\(\int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\) [801]

Optimal result

Integrand size = 25, antiderivative size = 68 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

-a*arctanh(cos(d*x+c))/d+a*sec(d*x+c)/d+1/3*a*sec(d*x+c)^3/d+a*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2917, 2702, 308, 213, 3852} \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-((a*ArcTanh[Cos[c + d*x]])/d) + (a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) + (a*Tan[c + d*x])/d + (a*Tan[c
+ d*x]^3)/(3*d)

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2702

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

Rule 2917

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = a \int \sec ^4(c+d x) \, dx+a \int \csc (c+d x) \sec ^4(c+d x) \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.25 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \]

[In]

Integrate[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(64\)
default	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(64\)
parallelrisch	\(\frac {\left (\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}-\frac {8}{3}\right ) a}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(93\)
risch	\(\frac {2 a \left (-6 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i+{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(106\)
norman	\(\frac {-\frac {8 a}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(155\)

[In]

int(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.85 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} + 3 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, a \sin \left (d x + c\right ) + 4 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(4*a*cos(d*x + c)^2 + 3*(a*cos(d*x + c)*sin(d*x + c) - a*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a
*cos(d*x + c)*sin(d*x + c) - a*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*a*sin(d*x + c) + 4*a)/(d*cos(d*x
+ c)*sin(d*x + c) - d*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a + a {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*a + a*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x + c) +
1) + 3*log(cos(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*a*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a/(tan(1/2*d*x + 1/2*c) + 1) - (15*a*tan(1/2*d*x + 1/2*c)^2 - 24*a
*tan(1/2*d*x + 1/2*c) + 13*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 11.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.32 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {10\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {8\,a}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

int((a + a*sin(c + d*x))/(cos(c + d*x)^4*sin(c + d*x)),x)

[Out]

(a*log(tan(c/2 + (d*x)/2)))/d - ((8*a)/3 - (10*a*tan(c/2 + (d*x)/2))/3 + 2*a*tan(c/2 + (d*x)/2)^3)/(d*(2*tan(c
/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 - 1))