3.1.62 \(\int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [62]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 146, 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 {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} \end {gather*}

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 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))^2} \, dx &=\frac {\text {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {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 \text {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*}

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Mathematica [A]
time = 0.30, size = 91, normalized size = 0.62 \begin {gather*} \frac {15 \tanh ^{-1}(\sin (c+d x))+\frac {-16-47 \sin (c+d x)-14 \sin ^2(c+d x)+74 \sin ^3(c+d x)+66 \sin ^4(c+d x)-15 \sin ^5(c+d x)}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^4}}{192 a^2 d} \end {gather*}

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 = 0.27, size = 103, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{128}}{d \,a^{2}}\)	\(103\)
default	\(\frac {\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{128}}{d \,a^{2}}\)	\(103\)
risch	\(-\frac {i \left (-56 i {\mathrm e}^{6 i \left (d x +c \right )}-221 \,{\mathrm e}^{3 i \left (d x +c \right )}+416 i {\mathrm e}^{4 i \left (d x +c \right )}-14 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-132 i {\mathrm e}^{2 i \left (d x +c \right )}+14 \,{\mathrm e}^{7 i \left (d x +c \right )}+416 i {\mathrm e}^{8 i \left (d x +c \right )}+221 \,{\mathrm e}^{9 i \left (d x +c \right )}-132 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}\right )}{96 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d \,a^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{64 a^{2} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{64 a^{2} d}\)	\(208\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 167, normalized size = 1.14 \begin {gather*} -\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 {gather*}

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 = 0.37, size = 198, normalized size = 1.36 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 8.19, size = 126, normalized size = 0.86 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 10.50, size = 361, normalized size = 2.47 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,a^2\,d}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{32}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{8}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{96}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}-\frac {121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}-\frac {119\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{12}-\frac {121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{48}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )} \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))^2,x)

[Out]

(5*atanh(tan(c/2 + (d*x)/2)))/(32*a^2*d) - ((5*tan(c/2 + (d*x)/2))/32 + (5*tan(c/2 + (d*x)/2)^2)/8 + (35*tan(c
/2 + (d*x)/2)^3)/96 - (5*tan(c/2 + (d*x)/2)^4)/3 - (121*tan(c/2 + (d*x)/2)^5)/48 - (119*tan(c/2 + (d*x)/2)^6)/
12 - (121*tan(c/2 + (d*x)/2)^7)/48 - (5*tan(c/2 + (d*x)/2)^8)/3 + (35*tan(c/2 + (d*x)/2)^9)/96 + (5*tan(c/2 +
(d*x)/2)^10)/8 + (5*tan(c/2 + (d*x)/2)^11)/32)/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 - 12*a^2*tan(c/2 + (d*x)/2)^3 -
17*a^2*tan(c/2 + (d*x)/2)^4 + 8*a^2*tan(c/2 + (d*x)/2)^5 + 28*a^2*tan(c/2 + (d*x)/2)^6 + 8*a^2*tan(c/2 + (d*x)
/2)^7 - 17*a^2*tan(c/2 + (d*x)/2)^8 - 12*a^2*tan(c/2 + (d*x)/2)^9 + 2*a^2*tan(c/2 + (d*x)/2)^10 + 4*a^2*tan(c/
2 + (d*x)/2)^11 + a^2*tan(c/2 + (d*x)/2)^12 + a^2 + 4*a^2*tan(c/2 + (d*x)/2)))

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