3.32 \(\int \frac{\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=86 \[ \frac{\sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{f (a+b)^2}-\frac{(a-b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^2}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.099954, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4133, 471, 522, 206, 205} \[ \frac{\sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{f (a+b)^2}-\frac{(a-b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^2}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]

[Out]

(Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/((a + b)^2*f) - ((a - b)*ArcTanh[Cos[e + f*x]])/(2*(a
+ b)^2*f) - (Cot[e + f*x]*Csc[e + f*x])/(2*(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 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 ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1-x^2\right )^2 \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{b-a x^2}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{2 (a+b) f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f}-\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{2 (a+b)^2 f}+\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{(a+b)^2 f}\\ &=\frac{\sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{(a+b)^2 f}-\frac{(a-b) \tanh ^{-1}(\cos (e+f x))}{2 (a+b)^2 f}-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 1.86616, size = 371, normalized size = 4.31 \[ -\frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-8 \sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )-8 \sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+a \csc ^2\left (\frac{1}{2} (e+f x)\right )-a \sec ^2\left (\frac{1}{2} (e+f x)\right )-4 a \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )+4 a \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+b \csc ^2\left (\frac{1}{2} (e+f x)\right )-b \sec ^2\left (\frac{1}{2} (e+f x)\right )+4 b \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{16 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(-8*Sqrt[a]*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]]
- 8*Sqrt[a]*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]] + a*Csc[(e + f*x)/2]^2 + b*Csc[(e
+ f*x)/2]^2 + 4*a*Log[Cos[(e + f*x)/2]] - 4*b*Log[Cos[(e + f*x)/2]] - 4*a*Log[Sin[(e + f*x)/2]] + 4*b*Log[Sin
[(e + f*x)/2]] - a*Sec[(e + f*x)/2]^2 - b*Sec[(e + f*x)/2]^2)*Sec[e + f*x]^2)/(16*(a + b)^2*f*(a + b*Sec[e + f
*x]^2))

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Maple [B]  time = 0.081, size = 158, normalized size = 1.8 \begin{align*}{\frac{1}{f \left ( 4\,a+4\,b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) a}{4\,f \left ( a+b \right ) ^{2}}}+{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{2}}}+{\frac{ab}{f \left ( a+b \right ) ^{2}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{f \left ( 4\,a+4\,b \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) a}{4\,f \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x)

[Out]

1/f/(4*a+4*b)/(1+cos(f*x+e))-1/4/f/(a+b)^2*ln(1+cos(f*x+e))*a+1/4/f/(a+b)^2*ln(1+cos(f*x+e))*b+1/f*a*b/(a+b)^2
/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+1/f/(4*a+4*b)/(-1+cos(f*x+e))+1/4/f/(a+b)^2*ln(-1+cos(f*x+e))*a-
1/4/f/(a+b)^2*ln(-1+cos(f*x+e))*b

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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)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.674584, size = 848, normalized size = 9.86 \begin{align*} \left [\frac{2 \, \sqrt{-a b}{\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-a b} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \,{\left (a + b\right )} \cos \left (f x + e\right ) -{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) +{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}, \frac{4 \, \sqrt{a b}{\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac{\sqrt{a b} \cos \left (f x + e\right )}{b}\right ) + 2 \,{\left (a + b\right )} \cos \left (f x + e\right ) -{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) +{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(-a*b)*(cos(f*x + e)^2 - 1)*log(-(a*cos(f*x + e)^2 + 2*sqrt(-a*b)*cos(f*x + e) - b)/(a*cos(f*x + e
)^2 + b)) + 2*(a + b)*cos(f*x + e) - ((a - b)*cos(f*x + e)^2 - a + b)*log(1/2*cos(f*x + e) + 1/2) + ((a - b)*c
os(f*x + e)^2 - a + b)*log(-1/2*cos(f*x + e) + 1/2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^
2)*f), 1/4*(4*sqrt(a*b)*(cos(f*x + e)^2 - 1)*arctan(sqrt(a*b)*cos(f*x + e)/b) + 2*(a + b)*cos(f*x + e) - ((a -
b)*cos(f*x + e)^2 - a + b)*log(1/2*cos(f*x + e) + 1/2) + ((a - b)*cos(f*x + e)^2 - a + b)*log(-1/2*cos(f*x +
e) + 1/2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(csc(e + f*x)**3/(a + b*sec(e + f*x)**2), x)

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Giac [B]  time = 1.22227, size = 294, normalized size = 3.42 \begin{align*} -\frac{\frac{8 \, a 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^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b}} - \frac{2 \,{\left (a - b\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{{\left (a + b - \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{2 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\cos \left (f x + e\right ) - 1\right )}} + \frac{\cos \left (f x + e\right ) - 1}{{\left (a + b\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*(8*a*b*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/((a^2 + 2*a*b + b^2)*sqrt(a*b))
- 2*(a - b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^2 + 2*a*b + b^2) - (a + b - 2*a*(cos(f*x + e) - 1)
/(cos(f*x + e) + 1) + 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))*(cos(f*x + e) + 1)/((a^2 + 2*a*b + b^2)*(cos(
f*x + e) - 1)) + (cos(f*x + e) - 1)/((a + b)*(cos(f*x + e) + 1)))/f