3.207 \(\int \frac{\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}+\frac{\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}-\frac{2 a^3 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

(-2*a^3*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) + ((3*a^2*b - a
*(2*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x])/(3*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c + d*x]^3)/(3*(a^2
- b^2)*d)

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Rubi [A]  time = 0.306471, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2866, 12, 2659, 208} \[ \frac{\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}+\frac{\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}-\frac{2 a^3 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

(-2*a^3*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) + ((3*a^2*b - a
*(2*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x])/(3*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c + d*x]^3)/(3*(a^2
- b^2)*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 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2659

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.891265, size = 162, normalized size = 1.16 \[ \frac{\sqrt{a^2-b^2} \csc ^3(c+d x) \left (\left (3 a b^2-6 a^3\right ) \cos (c+d x)-6 a^2 b \cos (2 (c+d x))+10 a^2 b+2 a^3 \cos (3 (c+d x))+a b^2 \cos (3 (c+d x))-4 b^3\right )+24 a^3 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{12 d (a-b)^2 (a+b)^2 \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

(24*a^3*b*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]*(10*a^2*b - 4*b^3 + (-6*a^3 +
3*a*b^2)*Cos[c + d*x] - 6*a^2*b*Cos[2*(c + d*x)] + 2*a^3*Cos[3*(c + d*x)] + a*b^2*Cos[3*(c + d*x)])*Csc[c + d
*x]^3)/(12*(a - b)^2*(a + b)^2*Sqrt[a^2 - b^2]*d)

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Maple [A]  time = 0.069, size = 165, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{1}{8\, \left ( a-b \right ) ^{2}} \left ({\frac{a}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+3\,a\tan \left ( 1/2\,dx+c/2 \right ) -b\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{3}b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{24\,a+24\,b} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{3\,a+b}{8\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+b*sec(d*x+c)),x)

[Out]

1/d*(1/8/(a-b)^2*(1/3*tan(1/2*d*x+1/2*c)^3*a-1/3*b*tan(1/2*d*x+1/2*c)^3+3*a*tan(1/2*d*x+1/2*c)-b*tan(1/2*d*x+1
/2*c))-2/(a-b)^2/(a+b)^2*a^3*b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/24/
(a+b)/tan(1/2*d*x+1/2*c)^3-1/8*(3*a+b)/(a+b)^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)^4/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92126, size = 1214, normalized size = 8.67 \begin{align*} \left [-\frac{8 \, a^{4} b - 10 \, a^{2} b^{3} + 2 \, b^{5} + 2 \,{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{6 \,{\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}, -\frac{4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} +{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{3 \,{\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[-1/6*(8*a^4*b - 10*a^2*b^3 + 2*b^5 + 2*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 3*(a^3*b*cos(d*x + c)^2 - a
^3*b)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x +
c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 6*(a^4*b -
a^2*b^3)*cos(d*x + c)^2 - 6*(a^5 - a^3*b^2)*cos(d*x + c))/(((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c
)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)*sin(d*x + c)), -1/3*(4*a^4*b - 5*a^2*b^3 + b^5 + (2*a^5 - a^3*b^2
- a*b^4)*cos(d*x + c)^3 + 3*(a^3*b*cos(d*x + c)^2 - a^3*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d
*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 3*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 3*(a^5 - a^3*b^2)
*cos(d*x + c))/(((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)
*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+b*sec(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**4/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 1.40393, size = 363, normalized size = 2.59 \begin{align*} \frac{\frac{48 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(48*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1
/2*c))/sqrt(-a^2 + b^2)))*a^3*b/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + (a^2*tan(1/2*d*x + 1/2*c)^3 - 2*a
*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*tan(1/2*d*x + 1/2*c) - 12*a*b*tan(1/2*d*x + 1/2
*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (9*a*tan(1/2*d*x + 1/2*c)^2 + 3*b*tan(1/2*
d*x + 1/2*c)^2 + a + b)/((a^2 + 2*a*b + b^2)*tan(1/2*d*x + 1/2*c)^3))/d