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

Optimal. Leaf size=146 \[ \frac{a^2}{32 d (a \sin (c+d x)+a)^4}-\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac{7 a}{48 d (a \sin (c+d x)+a)^3}+\frac{1}{64 d (a-a \sin (c+d x))^2}+\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + 1/(64*d*(a - a*Sin[c + d*x])^2) + a^2/(32*d*(a + a*Sin[c + d*x])^4) - (
7*a)/(48*d*(a + a*Sin[c + d*x])^3) + 1/(4*d*(a + a*Sin[c + d*x])^2) - 5/(64*d*(a^2 - a^2*Sin[c + d*x])) - 5/(3
2*d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.107958, antiderivative size = 146, 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} \[ \frac{a^2}{32 d (a \sin (c+d x)+a)^4}-\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac{7 a}{48 d (a \sin (c+d x)+a)^3}+\frac{1}{64 d (a-a \sin (c+d x))^2}+\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + 1/(64*d*(a - a*Sin[c + d*x])^2) + a^2/(32*d*(a + a*Sin[c + d*x])^4) - (
7*a)/(48*d*(a + a*Sin[c + d*x])^3) + 1/(4*d*(a + a*Sin[c + d*x])^2) - 5/(64*d*(a^2 - a^2*Sin[c + d*x])) - 5/(3
2*d*(a^2 + a^2*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))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{32 (a-x)^3}-\frac{5}{64 a (a-x)^2}-\frac{a^2}{8 (a+x)^5}+\frac{7 a}{16 (a+x)^4}-\frac{1}{2 (a+x)^3}+\frac{5}{32 a (a+x)^2}+\frac{5}{64 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{64 d (a-a \sin (c+d x))^2}+\frac{a^2}{32 d (a+a \sin (c+d x))^4}-\frac{7 a}{48 d (a+a \sin (c+d x))^3}+\frac{1}{4 d (a+a \sin (c+d x))^2}-\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d}\\ &=\frac{5 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}+\frac{1}{64 d (a-a \sin (c+d x))^2}+\frac{a^2}{32 d (a+a \sin (c+d x))^4}-\frac{7 a}{48 d (a+a \sin (c+d x))^3}+\frac{1}{4 d (a+a \sin (c+d x))^2}-\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.449695, size = 91, normalized size = 0.62 \[ \frac{\frac{-15 \sin ^5(c+d x)+66 \sin ^4(c+d x)+74 \sin ^3(c+d x)-14 \sin ^2(c+d x)-47 \sin (c+d x)-16}{(\sin (c+d x)-1)^2 (\sin (c+d x)+1)^4}+15 \tanh ^{-1}(\sin (c+d x))}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[Sin[c + d*x]] + (-16 - 47*Sin[c + d*x] - 14*Sin[c + d*x]^2 + 74*Sin[c + d*x]^3 + 66*Sin[c + d*x]^4
- 15*Sin[c + d*x]^5)/((-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^4))/(192*a^2*d)

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Maple [A]  time = 0.084, size = 144, normalized size = 1. \begin{align*}{\frac{1}{64\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{5}{64\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{128\,d{a}^{2}}}+{\frac{1}{32\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{7}{48\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{1}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5}{32\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{128\,d{a}^{2}}} \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))^2,x)

[Out]

1/64/d/a^2/(sin(d*x+c)-1)^2+5/64/d/a^2/(sin(d*x+c)-1)-5/128/d/a^2*ln(sin(d*x+c)-1)+1/32/d/a^2/(1+sin(d*x+c))^4
-7/48/d/a^2/(1+sin(d*x+c))^3+1/4/d/a^2/(1+sin(d*x+c))^2-5/32/d/a^2/(1+sin(d*x+c))+5/128*ln(1+sin(d*x+c))/a^2/d

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Maxima [A]  time = 1.05414, size = 225, normalized size = 1.54 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 66 \, \sin \left (d x + c\right )^{4} - 74 \, \sin \left (d x + c\right )^{3} + 14 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac{15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, 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))^2,x, algorithm="maxima")

[Out]

-1/384*(2*(15*sin(d*x + c)^5 - 66*sin(d*x + c)^4 - 74*sin(d*x + c)^3 + 14*sin(d*x + c)^2 + 47*sin(d*x + c) + 1
6)/(a^2*sin(d*x + c)^6 + 2*a^2*sin(d*x + c)^5 - a^2*sin(d*x + c)^4 - 4*a^2*sin(d*x + c)^3 - a^2*sin(d*x + c)^2
+ 2*a^2*sin(d*x + c) + a^2) - 15*log(sin(d*x + c) + 1)/a^2 + 15*log(sin(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.52801, size = 532, normalized size = 3.64 \begin{align*} -\frac{132 \, \cos \left (d x + c\right )^{4} - 236 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \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 )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (15 \, \cos \left (d x + c\right )^{4} + 44 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) + 72}{384 \,{\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/384*(132*cos(d*x + c)^4 - 236*cos(d*x + c)^2 - 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d
*x + c)^4)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(
-sin(d*x + c) + 1) - 2*(15*cos(d*x + c)^4 + 44*cos(d*x + c)^2 - 12)*sin(d*x + c) + 72)/(a^2*d*cos(d*x + c)^6 -
2*a^2*d*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*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 ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

Integral(tan(c + d*x)**5/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 5.28893, size = 170, normalized size = 1.16 \begin{align*} \frac{\frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (15 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right ) - 1\right )}}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac{125 \, \sin \left (d x + c\right )^{4} + 740 \, \sin \left (d x + c\right )^{3} + 1086 \, \sin \left (d x + c\right )^{2} + 676 \, \sin \left (d x + c\right ) + 157}{a^{2}{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{1536 \, 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))^2,x, algorithm="giac")

[Out]

1/1536*(60*log(abs(sin(d*x + c) + 1))/a^2 - 60*log(abs(sin(d*x + c) - 1))/a^2 + 6*(15*sin(d*x + c)^2 - 10*sin(
d*x + c) - 1)/(a^2*(sin(d*x + c) - 1)^2) - (125*sin(d*x + c)^4 + 740*sin(d*x + c)^3 + 1086*sin(d*x + c)^2 + 67
6*sin(d*x + c) + 157)/(a^2*(sin(d*x + c) + 1)^4))/d