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

Optimal. Leaf size=64 \[ -\frac{a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d}-\frac{b \cot ^2(c+d x)}{2 d}+\frac{b \log (\tan (c+d x))}{d} \]

[Out]

-(a*ArcTanh[Cos[c + d*x]])/(2*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

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Rubi [A]  time = 0.103521, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3872, 2834, 2620, 14, 3768, 3770} \[ -\frac{a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d}-\frac{b \cot ^2(c+d x)}{2 d}+\frac{b \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*ArcTanh[Cos[c + d*x]])/(2*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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2834

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 2620

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 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 3768

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \csc ^3(c+d x) (a+b \sec (c+d x)) \, dx &=-\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 \operatorname{Subst}\left (\int \frac{1+x^2}{x^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d}+\frac{b \operatorname{Subst}\left (\int \left (\frac{1}{x^3}+\frac{1}{x}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{a \tanh ^{-1}(\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]  time = 0.52254, size = 114, normalized size = 1.78 \[ -\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{b \left (\csc ^2(c+d x)-2 \log (\sin (c+d x))+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.094, size = 68, normalized size = 1.1 \begin{align*} -{\frac{a\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{2\,d}}+{\frac{a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{b}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972545, size = 96, normalized size = 1.5 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 1.82602, size = 316, normalized size = 4.94 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right ) \csc ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.357, size = 228, normalized size = 3.56 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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