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

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

[Out]

-2*a^3*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d+1/3*(3*a^2*b-a*(2*a^2+b
^2)*cos(d*x+c))*csc(d*x+c)/(a^2-b^2)^2/d+1/3*(b-a*cos(d*x+c))*csc(d*x+c)^3/(a^2-b^2)/d

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Rubi [A]  time = 0.31, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 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]

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*}

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Mathematica [A]  time = 0.92, size = 162, normalized size = 1.16 \[ \frac {24 a^3 b \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} \csc ^3(c+d x) \left (\left (3 a b^2-6 a^3\right ) \cos (c+d x)+2 a^3 \cos (3 (c+d x))-6 a^2 b \cos (2 (c+d x))+10 a^2 b+a b^2 \cos (3 (c+d x))-4 b^3\right )}{12 d (a-b)^2 (a+b)^2 \sqrt {a^2-b^2}} \]

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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fricas [A]  time = 0.85, size = 558, normalized size = 3.99 \[ \left [-\frac {8 \, a^{4} b - 10 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{3 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

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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giac [B]  time = 0.27, size = 269, normalized size = 1.92 \[ \frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

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

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maple [A]  time = 0.54, size = 165, normalized size = 1.18 \[ \frac {\frac {\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{3}+3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a +b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

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*a*tan(1/2*d*x+1/2*c)^3-1/3*tan(1/2*d*x+1/2*c)^3*b+3*a*tan(1/2*d*x+1/2*c)-tan(1/2*d*x+1/2
*c)*b)-2/(a-b)^2/(a+b)^2*a^3*b/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.38, size = 219, normalized size = 1.56 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2}{8\,a-8\,b}+\frac {8\,a+8\,b}{{\left (8\,a-8\,b\right )}^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\left (8\,a-8\,b\right )}-\frac {\frac {a^2-2\,a\,b+b^2}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3-5\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^2-16\,a\,b+8\,b^2\right )}-\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^4*(a + b/cos(c + d*x))),x)

[Out]

(tan(c/2 + (d*x)/2)*(2/(8*a - 8*b) + (8*a + 8*b)/(8*a - 8*b)^2))/d + tan(c/2 + (d*x)/2)^3/(3*d*(8*a - 8*b)) -
((a^2 - 2*a*b + b^2)/(3*(a + b)) + (tan(c/2 + (d*x)/2)^2*(a*b^2 - 5*a^2*b + 3*a^3 + b^3))/(a + b)^2)/(d*tan(c/
2 + (d*x)/2)^3*(8*a^2 - 16*a*b + 8*b^2)) - (2*a^3*b*atanh((tan(c/2 + (d*x)/2)*(a^4 + b^4 - 2*a^2*b^2))/((a + b
)^(5/2)*(a - b)^(3/2))))/(d*(a + b)^(5/2)*(a - b)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

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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