3.884 \(\int \frac{\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=152 \[ \frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^6(c+d x)}{6 a d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{128 a d}-\frac{\tan ^5(c+d x) \sec ^3(c+d x)}{8 a d}+\frac{5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a d}-\frac{5 \tan (c+d x) \sec ^3(c+d x)}{64 a d}+\frac{5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

[Out]

(5*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]^6/(6*a*d) + Tan[c + d*x]^8/(8*a*d)

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Rubi [A]  time = 0.244631, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{\tan ^8(c+d x)}{8 a d}+\frac{\tan ^6(c+d x)}{6 a d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{128 a d}-\frac{\tan ^5(c+d x) \sec ^3(c+d x)}{8 a d}+\frac{5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a d}-\frac{5 \tan (c+d x) \sec ^3(c+d x)}{64 a d}+\frac{5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(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]^6/(6*a*d) + Tan[c + d*x]^8/(8*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 ^5(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^4(c+d x) \tan ^5(c+d x) \, dx}{a}-\frac{\int \sec ^3(c+d x) \tan ^6(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^5 \left (1+x^2\right ) \, 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{5 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{16 a}+\frac{\operatorname{Subst}\left (\int \left (x^5+x^7\right ) \, dx,x,\tan (c+d x)\right )}{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 ^6(c+d x)}{6 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 ^6(c+d x)}{6 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 ^6(c+d x)}{6 a d}+\frac{\tan ^8(c+d x)}{8 a d}\\ \end{align*}

Mathematica [A]  time = 1.1749, size = 92, normalized size = 0.61 \[ \frac{-\frac{15}{\sin (c+d x)-1}-\frac{15}{(\sin (c+d x)-1)^2}+\frac{30}{(\sin (c+d x)+1)^2}-\frac{4}{(\sin (c+d x)-1)^3}-\frac{24}{(\sin (c+d x)+1)^3}+\frac{6}{(\sin (c+d x)+1)^4}+15 \tanh ^{-1}(\sin (c+d x))}{384 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(15*ArcTanh[Sin[c + d*x]] - 4/(-1 + Sin[c + d*x])^3 - 15/(-1 + Sin[c + d*x])^2 - 15/(-1 + Sin[c + d*x]) + 6/(1
+ Sin[c + d*x])^4 - 24/(1 + Sin[c + d*x])^3 + 30/(1 + Sin[c + d*x])^2)/(384*a*d)

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Maple [A]  time = 0.089, size = 144, normalized size = 1. \begin{align*} -{\frac{1}{96\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}-{\frac{5}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{5}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,da}}+{\frac{1}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{16\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{5}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x)

[Out]

-1/96/d/a/(sin(d*x+c)-1)^3-5/128/d/a/(sin(d*x+c)-1)^2-5/128/a/d/(sin(d*x+c)-1)-5/256/a/d*ln(sin(d*x+c)-1)+1/64
/d/a/(1+sin(d*x+c))^4-1/16/d/a/(1+sin(d*x+c))^3+5/64/a/d/(1+sin(d*x+c))^2+5/256*ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.02561, size = 234, normalized size = 1.54 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} + 88 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} - 63 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) + 16\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/768*(2*(15*sin(d*x + c)^6 + 15*sin(d*x + c)^5 + 88*sin(d*x + c)^4 - 8*sin(d*x + c)^3 - 63*sin(d*x + c)^2 +
sin(d*x + c) + 16)/(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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Fricas [A]  time = 1.63004, size = 460, normalized size = 3.03 \begin{align*} -\frac{30 \, \cos \left (d x + c\right )^{6} - 266 \, \cos \left (d x + c\right )^{4} + 316 \, \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 (15 \, \cos \left (d x + c\right )^{4} - 22 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 112}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/768*(30*cos(d*x + c)^6 - 266*cos(d*x + c)^4 + 316*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
*(15*cos(d*x + c)^4 - 22*cos(d*x + c)^2 + 8)*sin(d*x + c) - 112)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*
x + c)^6)

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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)**7*sin(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.36729, size = 184, 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{2 \,{\left (55 \, \sin \left (d x + c\right )^{3} - 225 \, \sin \left (d x + c\right )^{2} + 225 \, \sin \left (d x + c\right ) - 71\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac{125 \, \sin \left (d x + c\right )^{4} + 500 \, \sin \left (d x + c\right )^{3} + 510 \, \sin \left (d x + c\right )^{2} + 212 \, \sin \left (d x + c\right ) + 29}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(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 - 225*sin(d*x
+ c)^2 + 225*sin(d*x + c) - 71)/(a*(sin(d*x + c) - 1)^3) - (125*sin(d*x + c)^4 + 500*sin(d*x + c)^3 + 510*sin
(d*x + c)^2 + 212*sin(d*x + c) + 29)/(a*(sin(d*x + c) + 1)^4))/d