3.1.82 \(\int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [82]

Optimal. Leaf size=195 \[ -\frac {\tanh ^{-1}(\sin (c+d x))}{128 a^4 d}+\frac {a^2}{48 d (a+a \sin (c+d x))^6}-\frac {7 a}{80 d (a+a \sin (c+d x))^5}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {5}{96 a d (a+a \sin (c+d x))^3}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

-1/128*arctanh(sin(d*x+c))/a^4/d+1/48*a^2/d/(a+a*sin(d*x+c))^6-7/80*a/d/(a+a*sin(d*x+c))^5+1/8/d/(a+a*sin(d*x+
c))^4-5/96/a/d/(a+a*sin(d*x+c))^3+1/256/d/(a^2-a^2*sin(d*x+c))^2-5/256/d/(a^2+a^2*sin(d*x+c))^2-3/256/d/(a^4-a
^4*sin(d*x+c))-1/256/d/(a^4+a^4*sin(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 195, normalized size of antiderivative = 1.00, 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 = {2786, 90, 212} \begin {gather*} -\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\tanh ^{-1}(\sin (c+d x))}{128 a^4 d}+\frac {a^2}{48 d (a \sin (c+d x)+a)^6}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {7 a}{80 d (a \sin (c+d x)+a)^5}+\frac {1}{8 d (a \sin (c+d x)+a)^4}-\frac {5}{96 a d (a \sin (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/128*ArcTanh[Sin[c + d*x]]/(a^4*d) + a^2/(48*d*(a + a*Sin[c + d*x])^6) - (7*a)/(80*d*(a + a*Sin[c + d*x])^5)
+ 1/(8*d*(a + a*Sin[c + d*x])^4) - 5/(96*a*d*(a + a*Sin[c + d*x])^3) + 1/(256*d*(a^2 - a^2*Sin[c + d*x])^2) -
5/(256*d*(a^2 + a^2*Sin[c + d*x])^2) - 3/(256*d*(a^4 - a^4*Sin[c + d*x])) - 1/(256*d*(a^4 + a^4*Sin[c + d*x])
)

Rule 90

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 212

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

Rule 2786

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]

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\text {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^7} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{128 a^2 (a-x)^3}-\frac {3}{256 a^3 (a-x)^2}-\frac {a^2}{8 (a+x)^7}+\frac {7 a}{16 (a+x)^6}-\frac {1}{2 (a+x)^5}+\frac {5}{32 a (a+x)^4}+\frac {5}{128 a^2 (a+x)^3}+\frac {1}{256 a^3 (a+x)^2}-\frac {1}{128 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^2}{48 d (a+a \sin (c+d x))^6}-\frac {7 a}{80 d (a+a \sin (c+d x))^5}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {5}{96 a d (a+a \sin (c+d x))^3}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a^3 d}\\ &=-\frac {\tanh ^{-1}(\sin (c+d x))}{128 a^4 d}+\frac {a^2}{48 d (a+a \sin (c+d x))^6}-\frac {7 a}{80 d (a+a \sin (c+d x))^5}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {5}{96 a d (a+a \sin (c+d x))^3}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.95, size = 112, normalized size = 0.57 \begin {gather*} -\frac {30 \tanh ^{-1}(\sin (c+d x))-\frac {2 \left (-48-177 \sin (c+d x)-132 \sin ^2(c+d x)+257 \sin ^3(c+d x)+440 \sin ^4(c+d x)+65 \sin ^5(c+d x)+60 \sin ^6(c+d x)+15 \sin ^7(c+d x)\right )}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^6}}{3840 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/3840*(30*ArcTanh[Sin[c + d*x]] - (2*(-48 - 177*Sin[c + d*x] - 132*Sin[c + d*x]^2 + 257*Sin[c + d*x]^3 + 440
*Sin[c + d*x]^4 + 65*Sin[c + d*x]^5 + 60*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^2*(1 + Sin[
c + d*x])^6))/(a^4*d)

