\(\int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx\) [165]

Optimal result

Integrand size = 19, antiderivative size = 64 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{2 d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d}+\frac {b \log (\tan (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3957, 2913, 2700, 14, 3853, 3855} \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d}-\frac {b \cot ^2(c+d x)}{2 d}+\frac {b \log (\tan (c+d x))}{d} \]

[In]

Int[Csc[c + d*x]^3*(a + b*Sec[c + d*x]),x]

[Out]

-1/2*(a*ArcTanh[Cos[c + d*x]])/d - (b*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (b*Log[Tan
[c + d*x]])/d

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.88 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {b \csc ^2(c+d x)}{2 d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {b \log (\cos (c+d x))}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {b \log (\sin (c+d x))}{d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d} \]

[In]

Integrate[Csc[c + d*x]^3*(a + b*Sec[c + d*x]),x]

[Out]

-1/8*(a*Csc[(c + d*x)/2]^2)/d - (b*Csc[c + d*x]^2)/(2*d) - (a*Log[Cos[(c + d*x)/2]])/(2*d) - (b*Log[Cos[c + d*
x]])/d + (a*Log[Sin[(c + d*x)/2]])/(2*d) + (b*Log[Sin[c + d*x]])/d + (a*Sec[(c + d*x)/2]^2)/(8*d)

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(61\)
default	\(\frac {a \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(61\)
parallelrisch	\(\frac {-8 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-8 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (4 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a -b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )}{8 d}\)	\(91\)
norman	\(\frac {-\frac {a +b}{8 d}+\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(100\)
risch	\(\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(142\)

[In]

int(csc(d*x+c)^3*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (58) = 116\).

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.92 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {2 \, a \cos \left (d x + c\right ) - 4 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{2} - a - 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, b}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \csc ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**3*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*csc(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 4 \, b \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (a \cos \left (d x + c\right ) + b\right )}}{\cos \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (58) = 116\).

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.64 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {2 \, {\left (a + 2 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 8 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {{\left (a + b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/8*(2*(a + 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 8*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - 1)) + (a + b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) +
1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(co
s(d*x + c) + 1))/d

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.19 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {\frac {\frac {b}{2}+\frac {a\,\cos \left (c+d\,x\right )}{2}}{{\cos \left (c+d\,x\right )}^2-1}+\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {a}{4}+\frac {b}{2}\right )-\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (\frac {a}{4}-\frac {b}{2}\right )-b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

int((a + b/cos(c + d*x))/sin(c + d*x)^3,x)

[Out]

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