3.1.45 \(\int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx\) [45]

Optimal. Leaf size=106 \[ \frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac {5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {\tan ^6(c+d x)}{6 a d} \]

[Out]

5/16*arctanh(sin(d*x+c))/a/d-5/16*sec(d*x+c)*tan(d*x+c)/a/d+5/24*sec(d*x+c)*tan(d*x+c)^3/a/d-1/6*sec(d*x+c)*ta
n(d*x+c)^5/a/d+1/6*tan(d*x+c)^6/a/d

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Rubi [A]
time = 0.10, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2785, 2687, 30, 2691, 3855} \begin {gather*} \frac {\tan ^6(c+d x)}{6 a d}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac {\tan ^5(c+d x) \sec (c+d x)}{6 a d}+\frac {5 \tan ^3(c+d x) \sec (c+d x)}{24 a d}-\frac {5 \tan (c+d x) \sec (c+d x)}{16 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(16*a*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(16*a*d) + (5*Sec[c + d*x]*Tan[c + d*x]^3)/
(24*a*d) - (Sec[c + d*x]*Tan[c + d*x]^5)/(6*a*d) + Tan[c + d*x]^6/(6*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^2(c+d x) \tan ^5(c+d x) \, dx}{a}-\frac {\int \sec (c+d x) \tan ^6(c+d x) \, dx}{a}\\ &=-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {5 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{6 a}+\frac {\text {Subst}\left (\int x^5 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {\tan ^6(c+d x)}{6 a d}-\frac {5 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{8 a}\\ &=-\frac {5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac {5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {5 \int \sec (c+d x) \, dx}{16 a}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac {5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {\tan ^6(c+d x)}{6 a d}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 84, normalized size = 0.79 \begin {gather*} \frac {30 \tanh ^{-1}(\sin (c+d x))+\frac {3}{(1-\sin (c+d x))^2}-\frac {18}{1-\sin (c+d x)}+\frac {4}{(1+\sin (c+d x))^3}-\frac {21}{(1+\sin (c+d x))^2}+\frac {48}{1+\sin (c+d x)}}{96 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(30*ArcTanh[Sin[c + d*x]] + 3/(1 - Sin[c + d*x])^2 - 18/(1 - Sin[c + d*x]) + 4/(1 + Sin[c + d*x])^3 - 21/(1 +
Sin[c + d*x])^2 + 48/(1 + Sin[c + d*x]))/(96*a*d)

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Maple [A]
time = 0.21, size = 91, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {7}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\)	\(91\)
default	\(\frac {\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {7}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\)	\(91\)
risch	\(\frac {i \left (-8 \,{\mathrm e}^{7 i \left (d x +c \right )}+2 i {\mathrm e}^{6 i \left (d x +c \right )}+78 \,{\mathrm e}^{5 i \left (d x +c \right )}+18 i {\mathrm e}^{8 i \left (d x +c \right )}+33 \,{\mathrm e}^{9 i \left (d x +c \right )}-2 i {\mathrm e}^{4 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+33 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a d}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*(1/24/(1+sin(d*x+c))^3-7/32/(1+sin(d*x+c))^2+1/2/(1+sin(d*x+c))+5/32*ln(1+sin(d*x+c))+1/32/(sin(d*x+c)-1
)^2+3/16/(sin(d*x+c)-1)-5/32*ln(sin(d*x+c)-1))

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Maxima [A]
time = 0.30, size = 130, normalized size = 1.23 \begin {gather*} \frac {\frac {2 \, {\left (33 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 31 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, 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)),x, algorithm="maxima")

[Out]

1/96*(2*(33*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 31*sin(d*x + c)^2 - 7*sin(d*x + c) + 8)/(a*sin(d*x + c)^5 + a*
sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) + 15*log(sin(d*x + c) + 1)/a -
15*log(sin(d*x + c) - 1)/a)/d

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Fricas [A]
time = 0.38, size = 147, normalized size = 1.39 \begin {gather*} \frac {66 \, \cos \left (d x + c\right )^{4} - 70 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \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 )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 20}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a 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)),x, algorithm="fricas")

[Out]

1/96*(66*cos(d*x + c)^4 - 70*cos(d*x + c)^2 + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x +
c) + 1) - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(9*cos(d*x + c)^2 - 2)*
sin(d*x + c) + 20)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*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 {\left (c + d x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [A]
time = 12.55, size = 116, normalized size = 1.09 \begin {gather*} \frac {\frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 18 \, \sin \left (d x + c\right ) + 5\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, \sin \left (d x + c\right )^{3} + 69 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 7}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, 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)),x, algorithm="giac")

[Out]

1/192*(30*log(abs(sin(d*x + c) + 1))/a - 30*log(abs(sin(d*x + c) - 1))/a + 3*(15*sin(d*x + c)^2 - 18*sin(d*x +
c) + 5)/(a*(sin(d*x + c) - 1)^2) - (55*sin(d*x + c)^3 + 69*sin(d*x + c)^2 + 15*sin(d*x + c) - 7)/(a*(sin(d*x
+ c) + 1)^3))/d

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Mupad [B]
time = 10.43, size = 281, normalized size = 2.65 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {55\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{12}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {55\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\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)),x)

[Out]

(5*atanh(tan(c/2 + (d*x)/2)))/(8*a*d) - ((5*tan(c/2 + (d*x)/2))/8 + (5*tan(c/2 + (d*x)/2)^2)/4 - (5*tan(c/2 +
(d*x)/2)^3)/3 - (55*tan(c/2 + (d*x)/2)^4)/12 + (3*tan(c/2 + (d*x)/2)^5)/4 - (55*tan(c/2 + (d*x)/2)^6)/12 - (5*
tan(c/2 + (d*x)/2)^7)/3 + (5*tan(c/2 + (d*x)/2)^8)/4 + (5*tan(c/2 + (d*x)/2)^9)/8)/(d*(a + 2*a*tan(c/2 + (d*x)
/2) - 3*a*tan(c/2 + (d*x)/2)^2 - 8*a*tan(c/2 + (d*x)/2)^3 + 2*a*tan(c/2 + (d*x)/2)^4 + 12*a*tan(c/2 + (d*x)/2)
^5 + 2*a*tan(c/2 + (d*x)/2)^6 - 8*a*tan(c/2 + (d*x)/2)^7 - 3*a*tan(c/2 + (d*x)/2)^8 + 2*a*tan(c/2 + (d*x)/2)^9
+ a*tan(c/2 + (d*x)/2)^10))

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