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

Optimal. Leaf size=174 \[ \frac{\sec ^{10}(c+d x)}{10 a d}-\frac{\sec ^8(c+d x)}{8 a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^3(c+d x) \sec ^7(c+d x)}{10 a d}+\frac{3 \tan (c+d x) \sec ^7(c+d x)}{80 a d}-\frac{\tan (c+d x) \sec ^5(c+d x)}{160 a d}-\frac{\tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac{3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(256*a*d) - Sec[c + d*x]^8/(8*a*d) + Sec[c + d*x]^10/(10*a*d) - (3*Sec[c + d*x]*Tan
[c + d*x])/(256*a*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(128*a*d) - (Sec[c + d*x]^5*Tan[c + d*x])/(160*a*d) + (3*
Sec[c + d*x]^7*Tan[c + d*x])/(80*a*d) - (Sec[c + d*x]^7*Tan[c + d*x]^3)/(10*a*d)

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Rubi [A]  time = 0.241546, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2835, 2606, 14, 2611, 3768, 3770} \[ \frac{\sec ^{10}(c+d x)}{10 a d}-\frac{\sec ^8(c+d x)}{8 a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^3(c+d x) \sec ^7(c+d x)}{10 a d}+\frac{3 \tan (c+d x) \sec ^7(c+d x)}{80 a d}-\frac{\tan (c+d x) \sec ^5(c+d x)}{160 a d}-\frac{\tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac{3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(256*a*d) - Sec[c + d*x]^8/(8*a*d) + Sec[c + d*x]^10/(10*a*d) - (3*Sec[c + d*x]*Tan
[c + d*x])/(256*a*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(128*a*d) - (Sec[c + d*x]^5*Tan[c + d*x])/(160*a*d) + (3*
Sec[c + d*x]^7*Tan[c + d*x])/(80*a*d) - (Sec[c + d*x]^7*Tan[c + d*x]^3)/(10*a*d)

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 2606

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^8(c+d x) \tan ^3(c+d x) \, dx}{a}-\frac{\int \sec ^7(c+d x) \tan ^4(c+d x) \, dx}{a}\\ &=-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}+\frac{3 \int \sec ^7(c+d x) \tan ^2(c+d x) \, dx}{10 a}+\frac{\operatorname{Subst}\left (\int x^7 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac{3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}-\frac{3 \int \sec ^7(c+d x) \, dx}{80 a}+\frac{\operatorname{Subst}\left (\int \left (-x^7+x^9\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{\sec ^8(c+d x)}{8 a d}+\frac{\sec ^{10}(c+d x)}{10 a d}-\frac{\sec ^5(c+d x) \tan (c+d x)}{160 a d}+\frac{3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}-\frac{\int \sec ^5(c+d x) \, dx}{32 a}\\ &=-\frac{\sec ^8(c+d x)}{8 a d}+\frac{\sec ^{10}(c+d x)}{10 a d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{\sec ^5(c+d x) \tan (c+d x)}{160 a d}+\frac{3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}-\frac{3 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=-\frac{\sec ^8(c+d x)}{8 a d}+\frac{\sec ^{10}(c+d x)}{10 a d}-\frac{3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{\sec ^5(c+d x) \tan (c+d x)}{160 a d}+\frac{3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}-\frac{3 \int \sec (c+d x) \, dx}{256 a}\\ &=-\frac{3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\sec ^8(c+d x)}{8 a d}+\frac{\sec ^{10}(c+d x)}{10 a d}-\frac{3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{\sec ^5(c+d x) \tan (c+d x)}{160 a d}+\frac{3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 2.90546, size = 104, normalized size = 0.6 \[ -\frac{-\frac{30}{\sin (c+d x)-1}+\frac{15}{(\sin (c+d x)-1)^2}+\frac{15}{(\sin (c+d x)+1)^2}+\frac{20}{(\sin (c+d x)+1)^3}-\frac{10}{(\sin (c+d x)-1)^4}+\frac{10}{(\sin (c+d x)+1)^4}-\frac{16}{(\sin (c+d x)+1)^5}+30 \tanh ^{-1}(\sin (c+d x))}{2560 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(30*ArcTanh[Sin[c + d*x]] - 10/(-1 + Sin[c + d*x])^4 + 15/(-1 + Sin[c + d*x])^2 - 30/(-1 + Sin[c + d*x]) - 16
/(1 + Sin[c + d*x])^5 + 10/(1 + Sin[c + d*x])^4 + 20/(1 + Sin[c + d*x])^3 + 15/(1 + Sin[c + d*x])^2)/(2560*a*d
)

