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

Optimal. Leaf size=133 \[ -\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{3 a b^2 \cot (c+d x)}{d}-\frac{3 b^3 \csc (c+d x)}{2 d}+\frac{3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.272508, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {3872, 2912, 3767, 8, 2621, 321, 207, 2620, 14, 288} \[ -\frac{3 a^2 b \csc (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}-\frac{3 a b^2 \cot (c+d x)}{d}-\frac{3 b^3 \csc (c+d x)}{2 d}+\frac{3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (3*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (3*a*b^2*Cot[
c + d*x])/d - (3*a^2*b*Csc[c + d*x])/d - (3*b^3*Csc[c + d*x])/(2*d) + (b^3*Csc[c + d*x]*Sec[c + d*x]^2)/(2*d)
+ (3*a*b^2*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 2912

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

Mathematica [B]  time = 0.640034, size = 406, normalized size = 3.05 \[ -\frac{\csc ^5\left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (6 a \left (a^2+2 b^2\right ) \cos (c+d x)+6 \left (2 a^2 b+b^3\right ) \cos (2 (c+d x))+6 a^2 b \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 a^2 b \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+6 a^2 b \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 a^2 b \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 b+2 a^3 \cos (3 (c+d x))+12 a b^2 \cos (3 (c+d x))+3 b^3 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 b^3 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 b^3 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 b^3 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 b^3\right )}{16 d \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]*(12*a^2*b + 2*b^3 + 6*a*(a^2 + 2*b^2)*Cos[c + d*x] + 6*(2*a^2*b + b^3)*C
os[2*(c + d*x)] + 2*a^3*Cos[3*(c + d*x)] + 12*a*b^2*Cos[3*(c + d*x)] + 6*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]]*Sin[c + d*x] + 3*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 6*a^2*b*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 3*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 6*a^2*b
*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 3*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Si
n[3*(c + d*x)] - 6*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 3*b^3*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)]))/(16*d*(-1 + Cot[(c + d*x)/2]^2)^2)

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Maple [A]  time = 0.042, size = 158, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{2}b}{d\sin \left ( dx+c \right ) }}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-6\,{\frac{a{b}^{2}\cot \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}}{2\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{b}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{3\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \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)^2*(a+b*sec(d*x+c))^3,x)

[Out]

-a^3*cot(d*x+c)/d-3/d*a^2*b/sin(d*x+c)+3/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/d*a*b^2/sin(d*x+c)/cos(d*x+c)-6*a
*b^2*cot(d*x+c)/d+1/2/d*b^3/sin(d*x+c)/cos(d*x+c)^2-3/2/d*b^3/sin(d*x+c)+3/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.963803, size = 188, normalized size = 1.41 \begin{align*} -\frac{b^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} 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 )} + 12 \, a b^{2}{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac{4 \, a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.82114, size = 378, normalized size = 2.84 \begin{align*} \frac{3 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 12 \, a b^{2} \cos \left (d x + c\right ) - 4 \,{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 6 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}}{4 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(3*(2*a^2*b + b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1)*sin(d*x + c) - 3*(2*a^2*b + b^3)*cos(d*x + c)^2*lo
g(-sin(d*x + c) + 1)*sin(d*x + c) + 12*a*b^2*cos(d*x + c) - 4*(a^3 + 6*a*b^2)*cos(d*x + c)^3 + 2*b^3 - 6*(2*a^
2*b + b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^2*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.301, size = 304, normalized size = 2.29 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \,{\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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