3.19 \(\int \csc ^3(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=104 \[ -\frac{\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac{b (6 a+5 b) \sec (e+f x)}{3 f}-\frac{(a+b) (a+5 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f} \]

[Out]

-((a + b)*(a + 5*b)*ArcTanh[Cos[e + f*x]])/(2*f) - ((3*a^2 + 6*a*b + 5*b^2)*Cot[e + f*x]*Csc[e + f*x])/(6*f) +
(b*(6*a + 5*b)*Sec[e + f*x])/(3*f) + (b^2*Csc[e + f*x]^2*Sec[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.110187, antiderivative size = 104, 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, 462, 456, 453, 206} \[ -\frac{\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac{b (6 a+5 b) \sec (e+f x)}{3 f}-\frac{(a+b) (a+5 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)*(a + 5*b)*ArcTanh[Cos[e + f*x]])/(2*f) - ((3*a^2 + 6*a*b + 5*b^2)*Cot[e + f*x]*Csc[e + f*x])/(6*f) +
(b*(6*a + 5*b)*Sec[e + f*x])/(3*f) + (b^2*Csc[e + f*x]^2*Sec[e + f*x]^3)/(3*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 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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])

Rubi steps

\begin{align*} \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (b+a x^2\right )^2}{x^4 \left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f}-\frac{\operatorname{Subst}\left (\int \frac{b (6 a+5 b)+3 a^2 x^2}{x^2 \left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{3 f}\\ &=-\frac{\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f}+\frac{\operatorname{Subst}\left (\int \frac{-2 b (6 a+5 b)-\left (3 a^2+6 a b+5 b^2\right ) x^2}{x^2 \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{6 f}\\ &=-\frac{\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac{b (6 a+5 b) \sec (e+f x)}{3 f}+\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f}-\frac{((a+b) (a+5 b)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{2 f}\\ &=-\frac{(a+b) (a+5 b) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac{b (6 a+5 b) \sec (e+f x)}{3 f}+\frac{b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [B]  time = 6.57456, size = 1021, normalized size = 9.82 \[ \frac{\left (-a^2-2 b a-b^2\right ) \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{2 f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac{\left (a^2+2 b a+b^2\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{2 f (\cos (2 e+2 f x) a+a+2 b)^2}-\frac{2 \left (a^2+6 b a+5 b^2\right ) \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac{2 \left (a^2+6 b a+5 b^2\right ) \log \left (\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac{2 b (12 a+13 b) \sec (e) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac{2 \left (b \sec ^2(e+f x)+a\right )^2 \left (13 \sin \left (\frac{f x}{2}\right ) b^2+12 a \sin \left (\frac{f x}{2}\right ) b\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )-\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}-\frac{2 \left (b \sec ^2(e+f x)+a\right )^2 \left (13 \sin \left (\frac{f x}{2}\right ) b^2+12 a \sin \left (\frac{f x}{2}\right ) b\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )+\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}+\frac{\left (b \sec ^2(e+f x)+a\right )^2 \left (\cos \left (\frac{e}{2}\right ) b^2+\sin \left (\frac{e}{2}\right ) b^2\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )-\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )^2}+\frac{\left (b \sec ^2(e+f x)+a\right )^2 \left (b^2 \cos \left (\frac{e}{2}\right )-b^2 \sin \left (\frac{e}{2}\right )\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )+\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )^2}+\frac{2 b^2 \left (b \sec ^2(e+f x)+a\right )^2 \sin \left (\frac{f x}{2}\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )-\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )^3}-\frac{2 b^2 \left (b \sec ^2(e+f x)+a\right )^2 \sin \left (\frac{f x}{2}\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )+\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-a^2 - 2*a*b - b^2)*Cos[e + f*x]^4*Csc[e/2 + (f*x)/2]^2*(a + b*Sec[e + f*x]^2)^2)/(2*f*(a + 2*b + a*Cos[2*e
+ 2*f*x])^2) - (2*(a^2 + 6*a*b + 5*b^2)*Cos[e + f*x]^4*Log[Cos[e/2 + (f*x)/2]]*(a + b*Sec[e + f*x]^2)^2)/(f*(a
+ 2*b + a*Cos[2*e + 2*f*x])^2) + (2*(a^2 + 6*a*b + 5*b^2)*Cos[e + f*x]^4*Log[Sin[e/2 + (f*x)/2]]*(a + b*Sec[e
+ f*x]^2)^2)/(f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + (2*b*(12*a + 13*b)*Cos[e + f*x]^4*Sec[e]*(a + b*Sec[e + f
*x]^2)^2)/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + ((a^2 + 2*a*b + b^2)*Cos[e + f*x]^4*Sec[e/2 + (f*x)/2]^2*(a
+ b*Sec[e + f*x]^2)^2)/(2*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + (2*b^2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^
2*Sin[(f*x)/2])/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f
*x)/2])^3) + (Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(b^2*Cos[e/2] + b^2*Sin[e/2]))/(3*f*(a + 2*b + a*Cos[2*e
+ 2*f*x])^2*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^2) + (2*Cos[e + f*x]^4*(a + b*Sec
[e + f*x]^2)^2*(12*a*b*Sin[(f*x)/2] + 13*b^2*Sin[(f*x)/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] -
Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])) - (2*b^2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*Sin[(f*x
)/2])/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^3)
+ (Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(b^2*Cos[e/2] - b^2*Sin[e/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])
^2*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) - (2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^
2)^2*(12*a*b*Sin[(f*x)/2] + 13*b^2*Sin[(f*x)/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] + Sin[e/2])*
(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]))

