3.28 \(\int \csc ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.117606, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3517, 3768, 3770, 2621, 321, 207, 2622} \[ -\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rubi steps

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

Mathematica [B]  time = 1.95109, size = 250, normalized size = 2.63 \[ \frac{4 \left (a^2+2 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 \left (a^2+2 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-a^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )-8 a b \tan \left (\frac{1}{2} (c+d x)\right )-8 a b \cot \left (\frac{1}{2} (c+d x)\right )-16 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+16 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 b^2 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{8 b^2 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+8 b^2}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 120, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{ab}{d\sin \left ( dx+c \right ) }}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \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))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.10054, size = 165, normalized size = 1.74 \begin{align*} \frac{a^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, b^{2}{\left (\frac{2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 4 \, a b{\left (\frac{2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{4 \, 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))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.77791, size = 581, normalized size = 6.12 \begin{align*} \frac{8 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, b^{2} -{\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\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 )^{3} -{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 4 \,{\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \,{\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\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))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \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*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*csc(c + d*x)**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.63598, size = 232, normalized size = 2.44 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 16 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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 ) - \frac{16 \, b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \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} + 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\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))^2,x, algorithm="giac")

[Out]

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 16*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 16*a*b*log(abs(tan(1/2*d*x + 1/2
*c) - 1)) - 8*a*b*tan(1/2*d*x + 1/2*c) + 4*(a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - 16*b^2/(tan(1/2*d*x
+ 1/2*c)^2 - 1) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) +
a^2)/tan(1/2*d*x + 1/2*c)^2)/d