3.1.72 \(\int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [72]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 171, 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 {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 {a^2}{40 d (a \sin (c+d x)+a)^5}-\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} \end {gather*}

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 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))^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {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 {\text {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*}

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Mathematica [A]
time = 0.45, size = 102, normalized size = 0.60 \begin {gather*} \frac {15 \tanh ^{-1}(\sin (c+d x))-\frac {112+351 \sin (c+d x)+157 \sin ^2(c+d x)-540 \sin ^3(c+d x)-620 \sin ^4(c+d x)+45 \sin ^5(c+d x)+15 \sin ^6(c+d x)}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^5}}{1920 a^3 d} \end {gather*}

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 = 0.32, size = 115, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {\frac {1}{40 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {7}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{32 \sin \left (d x +c \right )-32}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{3}}\)	\(115\)
default	\(\frac {\frac {1}{40 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {7}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{32 \sin \left (d x +c \right )-32}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{3}}\)	\(115\)
risch	\(-\frac {i \left (15 \,{\mathrm e}^{i \left (d x +c \right )}-828 i {\mathrm e}^{8 i \left (d x +c \right )}+2390 \,{\mathrm e}^{3 i \left (d x +c \right )}-90 i {\mathrm e}^{2 i \left (d x +c \right )}-3870 i {\mathrm e}^{4 i \left (d x +c \right )}-7183 \,{\mathrm e}^{5 i \left (d x +c \right )}+828 i {\mathrm e}^{6 i \left (d x +c \right )}+90 i {\mathrm e}^{12 i \left (d x +c \right )}+3870 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}+2390 \,{\mathrm e}^{11 i \left (d x +c \right )}-7183 \,{\mathrm e}^{9 i \left (d x +c \right )}+2388 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a^{3} d}\)	\(231\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 188, normalized size = 1.10 \begin {gather*} -\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 {gather*}

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 = 0.37, size = 248, normalized size = 1.45 \begin {gather*} -\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 {gather*}

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

[Out]

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

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Giac [A]
time = 12.50, size = 136, normalized size = 0.80 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 10.05, size = 418, normalized size = 2.44 \begin {gather*} \frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{32}+\frac {527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{960}+\frac {901\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{80}+\frac {711\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{80}+\frac {901\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{80}+\frac {527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-38\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+72\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-38\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^3\,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))^3,x)

[Out]

(tan(c/2 + (d*x)/2)^4/32 - (3*tan(c/2 + (d*x)/2)^2)/32 - (17*tan(c/2 + (d*x)/2)^3)/96 - tan(c/2 + (d*x)/2)/64
+ (527*tan(c/2 + (d*x)/2)^5)/960 + (901*tan(c/2 + (d*x)/2)^6)/80 + (711*tan(c/2 + (d*x)/2)^7)/80 + (901*tan(c/
2 + (d*x)/2)^8)/80 + (527*tan(c/2 + (d*x)/2)^9)/960 + tan(c/2 + (d*x)/2)^10/32 - (17*tan(c/2 + (d*x)/2)^11)/96
- (3*tan(c/2 + (d*x)/2)^12)/32 - tan(c/2 + (d*x)/2)^13/64)/(d*(11*a^3*tan(c/2 + (d*x)/2)^2 - 4*a^3*tan(c/2 +
(d*x)/2)^3 - 39*a^3*tan(c/2 + (d*x)/2)^4 - 38*a^3*tan(c/2 + (d*x)/2)^5 + 27*a^3*tan(c/2 + (d*x)/2)^6 + 72*a^3*
tan(c/2 + (d*x)/2)^7 + 27*a^3*tan(c/2 + (d*x)/2)^8 - 38*a^3*tan(c/2 + (d*x)/2)^9 - 39*a^3*tan(c/2 + (d*x)/2)^1
0 - 4*a^3*tan(c/2 + (d*x)/2)^11 + 11*a^3*tan(c/2 + (d*x)/2)^12 + 6*a^3*tan(c/2 + (d*x)/2)^13 + a^3*tan(c/2 + (
d*x)/2)^14 + a^3 + 6*a^3*tan(c/2 + (d*x)/2))) + atanh(tan(c/2 + (d*x)/2))/(64*a^3*d)

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