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

Optimal. Leaf size=106 \[ \frac{\tan ^6(c+d x)}{6 a d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac{\tan ^5(c+d x) \sec (c+d x)}{6 a d}+\frac{5 \tan ^3(c+d x) \sec (c+d x)}{24 a d}-\frac{5 \tan (c+d x) \sec (c+d x)}{16 a d} \]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(16*a*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(16*a*d) + (5*Sec[c + d*x]*Tan[c + d*x]^3)/
(24*a*d) - (Sec[c + d*x]*Tan[c + d*x]^5)/(6*a*d) + Tan[c + d*x]^6/(6*a*d)

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Rubi [A]  time = 0.13628, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{\tan ^6(c+d x)}{6 a d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac{\tan ^5(c+d x) \sec (c+d x)}{6 a d}+\frac{5 \tan ^3(c+d x) \sec (c+d x)}{24 a d}-\frac{5 \tan (c+d x) \sec (c+d x)}{16 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(16*a*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(16*a*d) + (5*Sec[c + d*x]*Tan[c + d*x]^3)/
(24*a*d) - (Sec[c + d*x]*Tan[c + d*x]^5)/(6*a*d) + Tan[c + d*x]^6/(6*a*d)

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -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 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 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{\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^2(c+d x) \tan ^5(c+d x) \, dx}{a}-\frac{\int \sec (c+d x) \tan ^6(c+d x) \, dx}{a}\\ &=-\frac{\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac{5 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{6 a}+\frac{\operatorname{Subst}\left (\int x^5 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac{\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac{\tan ^6(c+d x)}{6 a d}-\frac{5 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{8 a}\\ &=-\frac{5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac{5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac{\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac{\tan ^6(c+d x)}{6 a d}+\frac{5 \int \sec (c+d x) \, dx}{16 a}\\ &=\frac{5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac{5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac{5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac{\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac{\tan ^6(c+d x)}{6 a d}\\ \end{align*}

Mathematica [A]  time = 0.308239, size = 84, normalized size = 0.79 \[ \frac{-\frac{18}{1-\sin (c+d x)}+\frac{48}{\sin (c+d x)+1}+\frac{3}{(1-\sin (c+d x))^2}-\frac{21}{(\sin (c+d x)+1)^2}+\frac{4}{(\sin (c+d x)+1)^3}+30 \tanh ^{-1}(\sin (c+d x))}{96 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*ArcTanh[Sin[c + d*x]] + 3/(1 - Sin[c + d*x])^2 - 18/(1 - Sin[c + d*x]) + 4/(1 + Sin[c + d*x])^3 - 21/(1 +
Sin[c + d*x])^2 + 48/(1 + Sin[c + d*x]))/(96*a*d)

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Maple [A]  time = 0.056, size = 126, normalized size = 1.2 \begin{align*}{\frac{1}{32\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{3}{16\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{32\,da}}+{\frac{1}{24\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{7}{32\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{32\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d/a/(sin(d*x+c)-1)^2+3/16/a/d/(sin(d*x+c)-1)-5/32/a/d*ln(sin(d*x+c)-1)+1/24/d/a/(1+sin(d*x+c))^3-7/32/a/d
/(1+sin(d*x+c))^2+1/2/a/d/(1+sin(d*x+c))+5/32*ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 0.999884, size = 176, normalized size = 1.66 \begin{align*} \frac{\frac{2 \,{\left (33 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 31 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, 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}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(2*(33*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 31*sin(d*x + c)^2 - 7*sin(d*x + c) + 8)/(a*sin(d*x + c)^5 + a*
sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*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 = 1.8359, size = 398, normalized size = 3.75 \begin{align*} \frac{66 \, \cos \left (d x + c\right )^{4} - 70 \, \cos \left (d x + c\right )^{2} + 15 \,{\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 20}{96 \,{\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(66*cos(d*x + c)^4 - 70*cos(d*x + c)^2 + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x +
c) + 1) - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(9*cos(d*x + c)^2 - 2)*
sin(d*x + c) + 20)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{5}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(c + d*x)**5/(sin(c + d*x) + 1), x)/a

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Giac [A]  time = 3.93359, size = 157, normalized size = 1.48 \begin{align*} \frac{\frac{30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{3 \,{\left (15 \, \sin \left (d x + c\right )^{2} - 18 \, \sin \left (d x + c\right ) + 5\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac{55 \, \sin \left (d x + c\right )^{3} + 69 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 7}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(30*log(abs(sin(d*x + c) + 1))/a - 30*log(abs(sin(d*x + c) - 1))/a + 3*(15*sin(d*x + c)^2 - 18*sin(d*x +
c) + 5)/(a*(sin(d*x + c) - 1)^2) - (55*sin(d*x + c)^3 + 69*sin(d*x + c)^2 + 15*sin(d*x + c) - 7)/(a*(sin(d*x
+ c) + 1)^3))/d