3.72 \(\int \frac{\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=171 \[ \frac{a^2}{40 d (a \sin (c+d x)+a)^5}-\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{5}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac{7 a}{64 d (a \sin (c+d x)+a)^4}+\frac{1}{6 d (a \sin (c+d x)+a)^3}+\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{5}{64 a d (a \sin (c+d x)+a)^2} \]

[Out]

ArcTanh[Sin[c + d*x]]/(128*a^3*d) + 1/(128*a*d*(a - a*Sin[c + d*x])^2) + a^2/(40*d*(a + a*Sin[c + d*x])^5) - (
7*a)/(64*d*(a + a*Sin[c + d*x])^4) + 1/(6*d*(a + a*Sin[c + d*x])^3) - 5/(64*a*d*(a + a*Sin[c + d*x])^2) - 1/(3
2*d*(a^3 - a^3*Sin[c + d*x])) - 5/(128*d*(a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.123351, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2707, 88, 206} \[ \frac{a^2}{40 d (a \sin (c+d x)+a)^5}-\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{5}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac{7 a}{64 d (a \sin (c+d x)+a)^4}+\frac{1}{6 d (a \sin (c+d x)+a)^3}+\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{5}{64 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

ArcTanh[Sin[c + d*x]]/(128*a^3*d) + 1/(128*a*d*(a - a*Sin[c + d*x])^2) + a^2/(40*d*(a + a*Sin[c + d*x])^5) - (
7*a)/(64*d*(a + a*Sin[c + d*x])^4) + 1/(6*d*(a + a*Sin[c + d*x])^3) - 5/(64*a*d*(a + a*Sin[c + d*x])^2) - 1/(3
2*d*(a^3 - a^3*Sin[c + d*x])) - 5/(128*d*(a^3 + a^3*Sin[c + d*x]))

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{(a-x)^3 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{64 a (a-x)^3}-\frac{1}{32 a^2 (a-x)^2}-\frac{a^2}{8 (a+x)^6}+\frac{7 a}{16 (a+x)^5}-\frac{1}{2 (a+x)^4}+\frac{5}{32 a (a+x)^3}+\frac{5}{128 a^2 (a+x)^2}+\frac{1}{128 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{128 a d (a-a \sin (c+d x))^2}+\frac{a^2}{40 d (a+a \sin (c+d x))^5}-\frac{7 a}{64 d (a+a \sin (c+d x))^4}+\frac{1}{6 d (a+a \sin (c+d x))^3}-\frac{5}{64 a d (a+a \sin (c+d x))^2}-\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{5}{128 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a^2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{128 a^3 d}+\frac{1}{128 a d (a-a \sin (c+d x))^2}+\frac{a^2}{40 d (a+a \sin (c+d x))^5}-\frac{7 a}{64 d (a+a \sin (c+d x))^4}+\frac{1}{6 d (a+a \sin (c+d x))^3}-\frac{5}{64 a d (a+a \sin (c+d x))^2}-\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{5}{128 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.925661, size = 102, normalized size = 0.6 \[ \frac{15 \tanh ^{-1}(\sin (c+d x))-\frac{15 \sin ^6(c+d x)+45 \sin ^5(c+d x)-620 \sin ^4(c+d x)-540 \sin ^3(c+d x)+157 \sin ^2(c+d x)+351 \sin (c+d x)+112}{(\sin (c+d x)-1)^2 (\sin (c+d x)+1)^5}}{1920 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

(15*ArcTanh[Sin[c + d*x]] - (112 + 351*Sin[c + d*x] + 157*Sin[c + d*x]^2 - 540*Sin[c + d*x]^3 - 620*Sin[c + d*
x]^4 + 45*Sin[c + d*x]^5 + 15*Sin[c + d*x]^6)/((-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^5))/(1920*a^3*d)

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Maple [A]  time = 0.106, size = 162, normalized size = 1. \begin{align*}{\frac{1}{128\,d{a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{1}{32\,d{a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,d{a}^{3}}}+{\frac{1}{40\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{7}{64\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{6\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5}{64\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5}{128\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,d{a}^{3}}} \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))^3,x)

[Out]

