\(\int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [1488]

Optimal result

Integrand size = 25, antiderivative size = 99 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \log (\tan (c+d x))}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^2(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \]

[Out]

3/8*b*arctanh(sin(d*x+c))/d+a*ln(tan(d*x+c))/d+3/8*b*sec(d*x+c)*tan(d*x+c)/d+1/4*b*sec(d*x+c)^3*tan(d*x+c)/d+a
*tan(d*x+c)^2/d+1/4*a*tan(d*x+c)^4/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2913, 2700, 272, 45, 3853, 3855} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan ^4(c+d x)}{4 d}+\frac {a \tan ^2(c+d x)}{d}+\frac {a \log (\tan (c+d x))}{d}+\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 b \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2700

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

Rule 2913

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

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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*} \text {integral}& = a \int \csc (c+d x) \sec ^5(c+d x) \, dx+b \int \sec ^5(c+d x) \, dx \\ & = \frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (3 b) \int \sec ^3(c+d x) \, dx+\frac {a \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (3 b) \int \sec (c+d x) \, dx+\frac {a \text {Subst}\left (\int \frac {(1+x)^2}{x} \, dx,x,\tan ^2(c+d x)\right )}{2 d} \\ & = \frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \text {Subst}\left (\int \left (2+\frac {1}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d} \\ & = \frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \log (\tan (c+d x))}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^2(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.16 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[In]

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

[Out]

(3*b*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*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(82\)
default	\(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(82\)
parallelrisch	\(\frac {-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {3 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right ) a +11 b \sin \left (d x +c \right )+3 b \sin \left (3 d x +3 c \right )+7 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(196\)
risch	\(-\frac {i \left (8 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{7 i \left (d x +c \right )}+32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+11 b \,{\mathrm e}^{5 i \left (d x +c \right )}+8 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-11 b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(202\)
norman	\(\frac {\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\)	\(214\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.26 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \, a \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 5 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

-1/16*((8*a - 3*b)*log(sin(d*x + c) + 1) + (8*a + 3*b)*log(sin(d*x + c) - 1) - 16*a*log(sin(d*x + c)) + 2*(3*b
*sin(d*x + c)^3 + 4*a*sin(d*x + c)^2 - 5*b*sin(d*x + c) - 6*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (8 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (6 \, a \sin \left (d x + c\right )^{4} - 3 \, b \sin \left (d x + c\right )^{3} - 16 \, a \sin \left (d x + c\right )^{2} + 5 \, b \sin \left (d x + c\right ) + 12 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*((8*a - 3*b)*log(abs(sin(d*x + c) + 1)) + (8*a + 3*b)*log(abs(sin(d*x + c) - 1)) - 16*a*log(abs(sin(d*x
+ c))) - 2*(6*a*sin(d*x + c)^4 - 3*b*sin(d*x + c)^3 - 16*a*sin(d*x + c)^2 + 5*b*sin(d*x + c) + 12*a)/(sin(d*x
+ c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-\frac {3\,b\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {5\,b\,\sin \left (c+d\,x\right )}{8}+\frac {3\,a}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a}{2}-\frac {3\,b}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a}{2}+\frac {3\,b}{16}\right )}{d}+\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]

[In]

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

[Out]

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