\(\int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx\) [852]

Optimal result

Integrand size = 19, antiderivative size = 115 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \]

[Out]

-23/16*a*ln(1-sin(d*x+c))/d+7/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d+1/8*a^3/d/(a-a*sin(d*x+c))^2-a^2/d/(a-a*s
in(d*x+c))+1/8*a^2/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2786, 90} \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]

[In]

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

[Out]

(-23*a*Log[1 - Sin[c + d*x]])/(16*d) + (7*a*Log[1 + Sin[c + d*x]])/(16*d) - (a*Sin[c + d*x])/d + a^3/(8*d*(a -
a*Sin[c + d*x])^2) - a^2/(d*(a - a*Sin[c + d*x])) + a^2/(8*d*(a + a*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 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-1+\frac {a^3}{4 (a-x)^3}-\frac {a^2}{(a-x)^2}+\frac {23 a}{16 (a-x)}-\frac {a^2}{8 (a+x)^2}+\frac {7 a}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.18 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {15 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]

[In]

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

[Out]

(15*a*ArcTanh[Sin[c + d*x]])/(8*d) + (15*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (15*a*Sec[c + d*x]^3*Tan[c + d*x
])/(4*d) + (5*a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (a*Sin[c + d*x]*Tan[c + d*x]^4)/d - (a*(4*Log[Cos[c + d*x]] +
2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {a \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(121\)
default	\(\frac {a \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(121\)
risch	\(i a x +\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}+\frac {i \left (2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a \,{\mathrm e}^{i \left (d x +c \right )}-2 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{3 i \left (d x +c \right )}+9 a \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}+\frac {7 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {23 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(182\)
parallelrisch	\(\frac {2 \left (\frac {5}{8}+\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {23 \left (-1-\cos \left (2 d x +2 c \right )+\frac {\sin \left (d x +c \right )}{2}+\frac {\sin \left (3 d x +3 c \right )}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {7 \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}-\frac {3 \cos \left (2 d x +2 c \right )}{8}-\frac {\cos \left (4 d x +4 c \right )}{4}-\frac {9 \sin \left (d x +c \right )}{8}-\frac {7 \sin \left (3 d x +3 c \right )}{8}\right ) a}{d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\)	\(217\)
norman	\(\frac {-\frac {15 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {10 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {9 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {10 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {15 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {6 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {23 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {7 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(240\)

[In]

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

[Out]

1/d*(a*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin
(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+a*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {16 \, a \cos \left (d x + c\right )^{4} + 2 \, a \cos \left (d x + c\right )^{2} + 7 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/16*(16*a*cos(d*x + c)^4 + 2*a*cos(d*x + c)^2 + 7*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(
d*x + c) + 1) - 23*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(8*a*cos(d*x
+ c)^2 + a)*sin(d*x + c) - 6*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \sin \left (d x + c\right ) + \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

[In]

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

[Out]

1/16*(7*a*log(sin(d*x + c) + 1) - 23*a*log(sin(d*x + c) - 1) - 16*a*sin(d*x + c) + 2*(9*a*sin(d*x + c)^2 - a*s
in(d*x + c) - 6*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {14 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 46 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 32 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (7 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {69 \, a \sin \left (d x + c\right )^{2} - 106 \, a \sin \left (d x + c\right ) + 41 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

[In]

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

[Out]

1/32*(14*a*log(abs(sin(d*x + c) + 1)) - 46*a*log(abs(sin(d*x + c) - 1)) - 32*a*sin(d*x + c) - 2*(7*a*sin(d*x +
c) + 5*a)/(sin(d*x + c) + 1) + (69*a*sin(d*x + c)^2 - 106*a*sin(d*x + c) + 41*a)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 10.35 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.04 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {23\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {7\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

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

[Out]

((11*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + (11*a*tan(c/2 + (d*x)/2)^3)/4 - 5*a*tan(c/2 + (
d*x)/2)^4 + (11*a*tan(c/2 + (d*x)/2)^5)/4 + (11*a*tan(c/2 + (d*x)/2)^6)/2 - (15*a*tan(c/2 + (d*x)/2)^7)/4)/(d*
(2*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^4 + 2*tan(c/2 + (d*x)/2)^5 - 2*tan(c/2 +
(d*x)/2)^7 + tan(c/2 + (d*x)/2)^8 + 1)) - (23*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + (7*a*log(tan(c/2 + (d*x)
/2) + 1))/(8*d) + (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d