3.906 \(\int \frac{\sec ^8(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ \frac{\sec ^{10}(c+d x)}{10 a d}+\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan (c+d x) \sec ^9(c+d x)}{10 a d}+\frac{\tan (c+d x) \sec ^7(c+d x)}{80 a d}+\frac{7 \tan (c+d x) \sec ^5(c+d x)}{480 a d}+\frac{7 \tan (c+d x) \sec ^3(c+d x)}{384 a d}+\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

(7*ArcTanh[Sin[c + d*x]])/(256*a*d) + Sec[c + d*x]^10/(10*a*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (7*
Sec[c + d*x]^3*Tan[c + d*x])/(384*a*d) + (7*Sec[c + d*x]^5*Tan[c + d*x])/(480*a*d) + (Sec[c + d*x]^7*Tan[c + d
*x])/(80*a*d) - (Sec[c + d*x]^9*Tan[c + d*x])/(10*a*d)

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Rubi [A]  time = 0.165216, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2835, 2606, 30, 2611, 3768, 3770} \[ \frac{\sec ^{10}(c+d x)}{10 a d}+\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan (c+d x) \sec ^9(c+d x)}{10 a d}+\frac{\tan (c+d x) \sec ^7(c+d x)}{80 a d}+\frac{7 \tan (c+d x) \sec ^5(c+d x)}{480 a d}+\frac{7 \tan (c+d x) \sec ^3(c+d x)}{384 a d}+\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^8*Tan[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(7*ArcTanh[Sin[c + d*x]])/(256*a*d) + Sec[c + d*x]^10/(10*a*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (7*
Sec[c + d*x]^3*Tan[c + d*x])/(384*a*d) + (7*Sec[c + d*x]^5*Tan[c + d*x])/(480*a*d) + (Sec[c + d*x]^7*Tan[c + d
*x])/(80*a*d) - (Sec[c + d*x]^9*Tan[c + d*x])/(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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[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 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 ^8(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^{10}(c+d x) \tan (c+d x) \, dx}{a}-\frac{\int \sec ^9(c+d x) \tan ^2(c+d x) \, dx}{a}\\ &=-\frac{\sec ^9(c+d x) \tan (c+d x)}{10 a d}+\frac{\int \sec ^9(c+d x) \, dx}{10 a}+\frac{\operatorname{Subst}\left (\int x^9 \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac{\sec ^{10}(c+d x)}{10 a d}+\frac{\sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^9(c+d x) \tan (c+d x)}{10 a d}+\frac{7 \int \sec ^7(c+d x) \, dx}{80 a}\\ &=\frac{\sec ^{10}(c+d x)}{10 a d}+\frac{7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}+\frac{\sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^9(c+d x) \tan (c+d x)}{10 a d}+\frac{7 \int \sec ^5(c+d x) \, dx}{96 a}\\ &=\frac{\sec ^{10}(c+d x)}{10 a d}+\frac{7 \sec ^3(c+d x) \tan (c+d x)}{384 a d}+\frac{7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}+\frac{\sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^9(c+d x) \tan (c+d x)}{10 a d}+\frac{7 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=\frac{\sec ^{10}(c+d x)}{10 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)}{384 a d}+\frac{7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}+\frac{\sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^9(c+d x) \tan (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{\sec ^{10}(c+d x)}{10 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)}{384 a d}+\frac{7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}+\frac{\sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac{\sec ^9(c+d x) \tan (c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 5.86991, size = 116, normalized size = 0.75 \[ \frac{-\frac{210}{\sin (c+d x)-1}+\frac{135}{(\sin (c+d x)-1)^2}+\frac{75}{(\sin (c+d x)+1)^2}-\frac{80}{(\sin (c+d x)-1)^3}+\frac{100}{(\sin (c+d x)+1)^3}+\frac{30}{(\sin (c+d x)-1)^4}+\frac{90}{(\sin (c+d x)+1)^4}+\frac{48}{(\sin (c+d x)+1)^5}+210 \tanh ^{-1}(\sin (c+d x))}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^8*Tan[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(210*ArcTanh[Sin[c + d*x]] + 30/(-1 + Sin[c + d*x])^4 - 80/(-1 + Sin[c + d*x])^3 + 135/(-1 + Sin[c + d*x])^2 -
210/(-1 + Sin[c + d*x]) + 48/(1 + Sin[c + d*x])^5 + 90/(1 + Sin[c + d*x])^4 + 100/(1 + Sin[c + d*x])^3 + 75/(
1 + Sin[c + d*x])^2)/(7680*a*d)

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Maple [A]  time = 0.071, size = 180, normalized size = 1.2 \begin{align*}{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}-{\frac{1}{96\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{9}{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{3}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5}{384\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{5}{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)/(a+a*sin(d*x+c)),x)

[Out]

1/256/d/a/(sin(d*x+c)-1)^4-1/96/d/a/(sin(d*x+c)-1)^3+9/512/d/a/(sin(d*x+c)-1)^2-7/256/a/d/(sin(d*x+c)-1)-7/512
/a/d*ln(sin(d*x+c)-1)+1/160/d/a/(1+sin(d*x+c))^5+3/256/d/a/(1+sin(d*x+c))^4+5/384/d/a/(1+sin(d*x+c))^3+5/512/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 = 1.07135, size = 289, normalized size = 1.88 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{7} - 385 \, \sin \left (d x + c\right )^{6} - 385 \, \sin \left (d x + c\right )^{5} + 511 \, \sin \left (d x + c\right )^{4} + 511 \, \sin \left (d x + c\right )^{3} - 279 \, \sin \left (d x + c\right )^{2} - 279 \, \sin \left (d x + c\right ) - 384\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)/(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 - 385*sin(d*x + c)^6 - 385*sin(d*x + c)^5 + 511*sin(d*x +
c)^4 + 511*sin(d*x + c)^3 - 279*sin(d*x + c)^2 - 279*sin(d*x + c) - 384)/(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 = 2.18468, size = 520, normalized size = 3.38 \begin{align*} -\frac{210 \, \cos \left (d x + c\right )^{8} - 70 \, \cos \left (d x + c\right )^{6} - 28 \, \cos \left (d x + c\right )^{4} - 16 \, \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} + 70 \, \cos \left (d x + c\right )^{4} + 56 \, \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)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/7680*(210*cos(d*x + c)^8 - 70*cos(d*x + c)^6 - 28*cos(d*x + c)^4 - 16*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 + 70*cos(d*x + c)^4 + 56*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)/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.31615, size = 211, normalized size = 1.37 \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} + 1662 \, \sin \left (d x + c\right )^{2} - 1484 \, \sin \left (d x + c\right ) + 539\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} + 9290 \, \sin \left (d x + c\right )^{3} + 8290 \, \sin \left (d x + c\right )^{2} + 2735 \, \sin \left (d x + c\right ) - 293}{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)/(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*sin
(d*x + c)^3 + 1662*sin(d*x + c)^2 - 1484*sin(d*x + c) + 539)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 +
4795*sin(d*x + c)^4 + 9290*sin(d*x + c)^3 + 8290*sin(d*x + c)^2 + 2735*sin(d*x + c) - 293)/(a*(sin(d*x + c) +
1)^5))/d