3.900 \(\int \frac{\sec ^2(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=178 \[ \frac{\tan ^{10}(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}-\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}+\frac{7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}-\frac{7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}+\frac{7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(256*a*d) - (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x
])/(128*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]^3)/(96*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x]^5)/(80*a*d) - (Sec[
c + d*x]^3*Tan[c + d*x]^7)/(10*a*d) + Tan[c + d*x]^8/(8*a*d) + Tan[c + d*x]^10/(10*a*d)

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Rubi [A]  time = 0.280451, antiderivative size = 178, 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, 2607, 14, 2611, 3768, 3770} \[ \frac{\tan ^{10}(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}-\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}+\frac{7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}-\frac{7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}+\frac{7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(256*a*d) - (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x
])/(128*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]^3)/(96*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x]^5)/(80*a*d) - (Sec[
c + d*x]^3*Tan[c + d*x]^7)/(10*a*d) + Tan[c + d*x]^8/(8*a*d) + Tan[c + d*x]^10/(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 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 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 ^2(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^4(c+d x) \tan ^7(c+d x) \, dx}{a}-\frac{\int \sec ^3(c+d x) \tan ^8(c+d x) \, dx}{a}\\ &=-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac{7 \int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{10 a}+\frac{\operatorname{Subst}\left (\int x^7 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{7 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{16 a}+\frac{\operatorname{Subst}\left (\int \left (x^7+x^9\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{32 a}\\ &=\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}-\frac{7 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=-\frac{7 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}-\frac{7 \int \sec (c+d x) \, dx}{256 a}\\ &=-\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{7 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 1.66669, size = 124, normalized size = 0.7 \[ -\frac{\frac{210}{1-\sin (c+d x)}-\frac{315}{(1-\sin (c+d x))^2}+\frac{525}{(\sin (c+d x)+1)^2}+\frac{160}{(1-\sin (c+d x))^3}-\frac{580}{(\sin (c+d x)+1)^3}-\frac{30}{(1-\sin (c+d x))^4}+\frac{270}{(\sin (c+d x)+1)^4}-\frac{48}{(\sin (c+d x)+1)^5}+210 \tanh ^{-1}(\sin (c+d x))}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

-(210*ArcTanh[Sin[c + d*x]] - 30/(1 - Sin[c + d*x])^4 + 160/(1 - Sin[c + d*x])^3 - 315/(1 - Sin[c + d*x])^2 +
210/(1 - Sin[c + d*x]) - 48/(1 + Sin[c + d*x])^5 + 270/(1 + Sin[c + d*x])^4 - 580/(1 + Sin[c + d*x])^3 + 525/(
1 + Sin[c + d*x])^2)/(7680*a*d)

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Maple [A]  time = 0.096, size = 180, normalized size = 1. \begin{align*}{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}+{\frac{1}{48\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{21}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{7}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{7\,\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{9}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{29}{384\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{35}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,\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)^7/(a+a*sin(d*x+c)),x)

[Out]

1/256/d/a/(sin(d*x+c)-1)^4+1/48/d/a/(sin(d*x+c)-1)^3+21/512/d/a/(sin(d*x+c)-1)^2+7/256/a/d/(sin(d*x+c)-1)+7/51
2/a/d*ln(sin(d*x+c)-1)+1/160/d/a/(1+sin(d*x+c))^5-9/256/d/a/(1+sin(d*x+c))^4+29/384/d/a/(1+sin(d*x+c))^3-35/51
2/a/d/(1+sin(d*x+c))^2-7/512*ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.04846, size = 289, normalized size = 1.62 \begin{align*} \frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{7} + 895 \, \sin \left (d x + c\right )^{6} - 65 \, \sin \left (d x + c\right )^{5} - 961 \, \sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} + 489 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) - 96\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{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/7680*(2*(105*sin(d*x + c)^8 + 105*sin(d*x + c)^7 + 895*sin(d*x + c)^6 - 65*sin(d*x + c)^5 - 961*sin(d*x + c)
^4 - sin(d*x + c)^3 + 489*sin(d*x + c)^2 + 9*sin(d*x + c) - 96)/(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) - 105*log(sin(d*x + c) + 1)/a + 105*log(sin(d*x + c) - 1)/a)/d

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Fricas [A]  time = 2.33712, size = 529, normalized size = 2.97 \begin{align*} \frac{210 \, \cos \left (d x + c\right )^{8} - 2630 \, \cos \left (d x + c\right )^{6} + 4708 \, \cos \left (d x + c\right )^{4} - 3344 \, \cos \left (d x + c\right )^{2} - 105 \,{\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 ) + 105 \,{\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 (105 \, \cos \left (d x + c\right )^{6} - 250 \, \cos \left (d x + c\right )^{4} + 184 \, \cos \left (d x + c\right )^{2} - 48\right )} \sin \left (d x + c\right ) + 864}{7680 \,{\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)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/7680*(210*cos(d*x + c)^8 - 2630*cos(d*x + c)^6 + 4708*cos(d*x + c)^4 - 3344*cos(d*x + c)^2 - 105*(cos(d*x +
c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)
*log(-sin(d*x + c) + 1) - 2*(105*cos(d*x + c)^6 - 250*cos(d*x + c)^4 + 184*cos(d*x + c)^2 - 48)*sin(d*x + c) +
864)/(a*d*cos(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)**7/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.36264, size = 211, normalized size = 1.19 \begin{align*} -\frac{\frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{5 \,{\left (175 \, \sin \left (d x + c\right )^{4} - 868 \, \sin \left (d x + c\right )^{3} + 1302 \, \sin \left (d x + c\right )^{2} - 828 \, \sin \left (d x + c\right ) + 195\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{959 \, \sin \left (d x + c\right )^{5} + 4795 \, \sin \left (d x + c\right )^{4} + 7490 \, \sin \left (d x + c\right )^{3} + 5610 \, \sin \left (d x + c\right )^{2} + 2055 \, \sin \left (d x + c\right ) + 291}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/30720*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1))/a + 5*(175*sin(d*x + c)^4 - 868*si
n(d*x + c)^3 + 1302*sin(d*x + c)^2 - 828*sin(d*x + c) + 195)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 +
4795*sin(d*x + c)^4 + 7490*sin(d*x + c)^3 + 5610*sin(d*x + c)^2 + 2055*sin(d*x + c) + 291)/(a*(sin(d*x + c) +
1)^5))/d