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Maple [A]
time = 0.32, size = 127, normalized size = 0.65

method	result	size
derivativedivides	\(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d \,a^{4}}\)	\(127\)
default	\(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d \,a^{4}}\)	\(127\)
risch	\(\frac {i \left (4133 \,{\mathrm e}^{7 i \left (d x +c \right )}+5727 \,{\mathrm e}^{11 i \left (d x +c \right )}+365 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-5727 \,{\mathrm e}^{5 i \left (d x +c \right )}-4240 i {\mathrm e}^{4 i \left (d x +c \right )}+11656 i {\mathrm e}^{6 i \left (d x +c \right )}-8928 i {\mathrm e}^{8 i \left (d x +c \right )}+11656 i {\mathrm e}^{10 i \left (d x +c \right )}+120 i {\mathrm e}^{14 i \left (d x +c \right )}-365 \,{\mathrm e}^{13 i \left (d x +c \right )}+120 i {\mathrm e}^{2 i \left (d x +c \right )}-4240 i {\mathrm e}^{12 i \left (d x +c \right )}+15 \,{\mathrm e}^{15 i \left (d x +c \right )}-4133 \,{\mathrm e}^{9 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{12} d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d \,a^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 d \,a^{4}}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d/a^4*(1/256/(sin(d*x+c)-1)^2+3/256/(sin(d*x+c)-1)+1/256*ln(sin(d*x+c)-1)+1/48/(1+sin(d*x+c))^6-7/80/(1+sin(
d*x+c))^5+1/8/(1+sin(d*x+c))^4-5/96/(1+sin(d*x+c))^3-5/256/(1+sin(d*x+c))^2-1/256/(1+sin(d*x+c))-1/256*ln(1+si
n(d*x+c)))

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Maxima [A]
time = 0.29, size = 213, normalized size = 1.09 \begin {gather*} \frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} + 60 \, \sin \left (d x + c\right )^{6} + 65 \, \sin \left (d x + c\right )^{5} + 440 \, \sin \left (d x + c\right )^{4} + 257 \, \sin \left (d x + c\right )^{3} - 132 \, \sin \left (d x + c\right )^{2} - 177 \, \sin \left (d x + c\right ) - 48\right )}}{a^{4} \sin \left (d x + c\right )^{8} + 4 \, a^{4} \sin \left (d x + c\right )^{7} + 4 \, a^{4} \sin \left (d x + c\right )^{6} - 4 \, a^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{4} - 4 \, a^{4} \sin \left (d x + c\right )^{3} + 4 \, a^{4} \sin \left (d x + c\right )^{2} + 4 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4}}}{3840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/3840*(2*(15*sin(d*x + c)^7 + 60*sin(d*x + c)^6 + 65*sin(d*x + c)^5 + 440*sin(d*x + c)^4 + 257*sin(d*x + c)^3
- 132*sin(d*x + c)^2 - 177*sin(d*x + c) - 48)/(a^4*sin(d*x + c)^8 + 4*a^4*sin(d*x + c)^7 + 4*a^4*sin(d*x + c)
^6 - 4*a^4*sin(d*x + c)^5 - 10*a^4*sin(d*x + c)^4 - 4*a^4*sin(d*x + c)^3 + 4*a^4*sin(d*x + c)^2 + 4*a^4*sin(d*
x + c) + a^4) - 15*log(sin(d*x + c) + 1)/a^4 + 15*log(sin(d*x + c) - 1)/a^4)/d

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Fricas [A]
time = 0.38, size = 290, normalized size = 1.49 \begin {gather*} -\frac {120 \, \cos \left (d x + c\right )^{6} - 1240 \, \cos \left (d x + c\right )^{4} + 1856 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \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 (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \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 ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 110 \, \cos \left (d x + c\right )^{4} + 432 \, \cos \left (d x + c\right )^{2} - 160\right )} \sin \left (d x + c\right ) - 640}{3840 \, {\left (a^{4} d \cos \left (d x + c\right )^{8} - 8 \, a^{4} d \cos \left (d x + c\right )^{6} + 8 \, a^{4} d \cos \left (d x + c\right )^{4} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 2 \, a^{4} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/3840*(120*cos(d*x + c)^6 - 1240*cos(d*x + c)^4 + 1856*cos(d*x + c)^2 + 15*(cos(d*x + c)^8 - 8*cos(d*x + c)^
6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c)^6 - 2*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*(cos(d*x
+ c)^8 - 8*cos(d*x + c)^6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c)^6 - 2*cos(d*x + c)^4)*sin(d*x + c))*log(-sin(d
*x + c) + 1) + 2*(15*cos(d*x + c)^6 - 110*cos(d*x + c)^4 + 432*cos(d*x + c)^2 - 160)*sin(d*x + c) - 640)/(a^4*
d*cos(d*x + c)^8 - 8*a^4*d*cos(d*x + c)^6 + 8*a^4*d*cos(d*x + c)^4 - 4*(a^4*d*cos(d*x + c)^6 - 2*a^4*d*cos(d*x
+ c)^4)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(tan(c + d*x)**5/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a*
*4

