\(\int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx\) [15]

Optimal result

Integrand size = 19, antiderivative size = 69 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d} \]

[Out]

a*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-1/3*a*cot(d*x+c)^3/d-a*csc(d*x+c)/d-1/3*a*csc(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3957, 2917, 2701, 308, 213, 3852} \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d} \]

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (a*Cot[c + d*x]^3)/(3*d) - (a*Csc[c + d*x])/d - (a*Csc[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 2701

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {2 a \cot (c+d x)}{3 d}-\frac {a \cot (c+d x) \csc ^2(c+d x)}{3 d}-\frac {a \csc ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\sin ^2(c+d x)\right )}{3 d} \]

[In]

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

[Out]

(-2*a*Cot[c + d*x])/(3*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) - (a*Csc[c + d*x]^3*Hypergeometric2F1[-3/2,
1, -1/2, Sin[c + d*x]^2])/(3*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {a \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\)	\(63\)
default	\(\frac {a \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\)	\(63\)
parallelrisch	\(-\frac {a \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}\)	\(69\)
norman	\(\frac {-\frac {a}{12 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(90\)
risch	\(-\frac {2 i a \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+2\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(107\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.57 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} - 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, a \cos \left (d x + c\right ) - 8 \, a}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \csc ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {12 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{12 \, d} \]

[In]

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

[Out]

1/12*(12*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 3*a*tan(1/2*d*x + 1/
2*c) - (12*a*tan(1/2*d*x + 1/2*c)^2 + a)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 14.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{12}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \]

[In]

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

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (a/12 + a*tan(c/2 + (d*x)/2)^2)/(d*tan(c/2 + (d*x)/2)^3) - (a*tan(c/2 + (d
*x)/2))/(4*d)