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

Optimal. Leaf size=147 \[ \frac{(a-b) \cos (e+f x)}{2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 \sqrt{a} f (a+b)^3}-\frac{(a-3 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^3}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.172685, antiderivative size = 147, 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, 470, 527, 522, 206, 205} \[ \frac{(a-b) \cos (e+f x)}{2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 \sqrt{a} f (a+b)^3}-\frac{(a-3 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^3}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 470

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

Mathematica [C]  time = 1.90282, size = 468, normalized size = 3.18 \[ \frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left ((a+b) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)-4 (a-3 b) \sec (e+f x) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-(a+b) \csc ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)+4 (a-3 b) \sec (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-\frac{4 \sqrt{b} (b-3 a) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{\sqrt{a}}-\frac{4 \sqrt{b} (b-3 a) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{\sqrt{a}}-8 b (a+b)\right )}{32 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(-8*b*(a + b) - (4*Sqrt[b]*(-3*a + b)*ArcTan[((-Sqrt[a] - I*Sqr
t[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 + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x])/Sqrt[a] - (4*Sqrt[b]*(-3*a + b)*A
rcTan[((-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 + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x])/Sqrt[a] -
(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[(e + f*x)/2]^2*Sec[e + f*x] - 4*(a - 3*b)*(a + 2*b + a*Cos[2*(e +
f*x)])*Log[Cos[(e + f*x)/2]]*Sec[e + f*x] + 4*(a - 3*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Sin[(e + f*x)/2]]*S
ec[e + f*x] + (a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^2*Sec[e + f*x]))/(32*(a + b)^3*f*(a + b*
Sec[e + f*x]^2)^2)

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Maple [A]  time = 0.11, size = 250, normalized size = 1.7 \begin{align*}{\frac{1}{4\,f \left ( a+b \right ) ^{2} \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 ) ^{3}}}+{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{3}}}-{\frac{b\cos \left ( fx+e \right ) a}{2\,f \left ( a+b \right ) ^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{2}\cos \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{3\,ab}{2\,f \left ( a+b \right ) ^{3}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{2}}{2\,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}{4\,f \left ( a+b \right ) ^{2} \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 ) ^{3}}}-{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{3}}} \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)^2,x)

[Out]

1/4/f/(a+b)^2/(1+cos(f*x+e))-1/4/f/(a+b)^3*ln(1+cos(f*x+e))*a+3/4/f/(a+b)^3*ln(1+cos(f*x+e))*b-1/2/f/(a+b)^3*b
*cos(f*x+e)/(b+a*cos(f*x+e)^2)*a-1/2/f/(a+b)^3*b^2*cos(f*x+e)/(b+a*cos(f*x+e)^2)+3/2/f/(a+b)^3*b/(a*b)^(1/2)*a
rctan(a*cos(f*x+e)/(a*b)^(1/2))*a-1/2/f/(a+b)^3*b^2/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+1/4/f/(a+b)^2
/(-1+cos(f*x+e))+1/4/f/(a+b)^3*ln(-1+cos(f*x+e))*a-3/4/f/(a+b)^3*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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.822236, size = 1613, normalized size = 10.97 \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)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.24147, size = 787, normalized size = 5.35 \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)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/24*(6*(a - 3*b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 12*(3*a*b - b^
2)*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)) + (3*a^2 + 6*a*b + 3*b^2 + 4*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 20*a*b*(cos(f*x + e) - 1)/(cos(f*
x + e) + 1) - 24*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 2
*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 15*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 2*a^2*(cos
(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 4*a*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 6*b^2*(cos(f*x + e)
- 1)^3/(cos(f*x + e) + 1)^3)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + b*(co
s(f*x + e) - 1)/(cos(f*x + e) + 1) + 2*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 2*b*(cos(f*x + e) - 1)^2/
(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)
^3)) - 3*(cos(f*x + e) - 1)/((a^2 + 2*a*b + b^2)*(cos(f*x + e) + 1)))/f