3.61 \(\int \frac{\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=189 \[ \frac{a^3}{80 d (a \sin (c+d x)+a)^5}-\frac{5 a^2}{64 d (a \sin (c+d x)+a)^4}+\frac{21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{35}{256 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac{a}{192 d (a-a \sin (c+d x))^3}+\frac{19 a}{96 d (a \sin (c+d x)+a)^3}-\frac{1}{32 d (a-a \sin (c+d x))^2}-\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(128*a^2*d) + a/(192*d*(a - a*Sin[c + d*x])^3) - 1/(32*d*(a - a*Sin[c + d*x])^2) +
a^3/(80*d*(a + a*Sin[c + d*x])^5) - (5*a^2)/(64*d*(a + a*Sin[c + d*x])^4) + (19*a)/(96*d*(a + a*Sin[c + d*x])^
3) - 1/(4*d*(a + a*Sin[c + d*x])^2) + 21/(256*d*(a^2 - a^2*Sin[c + d*x])) + 35/(256*d*(a^2 + a^2*Sin[c + d*x])
)

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Rubi [A]  time = 0.148605, antiderivative size = 189, 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} \[ \frac{a^3}{80 d (a \sin (c+d x)+a)^5}-\frac{5 a^2}{64 d (a \sin (c+d x)+a)^4}+\frac{21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{35}{256 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac{a}{192 d (a-a \sin (c+d x))^3}+\frac{19 a}{96 d (a \sin (c+d x)+a)^3}-\frac{1}{32 d (a-a \sin (c+d x))^2}-\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(128*a^2*d) + a/(192*d*(a - a*Sin[c + d*x])^3) - 1/(32*d*(a - a*Sin[c + d*x])^2) +
a^3/(80*d*(a + a*Sin[c + d*x])^5) - (5*a^2)/(64*d*(a + a*Sin[c + d*x])^4) + (19*a)/(96*d*(a + a*Sin[c + d*x])^
3) - 1/(4*d*(a + a*Sin[c + d*x])^2) + 21/(256*d*(a^2 - a^2*Sin[c + d*x])) + 35/(256*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 ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^7}{(a-x)^4 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a}{64 (a-x)^4}-\frac{1}{16 (a-x)^3}+\frac{21}{256 a (a-x)^2}-\frac{a^3}{16 (a+x)^6}+\frac{5 a^2}{16 (a+x)^5}-\frac{19 a}{32 (a+x)^4}+\frac{1}{2 (a+x)^3}-\frac{35}{256 a (a+x)^2}-\frac{7}{128 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a}{192 d (a-a \sin (c+d x))^3}-\frac{1}{32 d (a-a \sin (c+d x))^2}+\frac{a^3}{80 d (a+a \sin (c+d x))^5}-\frac{5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac{19 a}{96 d (a+a \sin (c+d x))^3}-\frac{1}{4 d (a+a \sin (c+d x))^2}+\frac{21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{35}{256 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a d}\\ &=-\frac{7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac{a}{192 d (a-a \sin (c+d x))^3}-\frac{1}{32 d (a-a \sin (c+d x))^2}+\frac{a^3}{80 d (a+a \sin (c+d x))^5}-\frac{5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac{19 a}{96 d (a+a \sin (c+d x))^3}-\frac{1}{4 d (a+a \sin (c+d x))^2}+\frac{21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{35}{256 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.69255, size = 112, normalized size = 0.59 \[ -\frac{210 \tanh ^{-1}(\sin (c+d x))-\frac{2 \left (105 \sin ^7(c+d x)-750 \sin ^6(c+d x)-815 \sin ^5(c+d x)+560 \sin ^4(c+d x)+1039 \sin ^3(c+d x)+78 \sin ^2(c+d x)-393 \sin (c+d x)-144\right )}{(\sin (c+d x)-1)^3 (\sin (c+d x)+1)^5}}{3840 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

-(210*ArcTanh[Sin[c + d*x]] - (2*(-144 - 393*Sin[c + d*x] + 78*Sin[c + d*x]^2 + 1039*Sin[c + d*x]^3 + 560*Sin[
c + d*x]^4 - 815*Sin[c + d*x]^5 - 750*Sin[c + d*x]^6 + 105*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^3*(1 + Sin[c
+ d*x])^5))/(3840*a^2*d)

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Maple [A]  time = 0.096, size = 180, normalized size = 1. \begin{align*} -{\frac{1}{192\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}-{\frac{1}{32\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{21}{256\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{7\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,d{a}^{2}}}+{\frac{1}{80\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{5}{64\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{19}{96\,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{35}{256\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{7\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x)

