3.33 \(\int \frac{\csc ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=129 \[ -\frac{\left (3 a^2-6 a b-b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^3}+\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{f (a+b)^3}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b)}-\frac{(3 a-b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2} \]

[Out]

(a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/((a + b)^3*f) - ((3*a^2 - 6*a*b - b^2)*ArcTanh[Cos[e
+ f*x]])/(8*(a + b)^3*f) - ((3*a - b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f) - (Cot[e + f*x]*Csc[e + f*x]^
3)/(4*(a + b)*f)

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Rubi [A]  time = 0.153726, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4133, 471, 527, 522, 206, 205} \[ -\frac{\left (3 a^2-6 a b-b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^3}+\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{f (a+b)^3}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b)}-\frac{(3 a-b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]

[Out]

(a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/((a + b)^3*f) - ((3*a^2 - 6*a*b - b^2)*ArcTanh[Cos[e
+ f*x]])/(8*(a + b)^3*f) - ((3*a - b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f) - (Cot[e + f*x]*Csc[e + f*x]^
3)/(4*(a + b)*f)

Rule 4133

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

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, 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(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1-x^2\right )^3 \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{b-3 a x^2}{\left (1-x^2\right )^2 \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(3 a-b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{b (5 a+b)-a (3 a-b) x^2}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{8 (a+b)^2 f}\\ &=-\frac{(3 a-b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f}+\frac{\left (a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{(a+b)^3 f}-\frac{\left (3 a^2-6 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{8 (a+b)^3 f}\\ &=\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{(a+b)^3 f}-\frac{\left (3 a^2-6 a b-b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 (a+b)^3 f}-\frac{(3 a-b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 5.59755, size = 549, normalized size = 4.26 \[ -\frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-64 a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )-64 a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+a^2 \csc ^4\left (\frac{1}{2} (e+f x)\right )+6 a^2 \csc ^2\left (\frac{1}{2} (e+f x)\right )-a^2 \sec ^4\left (\frac{1}{2} (e+f x)\right )-6 a^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )-24 a^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )+24 a^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+2 a b \csc ^4\left (\frac{1}{2} (e+f x)\right )+4 a b \csc ^2\left (\frac{1}{2} (e+f x)\right )-2 a b \sec ^4\left (\frac{1}{2} (e+f x)\right )-4 a b \sec ^2\left (\frac{1}{2} (e+f x)\right )+48 a b \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-48 a b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+b^2 \csc ^4\left (\frac{1}{2} (e+f x)\right )-2 b^2 \csc ^2\left (\frac{1}{2} (e+f x)\right )-b^2 \sec ^4\left (\frac{1}{2} (e+f x)\right )+2 b^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )+8 b^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-8 b^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{128 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(-64*a^(3/2)*Sqrt[b]*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e
])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]
- 64*a^(3/2)*Sqrt[b]*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos
[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] + 6*a^2*Csc[(e + f*x)/2]^2 + 4*
a*b*Csc[(e + f*x)/2]^2 - 2*b^2*Csc[(e + f*x)/2]^2 + a^2*Csc[(e + f*x)/2]^4 + 2*a*b*Csc[(e + f*x)/2]^4 + b^2*Cs
c[(e + f*x)/2]^4 + 24*a^2*Log[Cos[(e + f*x)/2]] - 48*a*b*Log[Cos[(e + f*x)/2]] - 8*b^2*Log[Cos[(e + f*x)/2]] -
24*a^2*Log[Sin[(e + f*x)/2]] + 48*a*b*Log[Sin[(e + f*x)/2]] + 8*b^2*Log[Sin[(e + f*x)/2]] - 6*a^2*Sec[(e + f*
x)/2]^2 - 4*a*b*Sec[(e + f*x)/2]^2 + 2*b^2*Sec[(e + f*x)/2]^2 - a^2*Sec[(e + f*x)/2]^4 - 2*a*b*Sec[(e + f*x)/2
]^4 - b^2*Sec[(e + f*x)/2]^4)*Sec[e + f*x]^2)/(128*(a + b)^3*f*(a + b*Sec[e + f*x]^2))

