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

Optimal. Leaf size=197 \[ -\frac{3 \left (a^2-6 a b+b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^4}+\frac{3 a (a-3 b) \cos (e+f x)}{8 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{3 \sqrt{a} \sqrt{b} (a-b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 f (a+b)^4}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b) \left (a \cos ^2(e+f x)+b\right )}-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )} \]

[Out]

(3*Sqrt[a]*(a - b)*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*(a + b)^4*f) - (3*(a^2 - 6*a*b + b^2)*Ar
cTanh[Cos[e + f*x]])/(8*(a + b)^4*f) + (3*a*(a - 3*b)*Cos[e + f*x])/(8*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) - (
(a - 5*b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f*(b + a*Cos[e + f*x]^2)) - (Cot[e + f*x]*Csc[e + f*x]^3)/(4
*(a + b)*f*(b + a*Cos[e + f*x]^2))

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Rubi [A]  time = 0.246437, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{3 \left (a^2-6 a b+b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^4}+\frac{3 a (a-3 b) \cos (e+f x)}{8 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{3 \sqrt{a} \sqrt{b} (a-b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 f (a+b)^4}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b) \left (a \cos ^2(e+f x)+b\right )}-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[a]*(a - b)*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*(a + b)^4*f) - (3*(a^2 - 6*a*b + b^2)*Ar
cTanh[Cos[e + f*x]])/(8*(a + b)^4*f) + (3*a*(a - 3*b)*Cos[e + f*x])/(8*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) - (
(a - 5*b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f*(b + a*Cos[e + f*x]^2)) - (Cot[e + f*x]*Csc[e + f*x]^3)/(4
*(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 ^5(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 )^3 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{b+(-a+4 b) x^2}{\left (1-x^2\right )^2 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 (a-b) b-3 a (a-5 b) x^2}{\left (1-x^2\right ) \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{8 (a+b)^2 f}\\ &=\frac{3 a (a-3 b) \cos (e+f x)}{8 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-6 (3 a-b) b^2+6 a (a-3 b) b x^2}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{16 b (a+b)^3 f}\\ &=\frac{3 a (a-3 b) \cos (e+f x)}{8 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )}+\frac{(3 a (a-b) b) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{2 (a+b)^4 f}-\frac{\left (3 \left (a^2-6 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{8 (a+b)^4 f}\\ &=\frac{3 \sqrt{a} (a-b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 (a+b)^4 f}-\frac{3 \left (a^2-6 a b+b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 (a+b)^4 f}+\frac{3 a (a-3 b) \cos (e+f x)}{8 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{(a-5 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.32487, size = 450, normalized size = 2.28 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-24 \left (a^2-6 a b+b^2\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)+24 \left (a^2-6 a b+b^2\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-2 (a+b) \cot (e+f x) \csc ^3(e+f x) \left (4 \left (2 a^2-5 a b+5 b^2\right ) \cos (2 (e+f x))+11 a^2-3 a (a-3 b) \cos (4 (e+f x))+43 a b-4 b^2\right )+96 \sqrt{a} \sqrt{b} (a-b) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+96 \sqrt{a} \sqrt{b} (a-b) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )\right )}{256 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*(96*Sqrt[a]*(a - b)*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]))/Sq
rt[b]]*(a + 2*b + a*Cos[2*(e + f*x)]) + 96*Sqrt[a]*(a - b)*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 + 2*b + a*Cos[2*(e + f*x)]) - 2*(a + b)*(11*a^2 + 43*a*b - 4*b^2 + 4*(2*a^2 - 5*a*b + 5*b^2)*
Cos[2*(e + f*x)] - 3*a*(a - 3*b)*Cos[4*(e + f*x)])*Cot[e + f*x]*Csc[e + f*x]^3 - 24*(a^2 - 6*a*b + b^2)*(a + 2
*b + a*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2]] + 24*(a^2 - 6*a*b + b^2)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Sin
[(e + f*x)/2]])*Sec[e + f*x]^4)/(256*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [B]  time = 0.119, size = 390, normalized size = 2. \begin{align*}{\frac{1}{16\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{16\,f \left ( a+b \right ) ^{3} \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{5\,b}{16\,f \left ( a+b \right ) ^{3} \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 ) ^{4}}}+{\frac{9\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) ab}{8\,f \left ( a+b \right ) ^{4}}}-{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ){b}^{2}}{16\,f \left ( a+b \right ) ^{4}}}-{\frac{{a}^{2}b\cos \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{a{b}^{2}\cos \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{3\,{a}^{2}b}{2\,f \left ( a+b \right ) ^{4}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,a{b}^{2}}{2\,f \left ( a+b \right ) ^{4}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{16\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{16\,f \left ( a+b \right ) ^{3} \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{5\,b}{16\,f \left ( a+b \right ) ^{3} \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 ) ^{4}}}-{\frac{9\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) ab}{8\,f \left ( a+b \right ) ^{4}}}+{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ){b}^{2}}{16\,f \left ( a+b \right ) ^{4}}} \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)^2,x)

