3.16 \(\int \frac{\csc ^5(x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\cos (x))}{8 (a+b)^3}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}-\frac{(3 a+7 b) \cot (x) \csc (x)}{8 (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.137888, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3190, 414, 527, 522, 206, 205} \[ -\frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\cos (x))}{8 (a+b)^3}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}-\frac{(3 a+7 b) \cot (x) \csc (x)}{8 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(a + b*Cos[x]^2),x]

[Out]

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

Rule 3190

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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]

Rubi steps

\begin{align*} \int \frac{\csc ^5(x)}{a+b \cos ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\cos (x)\right )\\ &=-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}-\frac{\operatorname{Subst}\left (\int \frac{3 a+4 b+3 b x^2}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\cos (x)\right )}{4 (a+b)}\\ &=-\frac{(3 a+7 b) \cot (x) \csc (x)}{8 (a+b)^2}-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}-\frac{\operatorname{Subst}\left (\int \frac{3 a^2+7 a b+8 b^2+b (3 a+7 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\cos (x)\right )}{8 (a+b)^2}\\ &=-\frac{(3 a+7 b) \cot (x) \csc (x)}{8 (a+b)^2}-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\cos (x)\right )}{(a+b)^3}-\frac{\left (3 a^2+10 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (x)\right )}{8 (a+b)^3}\\ &=-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}-\frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\cos (x))}{8 (a+b)^3}-\frac{(3 a+7 b) \cot (x) \csc (x)}{8 (a+b)^2}-\frac{\cot (x) \csc ^3(x)}{4 (a+b)}\\ \end{align*}

Mathematica [B]  time = 1.23215, size = 204, normalized size = 2.17 \[ \frac{\sqrt{a} \left (-2 \left (3 a^2+10 a b+7 b^2\right ) \csc ^2\left (\frac{x}{2}\right )+2 \left (3 a^2+10 a b+7 b^2\right ) \sec ^2\left (\frac{x}{2}\right )-8 \left (3 a^2+10 a b+15 b^2\right ) \left (\log \left (\cos \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )\right )\right )-(a+b)^2 \csc ^4\left (\frac{x}{2}\right )+(a+b)^2 \sec ^4\left (\frac{x}{2}\right )\right )-64 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b}-\sqrt{a+b} \tan \left (\frac{x}{2}\right )}{\sqrt{a}}\right )-64 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{x}{2}\right )+\sqrt{b}}{\sqrt{a}}\right )}{64 \sqrt{a} (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(a + b*Cos[x]^2),x]

[Out]

(-64*b^(5/2)*ArcTan[(Sqrt[b] - Sqrt[a + b]*Tan[x/2])/Sqrt[a]] - 64*b^(5/2)*ArcTan[(Sqrt[b] + Sqrt[a + b]*Tan[x
/2])/Sqrt[a]] + Sqrt[a]*(-2*(3*a^2 + 10*a*b + 7*b^2)*Csc[x/2]^2 - (a + b)^2*Csc[x/2]^4 - 8*(3*a^2 + 10*a*b + 1
5*b^2)*(Log[Cos[x/2]] - Log[Sin[x/2]]) + 2*(3*a^2 + 10*a*b + 7*b^2)*Sec[x/2]^2 + (a + b)^2*Sec[x/2]^4))/(64*Sq
rt[a]*(a + b)^3)

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Maple [B]  time = 0.049, size = 205, normalized size = 2.2 \begin{align*}{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( 1+\cos \left ( x \right ) \right ) ^{2}}}+{\frac{3\,a}{16\, \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( x \right ) \right ) }}+{\frac{7\,b}{16\, \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( x \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\cos \left ( x \right ) \right ){a}^{2}}{16\, \left ( a+b \right ) ^{3}}}-{\frac{5\,\ln \left ( 1+\cos \left ( x \right ) \right ) ab}{8\, \left ( a+b \right ) ^{3}}}-{\frac{15\,\ln \left ( 1+\cos \left ( x \right ) \right ){b}^{2}}{16\, \left ( a+b \right ) ^{3}}}-{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \cos \left ( x \right ) -1 \right ) ^{2}}}+{\frac{3\,a}{16\, \left ( a+b \right ) ^{2} \left ( \cos \left ( x \right ) -1 \right ) }}+{\frac{7\,b}{16\, \left ( a+b \right ) ^{2} \left ( \cos \left ( x \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \cos \left ( x \right ) -1 \right ){a}^{2}}{16\, \left ( a+b \right ) ^{3}}}+{\frac{5\,\ln \left ( \cos \left ( x \right ) -1 \right ) ab}{8\, \left ( a+b \right ) ^{3}}}+{\frac{15\,\ln \left ( \cos \left ( x \right ) -1 \right ){b}^{2}}{16\, \left ( a+b \right ) ^{3}}}-{\frac{{b}^{3}}{ \left ( a+b \right ) ^{3}}\arctan \left ({b\cos \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^5/(a+b*cos(x)^2),x)