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Maple [B]  time = 0.063, size = 195, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2}\csc \left ( fx+e \right ) \cot \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-{\frac{ab}{f \left ( \sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) }}+3\,{\frac{ab}{f\cos \left ( fx+e \right ) }}+3\,{\frac{ab\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}+{\frac{{b}^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-{\frac{5\,{b}^{2}}{6\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) }}+{\frac{5\,{b}^{2}}{2\,f\cos \left ( fx+e \right ) }}+{\frac{5\,{b}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}} \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]

-1/2/f*a^2*csc(f*x+e)*cot(f*x+e)+1/2/f*a^2*ln(csc(f*x+e)-cot(f*x+e))-1/f*a*b/sin(f*x+e)^2/cos(f*x+e)+3/f*a*b/c
os(f*x+e)+3/f*a*b*ln(csc(f*x+e)-cot(f*x+e))+1/3/f*b^2/sin(f*x+e)^2/cos(f*x+e)^3-5/6/f*b^2/sin(f*x+e)^2/cos(f*x
+e)+5/2/f*b^2/cos(f*x+e)+5/2/f*b^2*ln(csc(f*x+e)-cot(f*x+e))

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Maxima [A]  time = 1.04664, size = 170, normalized size = 1.63 \begin{align*} -\frac{3 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (f x + e\right )^{5} - \cos \left (f x + e\right )^{3}}}{12 \, 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)^2,x, algorithm="maxima")

[Out]

-1/12*(3*(a^2 + 6*a*b + 5*b^2)*log(cos(f*x + e) + 1) - 3*(a^2 + 6*a*b + 5*b^2)*log(cos(f*x + e) - 1) - 2*(3*(a
^2 + 6*a*b + 5*b^2)*cos(f*x + e)^4 - 2*(6*a*b + 5*b^2)*cos(f*x + e)^2 - 2*b^2)/(cos(f*x + e)^5 - cos(f*x + e)^
3))/f

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Fricas [B]  time = 0.540695, size = 471, normalized size = 4.53 \begin{align*} \frac{6 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 4 \,{\left (6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, b^{2} - 3 \,{\left ({\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + 3 \,{\left ({\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{12 \,{\left (f \cos \left (f x + e\right )^{5} - f \cos \left (f x + e\right )^{3}\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)^2,x, algorithm="fricas")

[Out]

1/12*(6*(a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^4 - 4*(6*a*b + 5*b^2)*cos(f*x + e)^2 - 4*b^2 - 3*((a^2 + 6*a*b + 5*
b^2)*cos(f*x + e)^5 - (a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^3)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^2 + 6*a*b + 5*
b^2)*cos(f*x + e)^5 - (a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^3)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x + e)^5 -
f*cos(f*x + e)^3)

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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 = 1.29211, size = 491, normalized size = 4.72 \begin{align*} -\frac{\frac{3 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{6 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{3 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - 6 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - \frac{3 \,{\left (a^{2} + 2 \, a b + b^{2} - \frac{2 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{12 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{10 \, b^{2}{\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 )}}{\cos \left (f x + e\right ) - 1} - \frac{16 \,{\left (6 \, a b + 7 \, b^{2} + \frac{12 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{12 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{6 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{9 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}}}{24 \, 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)^2,x, algorithm="giac")

[Out]

-1/24*(3*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 6*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*b^2*(cos(
f*x + e) - 1)/(cos(f*x + e) + 1) - 6*(a^2 + 6*a*b + 5*b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1)) - 3*(a^
2 + 2*a*b + b^2 - 2*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 12*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) -
10*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))*(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - 16*(6*a*b + 7*b^2 + 12*
a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 12*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 6*a*b*(cos(f*x + e)
- 1)^2/(cos(f*x + e) + 1)^2 + 9*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/((cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + 1)^3)/f