[Out]

-1/192/d/a^2/(sin(d*x+c)-1)^3-1/32/d/a^2/(sin(d*x+c)-1)^2-21/256/d/a^2/(sin(d*x+c)-1)+7/256/d/a^2*ln(sin(d*x+c
)-1)+1/80/d/a^2/(1+sin(d*x+c))^5-5/64/d/a^2/(1+sin(d*x+c))^4+19/96/d/a^2/(1+sin(d*x+c))^3-1/4/d/a^2/(1+sin(d*x
+c))^2+35/256/d/a^2/(1+sin(d*x+c))-7/256*ln(1+sin(d*x+c))/a^2/d

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Maxima [A]  time = 1.10553, size = 273, normalized size = 1.44 \begin{align*} \frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{7} - 750 \, \sin \left (d x + c\right )^{6} - 815 \, \sin \left (d x + c\right )^{5} + 560 \, \sin \left (d x + c\right )^{4} + 1039 \, \sin \left (d x + c\right )^{3} + 78 \, \sin \left (d x + c\right )^{2} - 393 \, \sin \left (d x + c\right ) - 144\right )}}{a^{2} \sin \left (d x + c\right )^{8} + 2 \, a^{2} \sin \left (d x + c\right )^{7} - 2 \, a^{2} \sin \left (d x + c\right )^{6} - 6 \, a^{2} \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{3840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3840*(2*(105*sin(d*x + c)^7 - 750*sin(d*x + c)^6 - 815*sin(d*x + c)^5 + 560*sin(d*x + c)^4 + 1039*sin(d*x +
c)^3 + 78*sin(d*x + c)^2 - 393*sin(d*x + c) - 144)/(a^2*sin(d*x + c)^8 + 2*a^2*sin(d*x + c)^7 - 2*a^2*sin(d*x
+ c)^6 - 6*a^2*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*a^2*sin(d*x + c) - a^2) - 105*
log(sin(d*x + c) + 1)/a^2 + 105*log(sin(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.72815, size = 599, normalized size = 3.17 \begin{align*} \frac{1500 \, \cos \left (d x + c\right )^{6} - 3380 \, \cos \left (d x + c\right )^{4} + 2104 \, \cos \left (d x + c\right )^{2} - 105 \,{\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \,{\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (105 \, \cos \left (d x + c\right )^{6} + 500 \, \cos \left (d x + c\right )^{4} - 276 \, \cos \left (d x + c\right )^{2} + 64\right )} \sin \left (d x + c\right ) - 512}{3840 \,{\left (a^{2} d \cos \left (d x + c\right )^{8} - 2 \, a^{2} d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3840*(1500*cos(d*x + c)^6 - 3380*cos(d*x + c)^4 + 2104*cos(d*x + c)^2 - 105*(cos(d*x + c)^8 - 2*cos(d*x + c)
^6*sin(d*x + c) - 2*cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8 - 2*cos(d*x + c)^6*sin(d*x + c
) - 2*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(105*cos(d*x + c)^6 + 500*cos(d*x + c)^4 - 276*cos(d*x + c)^2
+ 64)*sin(d*x + c) - 512)/(a^2*d*cos(d*x + c)^8 - 2*a^2*d*cos(d*x + c)^6*sin(d*x + c) - 2*a^2*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(tan(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 9.84355, size = 197, normalized size = 1.04 \begin{align*} -\frac{\frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{10 \,{\left (77 \, \sin \left (d x + c\right )^{3} - 105 \, \sin \left (d x + c\right )^{2} + 27 \, \sin \left (d x + c\right ) + 9\right )}}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac{959 \, \sin \left (d x + c\right )^{5} + 6895 \, \sin \left (d x + c\right )^{4} + 14150 \, \sin \left (d x + c\right )^{3} + 13710 \, \sin \left (d x + c\right )^{2} + 6555 \, \sin \left (d x + c\right ) + 1251}{a^{2}{\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)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/15360*(420*log(abs(sin(d*x + c) + 1))/a^2 - 420*log(abs(sin(d*x + c) - 1))/a^2 + 10*(77*sin(d*x + c)^3 - 10
5*sin(d*x + c)^2 + 27*sin(d*x + c) + 9)/(a^2*(sin(d*x + c) - 1)^3) - (959*sin(d*x + c)^5 + 6895*sin(d*x + c)^4
+ 14150*sin(d*x + c)^3 + 13710*sin(d*x + c)^2 + 6555*sin(d*x + c) + 1251)/(a^2*(sin(d*x + c) + 1)^5))/d