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Giac [A]
time = 24.28, size = 146, normalized size = 0.75 \begin {gather*} -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4}} + \frac {30 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 7\right )}}{a^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {147 \, \sin \left (d x + c\right )^{6} + 822 \, \sin \left (d x + c\right )^{5} + 1605 \, \sin \left (d x + c\right )^{4} + 340 \, \sin \left (d x + c\right )^{3} - 675 \, \sin \left (d x + c\right )^{2} - 522 \, \sin \left (d x + c\right ) - 117}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{6}}}{15360 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/15360*(60*log(abs(sin(d*x + c) + 1))/a^4 - 60*log(abs(sin(d*x + c) - 1))/a^4 + 30*(3*sin(d*x + c)^2 - 12*si
n(d*x + c) + 7)/(a^4*(sin(d*x + c) - 1)^2) - (147*sin(d*x + c)^6 + 822*sin(d*x + c)^5 + 1605*sin(d*x + c)^4 +
340*sin(d*x + c)^3 - 675*sin(d*x + c)^2 - 522*sin(d*x + c) - 117)/(a^4*(sin(d*x + c) + 1)^6))/d

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Mupad [B]
time = 10.66, size = 476, normalized size = 2.44 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{8}+\frac {73\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{12}-\frac {139\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{320}+\frac {1073\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{120}+\frac {10277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{960}+\frac {237\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{10}+\frac {10277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{960}+\frac {1073\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{120}-\frac {139\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{320}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {73\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-36\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+198\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-36\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^4\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + a*sin(c + d*x))^4,x)

[Out]

(tan(c/2 + (d*x)/2)/64 + tan(c/2 + (d*x)/2)^2/8 + (73*tan(c/2 + (d*x)/2)^3)/192 + (5*tan(c/2 + (d*x)/2)^4)/12
- (139*tan(c/2 + (d*x)/2)^5)/320 + (1073*tan(c/2 + (d*x)/2)^6)/120 + (10277*tan(c/2 + (d*x)/2)^7)/960 + (237*t
an(c/2 + (d*x)/2)^8)/10 + (10277*tan(c/2 + (d*x)/2)^9)/960 + (1073*tan(c/2 + (d*x)/2)^10)/120 - (139*tan(c/2 +
(d*x)/2)^11)/320 + (5*tan(c/2 + (d*x)/2)^12)/12 + (73*tan(c/2 + (d*x)/2)^13)/192 + tan(c/2 + (d*x)/2)^14/8 +
tan(c/2 + (d*x)/2)^15/64)/(d*(24*a^4*tan(c/2 + (d*x)/2)^2 + 24*a^4*tan(c/2 + (d*x)/2)^3 - 36*a^4*tan(c/2 + (d*
x)/2)^4 - 120*a^4*tan(c/2 + (d*x)/2)^5 - 88*a^4*tan(c/2 + (d*x)/2)^6 + 88*a^4*tan(c/2 + (d*x)/2)^7 + 198*a^4*t
an(c/2 + (d*x)/2)^8 + 88*a^4*tan(c/2 + (d*x)/2)^9 - 88*a^4*tan(c/2 + (d*x)/2)^10 - 120*a^4*tan(c/2 + (d*x)/2)^
11 - 36*a^4*tan(c/2 + (d*x)/2)^12 + 24*a^4*tan(c/2 + (d*x)/2)^13 + 24*a^4*tan(c/2 + (d*x)/2)^14 + 8*a^4*tan(c/
2 + (d*x)/2)^15 + a^4*tan(c/2 + (d*x)/2)^16 + a^4 + 8*a^4*tan(c/2 + (d*x)/2))) - atanh(tan(c/2 + (d*x)/2))/(64
*a^4*d)

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