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Maple [B]  time = 0.084, size = 296, normalized size = 2.3 \begin{align*}{\frac{1}{2\,f \left ( 8\,a+8\,b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{16\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{b}{16\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ){a}^{2}}{16\,f \left ( a+b \right ) ^{3}}}+{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) ab}{8\,f \left ( a+b \right ) ^{3}}}+{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ){b}^{2}}{16\,f \left ( a+b \right ) ^{3}}}+{\frac{{a}^{2}b}{f \left ( a+b \right ) ^{3}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{2\,f \left ( 8\,a+8\,b \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{16\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{b}{16\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }}+{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ){a}^{2}}{16\,f \left ( a+b \right ) ^{3}}}-{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) ab}{8\,f \left ( a+b \right ) ^{3}}}-{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ){b}^{2}}{16\,f \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*sec(f*x+e)^2),x)

[Out]

1/2/f/(8*a+8*b)/(1+cos(f*x+e))^2+3/16/f/(a+b)^2/(1+cos(f*x+e))*a-1/16/f/(a+b)^2/(1+cos(f*x+e))*b-3/16/f/(a+b)^
3*ln(1+cos(f*x+e))*a^2+3/8/f/(a+b)^3*ln(1+cos(f*x+e))*a*b+1/16/f/(a+b)^3*ln(1+cos(f*x+e))*b^2+1/f*a^2*b/(a+b)^
3/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))-1/2/f/(8*a+8*b)/(-1+cos(f*x+e))^2+3/16/f/(a+b)^2/(-1+cos(f*x+e)
)*a-1/16/f/(a+b)^2/(-1+cos(f*x+e))*b+3/16/f/(a+b)^3*ln(-1+cos(f*x+e))*a^2-3/8/f/(a+b)^3*ln(-1+cos(f*x+e))*a*b-
1/16/f/(a+b)^3*ln(-1+cos(f*x+e))*b^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(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.799531, size = 1643, normalized size = 12.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^3 + 8*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^2 + a)*sqrt(-a*b)*log(-
(a*cos(f*x + e)^2 + 2*sqrt(-a*b)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) - 2*(5*a^2 + 6*a*b + b^2)*cos(f*x +
e) - ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 - 6*a*b - b^2)*lo
g(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x + e)^2 + 3
*a^2 - 6*a*b - b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 +
3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f), 1/16*(2*(3*a^2 + 2*a*b - b^2)
*cos(f*x + e)^3 + 16*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^2 + a)*sqrt(a*b)*arctan(sqrt(a*b)*cos(f*x + e)/b) -
2*(5*a^2 + 6*a*b + b^2)*cos(f*x + e) - ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x
+ e)^2 + 3*a^2 - 6*a*b - b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2
- 6*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 - 6*a*b - b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
f)]

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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(f*x+e)**5/(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.28211, size = 551, normalized size = 4.27 \begin{align*} -\frac{\frac{64 \, a^{2} b \arctan \left (-\frac{a \cos \left (f x + e\right ) - b}{\sqrt{a b} \cos \left (f x + e\right ) + \sqrt{a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b}} - \frac{4 \,{\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{\frac{8 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{a^{2} + 2 \, a b + b^{2}} + \frac{{\left (a^{2} + 2 \, a b + b^{2} - \frac{8 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{8 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{18 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{36 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{6 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(64*a^2*b*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/((a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*sqrt(a*b)) - 4*(3*a^2 - 6*a*b - b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^3 + 3*a^2*b + 3*a*b^
2 + b^3) + (8*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - b*(cos(f
*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^2 + 2*a*b + b^2) + (a^2 + 2*a*b + b^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(
f*x + e) + 1) - 8*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 18*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2
- 36*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 6*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*
x + e) + 1)^2/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(f*x + e) - 1)^2))/f