3.52 \(\int \frac{\csc ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac{\sqrt{b} (3 a-2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{7/2}}-\frac{a b \tan (e+f x)}{2 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\cot ^3(e+f x)}{3 f (a+b)^2}-\frac{(a-b) \cot (e+f x)}{f (a+b)^3} \]

[Out]

-((3*a - 2*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(7/2)*f) - ((a - b)*Cot[e + f*x])
/((a + b)^3*f) - Cot[e + f*x]^3/(3*(a + b)^2*f) - (a*b*Tan[e + f*x])/(2*(a + b)^3*f*(a + b + b*Tan[e + f*x]^2)
)

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Rubi [A]  time = 0.174616, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4132, 456, 1261, 205} \[ -\frac{\sqrt{b} (3 a-2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{7/2}}-\frac{a b \tan (e+f x)}{2 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\cot ^3(e+f x)}{3 f (a+b)^2}-\frac{(a-b) \cot (e+f x)}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-((3*a - 2*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(7/2)*f) - ((a - b)*Cot[e + f*x])
/((a + b)^3*f) - Cot[e + f*x]^3/(3*(a + b)^2*f) - (a*b*Tan[e + f*x])/(2*(a + b)^3*f*(a + b + b*Tan[e + f*x]^2)
)

Rule 4132

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{2}{b (a+b)}-\frac{2 a x^2}{b (a+b)^2}+\frac{a x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{2}{b (a+b)^2 x^4}-\frac{2 (a-b)}{b (a+b)^3 x^2}+\frac{3 a-2 b}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{(a-b) \cot (e+f x)}{(a+b)^3 f}-\frac{\cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a+b)^3 f}\\ &=-\frac{(3 a-2 b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 (a+b)^{7/2} f}-\frac{(a-b) \cot (e+f x)}{(a+b)^3 f}-\frac{\cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.76797, size = 637, normalized size = 5.18 \[ \frac{\sec ^4(e+f x) \left (a b \sin (2 e)-a b \sin (2 f x)+2 b^2 \sin (2 e)\right ) (a \cos (2 e+2 f x)+a+2 b)}{8 f (a+b)^3 (\cos (e)-\sin (e)) (\sin (e)+\cos (e)) \left (a+b \sec ^2(e+f x)\right )^2}-\frac{\cot (e) \csc ^2(e+f x) \sec ^4(e+f x) (a \cos (2 e+2 f x)+a+2 b)^2}{12 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^2}+\frac{\csc (e) \sin (f x) \csc ^3(e+f x) \sec ^4(e+f x) (a \cos (2 e+2 f x)+a+2 b)^2}{12 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^2}+\frac{\csc (e) \csc (e+f x) \sec ^4(e+f x) (a \sin (f x)-2 b \sin (f x)) (a \cos (2 e+2 f x)+a+2 b)^2}{6 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2}+\frac{(3 a-2 b) \sec ^4(e+f x) (a \cos (2 e+2 f x)+a+2 b)^2 \left (\frac{b \cos (2 e) \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (a \sin (2 e+f x)-a \sin (f x)-2 b \sin (f x))\right )}{8 f \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i b \sin (2 e) \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (a \sin (2 e+f x)-a \sin (f x)-2 b \sin (f x))\right )}{8 f \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right )}{(a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-((a + 2*b + a*Cos[2*e + 2*f*x])^2*Cot[e]*Csc[e + f*x]^2*Sec[e + f*x]^4)/(12*(a + b)^2*f*(a + b*Sec[e + f*x]^2
)^2) + ((3*a - 2*b)*(a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*((b*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a +
b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin
[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(8*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/8
)*b*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*
Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin[2*e])/(Sqrt[a + b]*f*
Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])))/((a + b)^3*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Cs
c[e]*Csc[e + f*x]^3*Sec[e + f*x]^4*Sin[f*x])/(12*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e
+ 2*f*x])^2*Csc[e]*Csc[e + f*x]*Sec[e + f*x]^4*(a*Sin[f*x] - 2*b*Sin[f*x]))/(6*(a + b)^3*f*(a + b*Sec[e + f*x
]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x])*Sec[e + f*x]^4*(a*b*Sin[2*e] + 2*b^2*Sin[2*e] - a*b*Sin[2*f*x]))/(8*
(a + b)^3*f*(a + b*Sec[e + f*x]^2)^2*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))

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Maple [A]  time = 0.106, size = 160, normalized size = 1.3 \begin{align*} -{\frac{1}{3\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{a}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}+{\frac{b}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}-{\frac{ab\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,ab}{2\,f \left ( a+b \right ) ^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{{b}^{2}}{f \left ( a+b \right ) ^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.688201, size = 1534, normalized size = 12.47 \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)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/24*(4*(4*a^2 - 11*a*b)*cos(f*x + e)^5 - 8*(3*a^2 - 8*a*b + 4*b^2)*cos(f*x + e)^3 + 3*((3*a^2 - 2*a*b)*cos(
f*x + e)^4 - (3*a^2 - 5*a*b + 2*b^2)*cos(f*x + e)^2 - 3*a*b + 2*b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^
2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*c
os(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x +
e) - 12*(3*a*b - 2*b^2)*cos(f*x + e))/(((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)*sin(f*x + e)), -1/12*(2*(4*a^2 - 1
1*a*b)*cos(f*x + e)^5 - 4*(3*a^2 - 8*a*b + 4*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - 2*a*b)*cos(f*x + e)^4 - (3*a^2
- 5*a*b + 2*b^2)*cos(f*x + e)^2 - 3*a*b + 2*b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqr
t(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(3*a*b - 2*b^2)*cos(f*x + e))/(((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)*sin(f*x + e))]

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

[Out]

Timed out

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Giac [A]  time = 1.28925, size = 259, normalized size = 2.11 \begin{align*} -\frac{\frac{3 \, a b \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}{\left (3 \, a b - 2 \, b^{2}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b + b^{2}}} + \frac{2 \,{\left (3 \, a \tan \left (f x + e\right )^{2} - 3 \, b \tan \left (f x + e\right )^{2} + a + b\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/6*(3*a*b*tan(f*x + e)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)) + 3*(pi*floor((f*x + e)/
pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3*a*b - 2*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sq
rt(a*b + b^2)) + 2*(3*a*tan(f*x + e)^2 - 3*b*tan(f*x + e)^2 + a + b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x
+ e)^3))/f