\(\int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [857]

Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {11 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}-\frac {5 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{2 d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \]

[Out]

-11/16*a*ln(1-sin(d*x+c))/d+a*ln(sin(d*x+c))/d-5/16*a*ln(1+sin(d*x+c))/d+1/8*a^3/d/(a-a*sin(d*x+c))^2+1/2*a^2/
d/(a-a*sin(d*x+c))+1/8*a^2/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2915, 12, 90} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{2 d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {11 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}-\frac {5 a \log (\sin (c+d x)+1)}{16 d} \]

[In]

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

[Out]

(-11*a*Log[1 - Sin[c + d*x]])/(16*d) + (a*Log[Sin[c + d*x]])/d - (5*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d
*(a - a*Sin[c + d*x])^2) + a^2/(2*d*(a - a*Sin[c + d*x])) + a^2/(8*d*(a + a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {a}{(a-x)^3 x (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \frac {1}{(a-x)^3 x (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \left (\frac {1}{4 a^3 (a-x)^3}+\frac {1}{2 a^4 (a-x)^2}+\frac {11}{16 a^5 (a-x)}+\frac {1}{a^5 x}-\frac {1}{8 a^4 (a+x)^2}-\frac {5}{16 a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {11 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}-\frac {5 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{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 = 0.02 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(82\)
default	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(82\)
risch	\(-\frac {i \left (2 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+18 a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a \,{\mathrm e}^{i \left (d x +c \right )}+3 a \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {11 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(154\)
parallelrisch	\(-\frac {11 \left (-\frac {3}{11}+\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {5 \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{11}+\frac {4 \left (\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )-2 \cos \left (2 d x +2 c \right )-2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{11}+\frac {3 \cos \left (2 d x +2 c \right )}{11}-\frac {\sin \left (d x +c \right )}{11}-\frac {3 \sin \left (3 d x +3 c \right )}{11}\right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\)	\(200\)
norman	\(\frac {\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {11 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {5 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(202\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.50 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{2} - 16 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 5 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a \sin \left (d x + c\right ) + 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-1/16*(6*a*cos(d*x + c)^2 - 16*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(1/2*sin(d*x + c)) + 5*(a
*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 11*(a*cos(d*x + c)^2*sin(d*x + c) - a
*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 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. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {5 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

[In]

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

[Out]

-1/16*(5*a*log(sin(d*x + c) + 1) + 11*a*log(sin(d*x + c) - 1) - 16*a*log(sin(d*x + c)) + 2*(3*a*sin(d*x + c)^2
+ a*sin(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 = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 22 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 32 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (5 \, a \sin \left (d x + c\right ) + 7 \, a\right )}}{\sin \left (d x + c\right ) + 1} - \frac {33 \, a \sin \left (d x + c\right )^{2} - 82 \, a \sin \left (d x + c\right ) + 53 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

[In]

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

[Out]

-1/32*(10*a*log(abs(sin(d*x + c) + 1)) + 22*a*log(abs(sin(d*x + c) - 1)) - 32*a*log(abs(sin(d*x + c))) - 2*(5*
a*sin(d*x + c) + 7*a)/(sin(d*x + c) + 1) - (33*a*sin(d*x + c)^2 - 82*a*sin(d*x + c) + 53*a)/(sin(d*x + c) - 1)
^2)/d

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {3\,a\,{\sin \left (c+d\,x\right )}^2}{8}+\frac {a\,\sin \left (c+d\,x\right )}{8}-\frac {3\,a}{4}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^3-\sin \left (c+d\,x\right )\right )}-\frac {11\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}-\frac {5\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d} \]

[In]

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

[Out]

(a*log(sin(c + d*x)))/d - ((a*sin(c + d*x))/8 - (3*a)/4 + (3*a*sin(c + d*x)^2)/8)/(d*(cos(c + d*x)^2 - sin(c +
d*x) + sin(c + d*x)^3)) - (11*a*log(sin(c + d*x) - 1))/(16*d) - (5*a*log(sin(c + d*x) + 1))/(16*d)