[Out]

1/2/(8*a+8*b)/(1+cos(x))^2+3/16/(a+b)^2/(1+cos(x))*a+7/16/(a+b)^2/(1+cos(x))*b-3/16/(a+b)^3*ln(1+cos(x))*a^2-5
/8/(a+b)^3*ln(1+cos(x))*a*b-15/16/(a+b)^3*ln(1+cos(x))*b^2-1/2/(8*a+8*b)/(cos(x)-1)^2+3/16/(a+b)^2/(cos(x)-1)*
a+7/16/(a+b)^2/(cos(x)-1)*b+3/16/(a+b)^3*ln(cos(x)-1)*a^2+5/8/(a+b)^3*ln(cos(x)-1)*a*b+15/16/(a+b)^3*ln(cos(x)
-1)*b^2-b^3/(a+b)^3/(a*b)^(1/2)*arctan(b*cos(x)/(a*b)^(1/2))

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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(x)^5/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.49897, size = 1490, normalized size = 15.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 + 10*a*b + 7*b^2)*cos(x)^3 + 8*(b^2*cos(x)^4 - 2*b^2*cos(x)^2 + b^2)*sqrt(-b/a)*log((b*cos(x)^
2 - 2*a*sqrt(-b/a)*cos(x) - a)/(b*cos(x)^2 + a)) - 2*(5*a^2 + 14*a*b + 9*b^2)*cos(x) - ((3*a^2 + 10*a*b + 15*b
^2)*cos(x)^4 - 2*(3*a^2 + 10*a*b + 15*b^2)*cos(x)^2 + 3*a^2 + 10*a*b + 15*b^2)*log(1/2*cos(x) + 1/2) + ((3*a^2
+ 10*a*b + 15*b^2)*cos(x)^4 - 2*(3*a^2 + 10*a*b + 15*b^2)*cos(x)^2 + 3*a^2 + 10*a*b + 15*b^2)*log(-1/2*cos(x)
+ 1/2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^
2 + b^3)*cos(x)^2), 1/16*(2*(3*a^2 + 10*a*b + 7*b^2)*cos(x)^3 - 16*(b^2*cos(x)^4 - 2*b^2*cos(x)^2 + b^2)*sqrt(
b/a)*arctan(sqrt(b/a)*cos(x)) - 2*(5*a^2 + 14*a*b + 9*b^2)*cos(x) - ((3*a^2 + 10*a*b + 15*b^2)*cos(x)^4 - 2*(3
*a^2 + 10*a*b + 15*b^2)*cos(x)^2 + 3*a^2 + 10*a*b + 15*b^2)*log(1/2*cos(x) + 1/2) + ((3*a^2 + 10*a*b + 15*b^2)
*cos(x)^4 - 2*(3*a^2 + 10*a*b + 15*b^2)*cos(x)^2 + 3*a^2 + 10*a*b + 15*b^2)*log(-1/2*cos(x) + 1/2))/((a^3 + 3*
a^2*b + 3*a*b^2 + b^3)*cos(x)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^2)]

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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(x)**5/(a+b*cos(x)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.15854, size = 240, normalized size = 2.55 \begin{align*} -\frac{b^{3} \arctan \left (\frac{b \cos \left (x\right )}{\sqrt{a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b}} - \frac{{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\cos \left (x\right ) + 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac{{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (-\cos \left (x\right ) + 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac{3 \, a \cos \left (x\right )^{3} + 7 \, b \cos \left (x\right )^{3} - 5 \, a \cos \left (x\right ) - 9 \, b \cos \left (x\right )}{8 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\cos \left (x\right )^{2} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+b*cos(x)^2),x, algorithm="giac")

[Out]

-b^3*arctan(b*cos(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*l
og(cos(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(-cos(x) + 1)/(a^3 + 3*a^2*
b + 3*a*b^2 + b^3) + 1/8*(3*a*cos(x)^3 + 7*b*cos(x)^3 - 5*a*cos(x) - 9*b*cos(x))/((a^2 + 2*a*b + b^2)*(cos(x)^
2 - 1)^2)