1/128/d/a^3/(sin(d*x+c)-1)^2+1/32/d/a^3/(sin(d*x+c)-1)-1/256/d/a^3*ln(sin(d*x+c)-1)+1/40/d/a^3/(1+sin(d*x+c))^
5-7/64/d/a^3/(1+sin(d*x+c))^4+1/6/d/a^3/(1+sin(d*x+c))^3-5/64/d/a^3/(1+sin(d*x+c))^2-5/128/d/a^3/(1+sin(d*x+c)
)+1/256*ln(1+sin(d*x+c))/a^3/d

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Maxima [A]  time = 1.96124, size = 254, normalized size = 1.49 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{6} + 45 \, \sin \left (d x + c\right )^{5} - 620 \, \sin \left (d x + c\right )^{4} - 540 \, \sin \left (d x + c\right )^{3} + 157 \, \sin \left (d x + c\right )^{2} + 351 \, \sin \left (d x + c\right ) + 112\right )}}{a^{3} \sin \left (d x + c\right )^{7} + 3 \, a^{3} \sin \left (d x + c\right )^{6} + a^{3} \sin \left (d x + c\right )^{5} - 5 \, a^{3} \sin \left (d x + c\right )^{4} - 5 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac{15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{3840 \, 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))^3,x, algorithm="maxima")

[Out]

-1/3840*(2*(15*sin(d*x + c)^6 + 45*sin(d*x + c)^5 - 620*sin(d*x + c)^4 - 540*sin(d*x + c)^3 + 157*sin(d*x + c)
^2 + 351*sin(d*x + c) + 112)/(a^3*sin(d*x + c)^7 + 3*a^3*sin(d*x + c)^6 + a^3*sin(d*x + c)^5 - 5*a^3*sin(d*x +
c)^4 - 5*a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x + c) + a^3) - 15*log(sin(d*x + c) + 1)/a^3 +
15*log(sin(d*x + c) - 1)/a^3)/d

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Fricas [A]  time = 1.67077, size = 657, normalized size = 3.84 \begin{align*} -\frac{30 \, \cos \left (d x + c\right )^{6} + 1150 \, \cos \left (d x + c\right )^{4} - 2076 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} +{\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} +{\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \,{\left (5 \, \cos \left (d x + c\right )^{4} + 50 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 672}{3840 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4} +{\left (a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\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))^3,x, algorithm="fricas")

[Out]

-1/3840*(30*cos(d*x + c)^6 + 1150*cos(d*x + c)^4 - 2076*cos(d*x + c)^2 - 15*(3*cos(d*x + c)^6 - 4*cos(d*x + c)
^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) + 15*(3*cos(d*x + c)^6 - 4*cos(d*
x + c)^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 18*(5*cos(d*x + c)^4 + 5
0*cos(d*x + c)^2 - 16)*sin(d*x + c) + 672)/(3*a^3*d*cos(d*x + c)^6 - 4*a^3*d*cos(d*x + c)^4 + (a^3*d*cos(d*x +
c)^6 - 4*a^3*d*cos(d*x + c)^4)*sin(d*x + c))

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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(tan(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 4.82376, size = 184, normalized size = 1.08 \begin{align*} \frac{\frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac{30 \,{\left (3 \, \sin \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) - 9\right )}}{a^{3}{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac{137 \, \sin \left (d x + c\right )^{5} + 1285 \, \sin \left (d x + c\right )^{4} + 4970 \, \sin \left (d x + c\right )^{3} + 6010 \, \sin \left (d x + c\right )^{2} + 3245 \, \sin \left (d x + c\right ) + 673}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{15360 \, 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))^3,x, algorithm="giac")

[Out]

1/15360*(60*log(abs(sin(d*x + c) + 1))/a^3 - 60*log(abs(sin(d*x + c) - 1))/a^3 + 30*(3*sin(d*x + c)^2 + 10*sin
(d*x + c) - 9)/(a^3*(sin(d*x + c) - 1)^2) - (137*sin(d*x + c)^5 + 1285*sin(d*x + c)^4 + 4970*sin(d*x + c)^3 +
6010*sin(d*x + c)^2 + 3245*sin(d*x + c) + 673)/(a^3*(sin(d*x + c) + 1)^5))/d