3.58 \(\int \frac{\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=122 \[ -\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d
*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(a^3*d) + (b*Csc[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*
d)

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Rubi [A]  time = 0.307234, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {3518, 3110, 3768, 3770, 2621, 321, 207, 2622, 3104, 3074, 206} \[ -\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d
*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(a^3*d) + (b*Csc[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*
d)

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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])

Rule 2622

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 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.826587, size = 179, normalized size = 1.47 \[ \frac{-16 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )+a^2 \left (-\csc ^2\left (\frac{1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )+4 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a b \tan \left (\frac{1}{2} (c+d x)\right )+4 a b \cot \left (\frac{1}{2} (c+d x)\right )+8 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + 4*a*b*Cot[(c + d*x)/2] - a^2*Csc[(
c + d*x)/2]^2 - 4*a^2*Log[Cos[(c + d*x)/2]] - 8*b^2*Log[Cos[(c + d*x)/2]] + 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^
2*Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x)/2]^2 + 4*a*b*Tan[(c + d*x)/2])/(8*a^3*d)

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Maple [A]  time = 0.069, size = 162, normalized size = 1.3 \begin{align*}{\frac{1}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{b}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{b\sqrt{{a}^{2}+{b}^{2}}}{d{a}^{3}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2+1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-2/d/a^3*b*(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2
*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/8/d/a/tan(1/2*d*x+1/2*c)^2+1/2/d/a*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1
/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tan(1/2*d*x+1/2*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.68281, size = 644, normalized size = 5.28 \begin{align*} \frac{2 \, a^{2} \cos \left (d x + c\right ) - 4 \, a b \sin \left (d x + c\right ) + 2 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) -{\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.35144, size = 282, normalized size = 2.31 \begin{align*} \frac{\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{4 \,{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{8 \,{\left (a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3}} - \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*tan(1/2*d*x + 1/2*c)^2 + 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))
)/a^3 + 8*(a^2*b + b^3)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*
c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/
2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d