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Maple [A]  time = 0.084, size = 162, normalized size = 0.9 \begin{align*}{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}-{\frac{3}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{3}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{512\,da}}+{\frac{1}{160\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{128\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{512\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

1/256/d/a/(sin(d*x+c)-1)^4-3/512/d/a/(sin(d*x+c)-1)^2+3/256/a/d/(sin(d*x+c)-1)+3/512/a/d*ln(sin(d*x+c)-1)+1/16
0/d/a/(1+sin(d*x+c))^5-1/256/d/a/(1+sin(d*x+c))^4-1/128/d/a/(1+sin(d*x+c))^3-3/512/a/d/(1+sin(d*x+c))^2-3/512*
ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.05806, size = 289, normalized size = 1.66 \begin{align*} \frac{\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{4} + 73 \, \sin \left (d x + c\right )^{3} + 143 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) - 32\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, 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}}{2560 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2560*(2*(15*sin(d*x + c)^8 + 15*sin(d*x + c)^7 - 55*sin(d*x + c)^6 - 55*sin(d*x + c)^5 + 73*sin(d*x + c)^4 +
73*sin(d*x + c)^3 + 143*sin(d*x + c)^2 - 17*sin(d*x + c) - 32)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 - 4*a*sin
(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^4 - 4*a*sin(d*x + c)^3 - 4*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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Fricas [A]  time = 2.15705, size = 512, normalized size = 2.94 \begin{align*} \frac{30 \, \cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} - 368 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (15 \, \cos \left (d x + c\right )^{6} + 10 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 288}{2560 \,{\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2560*(30*cos(d*x + c)^8 - 10*cos(d*x + c)^6 - 4*cos(d*x + c)^4 - 368*cos(d*x + c)^2 - 15*(cos(d*x + c)^8*sin
(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(-sin
(d*x + c) + 1) - 2*(15*cos(d*x + c)^6 + 10*cos(d*x + c)^4 + 8*cos(d*x + c)^2 - 16)*sin(d*x + c) + 288)/(a*d*co
s(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**9*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.35749, size = 211, normalized size = 1.21 \begin{align*} -\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{5 \,{\left (25 \, \sin \left (d x + c\right )^{4} - 124 \, \sin \left (d x + c\right )^{3} + 234 \, \sin \left (d x + c\right )^{2} - 196 \, \sin \left (d x + c\right ) + 53\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{137 \, \sin \left (d x + c\right )^{5} + 685 \, \sin \left (d x + c\right )^{4} + 1310 \, \sin \left (d x + c\right )^{3} + 1110 \, \sin \left (d x + c\right )^{2} + 305 \, \sin \left (d x + c\right ) + 21}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/10240*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 5*(25*sin(d*x + c)^4 - 124*sin(d
*x + c)^3 + 234*sin(d*x + c)^2 - 196*sin(d*x + c) + 53)/(a*(sin(d*x + c) - 1)^4) - (137*sin(d*x + c)^5 + 685*s
in(d*x + c)^4 + 1310*sin(d*x + c)^3 + 1110*sin(d*x + c)^2 + 305*sin(d*x + c) + 21)/(a*(sin(d*x + c) + 1)^5))/d