3.92 \(\int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=109 \[ -\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d \sqrt {a+b}}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^5(c+d x)}{5 a d} \]

[Out]

-(a^2-a*b+b^2)*cot(d*x+c)/a^3/d-1/3*(2*a-b)*cot(d*x+c)^3/a^2/d-1/5*cot(d*x+c)^5/a/d-b^3*arctan((a+b)^(1/2)*tan
(d*x+c)/a^(1/2))/a^(7/2)/d/(a+b)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3187, 461, 205} \[ -\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d \sqrt {a+b}}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

-((b^3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(7/2)*Sqrt[a + b]*d)) - ((a^2 - a*b + b^2)*Cot[c + d*x])
/(a^3*d) - ((2*a - b)*Cot[c + d*x]^3)/(3*a^2*d) - Cot[c + d*x]^5/(5*a*d)

Rule 205

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^6 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a x^6}+\frac {2 a-b}{a^2 x^4}+\frac {a^2-a b+b^2}{a^3 x^2}+\frac {b^3}{a^3 \left (-a-(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{-a-(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=-\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b} d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 147, normalized size = 1.35 \[ -\frac {\csc ^2(c+d x) (2 a-b \cos (2 (c+d x))+b) \left (\sqrt {a} \sqrt {a+b} \cot (c+d x) \left (3 a^2 \csc ^4(c+d x)+8 a^2+a (4 a-5 b) \csc ^2(c+d x)-10 a b+15 b^2\right )+15 b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )\right )}{30 a^{7/2} d \sqrt {a+b} \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

-1/30*((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(15*b^3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt
[a]*Sqrt[a + b]*Cot[c + d*x]*(8*a^2 - 10*a*b + 15*b^2 + a*(4*a - 5*b)*Csc[c + d*x]^2 + 3*a^2*Csc[c + d*x]^4)))
/(a^(7/2)*Sqrt[a + b]*d*(b + a*Csc[c + d*x]^2))

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fricas [B]  time = 0.46, size = 595, normalized size = 5.46 \[ \left [-\frac {4 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{60 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 30 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{30 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\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)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/60*(4*(8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(d*x + c)^5 - 20*(4*a^4 - a^3*b + a^2*b^2 + 6*a*b^3)*cos
(d*x + c)^3 + 15*(b^3*cos(d*x + c)^4 - 2*b^3*cos(d*x + c)^2 + b^3)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)
*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))
*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 +
2*a*b + b^2))*sin(d*x + c) + 60*(a^4 + a*b^3)*cos(d*x + c))/(((a^5 + a^4*b)*d*cos(d*x + c)^4 - 2*(a^5 + a^4*b
)*d*cos(d*x + c)^2 + (a^5 + a^4*b)*d)*sin(d*x + c)), -1/30*(2*(8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(d*x
+ c)^5 - 10*(4*a^4 - a^3*b + a^2*b^2 + 6*a*b^3)*cos(d*x + c)^3 - 15*(b^3*cos(d*x + c)^4 - 2*b^3*cos(d*x + c)^
2 + b^3)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x +
c)))*sin(d*x + c) + 30*(a^4 + a*b^3)*cos(d*x + c))/(((a^5 + a^4*b)*d*cos(d*x + c)^4 - 2*(a^5 + a^4*b)*d*cos(d
*x + c)^2 + (a^5 + a^4*b)*d)*sin(d*x + c))]

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giac [A]  time = 0.17, size = 155, normalized size = 1.42 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{3}}{\sqrt {a^{2} + a b} a^{3}} + \frac {15 \, a^{2} \tan \left (d x + c\right )^{4} - 15 \, a b \tan \left (d x + c\right )^{4} + 15 \, b^{2} \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{2} - 5 \, a b \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/15*(15*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a
*b)))*b^3/(sqrt(a^2 + a*b)*a^3) + (15*a^2*tan(d*x + c)^4 - 15*a*b*tan(d*x + c)^4 + 15*b^2*tan(d*x + c)^4 + 10*
a^2*tan(d*x + c)^2 - 5*a*b*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d

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maple [A]  time = 0.58, size = 138, normalized size = 1.27 \[ -\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \,a^{3} \sqrt {a \left (a +b \right )}}-\frac {1}{5 d a \tan \left (d x +c \right )^{5}}-\frac {2}{3 d a \tan \left (d x +c \right )^{3}}+\frac {b}{3 d \,a^{2} \tan \left (d x +c \right )^{3}}-\frac {1}{d a \tan \left (d x +c \right )}+\frac {b}{d \,a^{2} \tan \left (d x +c \right )}-\frac {b^{2}}{d \,a^{3} \tan \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6/(a+b*sin(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.46, size = 98, normalized size = 0.90 \[ -\frac {\frac {15 \, b^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {15 \, {\left (a^{2} - a b + b^{2}\right )} \tan \left (d x + c\right )^{4} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/15*(15*b^3*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a^3) + (15*(a^2 - a*b + b^2)*tan(d
*x + c)^4 + 5*(2*a^2 - a*b)*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d

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mupad [B]  time = 13.80, size = 95, normalized size = 0.87 \[ -\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2-a\,b+b^2\right )+\frac {a^2}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a\,b}{3}-\frac {2\,a^2}{3}\right )}{a^3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^5}-\frac {b^3\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{7/2}\,d\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^6*(a + b*sin(c + d*x)^2)),x)

[Out]

- (tan(c + d*x)^4*(a^2 - a*b + b^2) + a^2/5 - tan(c + d*x)^2*((a*b)/3 - (2*a^2)/3))/(a^3*d*tan(c + d*x)^5) - (
b^3*atan((tan(c + d*x)*(a + b)^(1/2))/a^(1/2)))/(a^(7/2)*d*(a + b)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**6/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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