(c+d x)}{8 a d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 a d}-\frac {5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a d}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{64 a d}-\frac {5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[Sin[c + d*x]])/(128*a*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(128*a*d) + (5*Sec[c + d*x]^3*Tan[c + d*x
])/(64*a*d) - (5*Sec[c + d*x]^3*Tan[c + d*x]^3)/(48*a*d) + (Sec[c + d*x]^3*Tan[c + d*x]^5)/(8*a*d) - Tan[c + d
*x]^8/(8*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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 (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{a}-\frac {\int \sec ^2(c+d x) \tan ^7(c+d x) \, dx}{a}\\ &=\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {5 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{8 a}-\frac {\operatorname {Subst}\left (\int x^7 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {\tan ^8(c+d x)}{8 a d}+\frac {5 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{16 a}\\ &=\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {5 \int \sec ^3(c+d x) \, dx}{64 a}\\ &=-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {5 \int \sec (c+d x) \, dx}{128 a}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {\tan ^8(c+d x)}{8 a d}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 101, normalized size = 0.75 \[ -\frac {\frac {-15 \sin ^6(c+d x)+177 \sin ^5(c+d x)+104 \sin ^4(c+d x)-184 \sin ^3(c+d x)-129 \sin ^2(c+d x)+63 \sin (c+d x)+48}{(\sin (c+d x)-1)^3 (\sin (c+d x)+1)^4}+15 \tanh ^{-1}(\sin (c+d x))}{384 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(15*ArcTanh[Sin[c + d*x]] + (48 + 63*Sin[c + d*x] - 129*Sin[c + d*x]^2 - 184*Sin[c + d*x]^3 + 104*Sin[c
+ d*x]^4 + 177*Sin[c + d*x]^5 - 15*Sin[c + d*x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(a*d)

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fricas [A]  time = 0.47, size = 167, normalized size = 1.25 \[ \frac {30 \, \cos \left (d x + c\right )^{6} + 118 \, \cos \left (d x + c\right )^{4} - 68 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (177 \, \cos \left (d x + c\right )^{4} - 170 \, \cos \left (d x + c\right )^{2} + 56\right )} \sin \left (d x + c\right ) + 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/768*(30*cos(d*x + c)^6 + 118*cos(d*x + c)^4 - 68*cos(d*x + c)^2 - 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x
+ c)^6)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) + 2*(
177*cos(d*x + c)^4 - 170*cos(d*x + c)^2 + 56)*sin(d*x + c) + 16)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*
x + c)^6)

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giac [A]  time = 0.31, size = 136, normalized size = 1.01 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (55 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 111 \, \sin \left (d x + c\right ) + 57\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 980 \, \sin \left (d x + c\right )^{3} + 1662 \, \sin \left (d x + c\right )^{2} + 1140 \, \sin \left (d x + c\right ) + 285}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3072*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 2*(55*sin(d*x + c)^3 + 15*sin(d*x
+ c)^2 - 111*sin(d*x + c) + 57)/(a*(sin(d*x + c) - 1)^3) - (125*sin(d*x + c)^4 + 980*sin(d*x + c)^3 + 1662*si
n(d*x + c)^2 + 1140*sin(d*x + c) + 285)/(a*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.40, size = 162, normalized size = 1.21 \[ -\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {7}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a d}-\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{12 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {11}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{32 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

-1/96/a/d/(sin(d*x+c)-1)^3-7/128/a/d/(sin(d*x+c)-1)^2-15/128/a/d/(sin(d*x+c)-1)+5/256/a/d*ln(sin(d*x+c)-1)-1/6
4/a/d/(1+sin(d*x+c))^4+1/12/a/d/(1+sin(d*x+c))^3-11/64/a/d/(1+sin(d*x+c))^2+5/32/a/d/(1+sin(d*x+c))-5/256*ln(1
+sin(d*x+c))/a/d

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maxima [A]  time = 0.32, size = 175, normalized size = 1.31 \[ \frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} - 177 \, \sin \left (d x + c\right )^{5} - 104 \, \sin \left (d x + c\right )^{4} + 184 \, \sin \left (d x + c\right )^{3} + 129 \, \sin \left (d x + c\right )^{2} - 63 \, \sin \left (d x + c\right ) - 48\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/768*(2*(15*sin(d*x + c)^6 - 177*sin(d*x + c)^5 - 104*sin(d*x + c)^4 + 184*sin(d*x + c)^3 + 129*sin(d*x + c)^
2 - 63*sin(d*x + c) - 48)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a
*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c)
- 1)/a)/d

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mupad [B]  time = 17.12, size = 388, normalized size = 2.90 \[ \frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}+\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{16}+\frac {289\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}+\frac {33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{16}+\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

((5*tan(c/2 + (d*x)/2))/64 + (5*tan(c/2 + (d*x)/2)^2)/32 - (35*tan(c/2 + (d*x)/2)^3)/96 - (85*tan(c/2 + (d*x)/
2)^4)/96 + (113*tan(c/2 + (d*x)/2)^5)/192 + (33*tan(c/2 + (d*x)/2)^6)/16 + (289*tan(c/2 + (d*x)/2)^7)/16 + (33
*tan(c/2 + (d*x)/2)^8)/16 + (113*tan(c/2 + (d*x)/2)^9)/192 - (85*tan(c/2 + (d*x)/2)^10)/96 - (35*tan(c/2 + (d*
x)/2)^11)/96 + (5*tan(c/2 + (d*x)/2)^12)/32 + (5*tan(c/2 + (d*x)/2)^13)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 5
*a*tan(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 9*a*tan(c/2 + (d*x)/2)^4 + 30*a*tan(c/2 + (d*x)/2)^5 - 5
*a*tan(c/2 + (d*x)/2)^6 - 40*a*tan(c/2 + (d*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (d*x)/2)^9 + 9
*a*tan(c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + (d*x)/2)^12 + 2*a*tan(c/2 + (d*x)/2)^13
+ a*tan(c/2 + (d*x)/2)^14)) - (5*atanh(tan(c/2 + (d*x)/2)))/(64*a*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*sin(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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