3.20 \(\int \csc ^5(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=98 \[ -\frac{3 a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(-3*a*ArcTanh[Cos[c + d*x]])/(8*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Csc[c + d*x])/d - (3*a*Cot[c + d*x]*Csc[
c + d*x])/(8*d) - (b*Csc[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.0922901, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3517, 3768, 3770, 2621, 302, 207} \[ -\frac{3 a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*ArcTanh[Cos[c + d*x]])/(8*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Csc[c + d*x])/d - (3*a*Cot[c + d*x]*Csc[
c + d*x])/(8*d) - (b*Csc[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

Rule 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

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]

Rule 2621

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 302

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 207

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

Rubi steps

\begin{align*} \int \csc ^5(c+d x) (a+b \tan (c+d x)) \, dx &=\int \left (a \csc ^5(c+d x)+b \csc ^4(c+d x) \sec (c+d x)\right ) \, dx\\ &=a \int \csc ^5(c+d x) \, dx+b \int \csc ^4(c+d x) \sec (c+d x) \, dx\\ &=-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{4} (3 a) \int \csc ^3(c+d x) \, dx-\frac{b \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{8} (3 a) \int \csc (c+d x) \, dx-\frac{b \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{b \csc (c+d x)}{d}-\frac{3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b \csc (c+d x)}{d}-\frac{3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{a \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.0286861, size = 151, normalized size = 1.54 \[ -\frac{a \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{3 a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{a \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{3 a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{3 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{3 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{b \csc ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\sin ^2(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) - (b*Csc[c + d*x]^3*Hypergeometric2F1[-3/2, 1
, -1/2, Sin[c + d*x]^2])/(3*d) - (3*a*Log[Cos[(c + d*x)/2]])/(8*d) + (3*a*Log[Sin[(c + d*x)/2]])/(8*d) + (3*a*
Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/2]^4)/(64*d)

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Maple [A]  time = 0.084, size = 109, normalized size = 1.1 \begin{align*} -{\frac{b}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{b}{d\sin \left ( dx+c \right ) }}+{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{\cot \left ( dx+c \right ) a \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,\cot \left ( dx+c \right ) a\csc \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+b*tan(d*x+c)),x)

[Out]

-1/3/d*b/sin(d*x+c)^3-1/d*b/sin(d*x+c)+1/d*b*ln(sec(d*x+c)+tan(d*x+c))-1/4*a*cot(d*x+c)*csc(d*x+c)^3/d-3/8*a*c
ot(d*x+c)*csc(d*x+c)/d+3/8/d*a*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.26038, size = 166, normalized size = 1.69 \begin{align*} \frac{3 \, a{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, b{\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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*a*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x + c)
+ 1) + 3*log(cos(d*x + c) - 1)) - 8*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*l
og(sin(d*x + c) - 1)))/d

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Fricas [B]  time = 2.72285, size = 585, normalized size = 5.97 \begin{align*} \frac{18 \, a \cos \left (d x + c\right )^{3} - 30 \, a \cos \left (d x + c\right ) - 9 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 24 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 24 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \,{\left (3 \, b \cos \left (d x + c\right )^{2} - 4 \, b\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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)^5*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(18*a*cos(d*x + c)^3 - 30*a*cos(d*x + c) - 9*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*cos(d*x
+ c) + 1/2) + 9*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(-1/2*cos(d*x + c) + 1/2) + 24*(b*cos(d*x + c)^
4 - 2*b*cos(d*x + c)^2 + b)*log(sin(d*x + c) + 1) - 24*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-sin(d*
x + c) + 1) + 16*(3*b*cos(d*x + c)^2 - 4*b)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.42746, size = 239, normalized size = 2.44 \begin{align*} \frac{3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 192 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 192 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 72 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 120 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{150 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 120 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a*tan(1/2*d*x + 1/2*c)^4 - 8*b*tan(1/2*d*x + 1/2*c)^3 + 24*a*tan(1/2*d*x + 1/2*c)^2 + 192*b*log(abs(t
an(1/2*d*x + 1/2*c) + 1)) - 192*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 72*a*log(abs(tan(1/2*d*x + 1/2*c))) - 1
20*b*tan(1/2*d*x + 1/2*c) - (150*a*tan(1/2*d*x + 1/2*c)^4 + 120*b*tan(1/2*d*x + 1/2*c)^3 + 24*a*tan(1/2*d*x +
1/2*c)^2 + 8*b*tan(1/2*d*x + 1/2*c) + 3*a)/tan(1/2*d*x + 1/2*c)^4)/d