[Out]

1/16/f/(a+b)^2/(1+cos(f*x+e))^2+3/16/f/(a+b)^3/(1+cos(f*x+e))*a-5/16/f/(a+b)^3/(1+cos(f*x+e))*b-3/16/f/(a+b)^4
*ln(1+cos(f*x+e))*a^2+9/8/f/(a+b)^4*ln(1+cos(f*x+e))*a*b-3/16/f/(a+b)^4*ln(1+cos(f*x+e))*b^2-1/2/f*a^2/(a+b)^4
*b*cos(f*x+e)/(b+a*cos(f*x+e)^2)-1/2/f*a/(a+b)^4*b^2*cos(f*x+e)/(b+a*cos(f*x+e)^2)+3/2/f*a^2/(a+b)^4*b/(a*b)^(
1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))-3/2/f*a/(a+b)^4*b^2/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))-1/16/f/
(a+b)^2/(-1+cos(f*x+e))^2+3/16/f/(a+b)^3/(-1+cos(f*x+e))*a-5/16/f/(a+b)^3/(-1+cos(f*x+e))*b+3/16/f/(a+b)^4*ln(
-1+cos(f*x+e))*a^2-9/8/f/(a+b)^4*ln(-1+cos(f*x+e))*a*b+3/16/f/(a+b)^4*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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.973963, size = 2747, normalized size = 13.94 \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)^2,x, algorithm="fricas")

[Out]

[1/16*(6*(a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3 - 9*a^2*b - 9*a*b^2 + 5*b^3)*cos(f*x + e)^3 - 12*
((a^2 - a*b)*cos(f*x + e)^6 - (2*a^2 - 3*a*b + b^2)*cos(f*x + e)^4 + (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + a*
b - b^2)*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)) - 6*(3*a^2*
b + 2*a*b^2 - b^3)*cos(f*x + e) - 3*((a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 - b^
3)*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x
+ e) + 1/2) + 3*((a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 - b^3)*cos(f*x + e)^4 +
a^2*b - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^
5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a
*b^4 - b^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*cos(f*x + e)^2 + (a
^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f), 1/16*(6*(a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3
- 9*a^2*b - 9*a*b^2 + 5*b^3)*cos(f*x + e)^3 + 24*((a^2 - a*b)*cos(f*x + e)^6 - (2*a^2 - 3*a*b + b^2)*cos(f*x +
e)^4 + (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + a*b - b^2)*sqrt(a*b)*arctan(sqrt(a*b)*cos(f*x + e)/b) - 6*(3*a^
2*b + 2*a*b^2 - b^3)*cos(f*x + e) - 3*((a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 -
b^3)*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3)*cos(f*x + e)^2)*log(1/2*cos(f
*x + e) + 1/2) + 3*((a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 - b^3)*cos(f*x + e)^4
+ a^2*b - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((
a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2
*a*b^4 - b^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*cos(f*x + e)^2 +
(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.32657, size = 950, normalized size = 4.82 \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)^2,x, algorithm="giac")

[Out]

1/64*(12*(a^2 - 6*a*b + b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3
+ b^4) - 96*(a^2*b - a*b^2)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/((a^4 + 4*a^3*b
+ 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a*b)) - (8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*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 - b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (a^
2 + 2*a*b + b^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*b^2*(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 - 108*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 18*b^2*
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(
cos(f*x + e) - 1)^2) - 64*(a^2*b + a*b^2 + a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a*b^2*(cos(f*x + e) -
1)/(cos(f*x + e) + 1))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(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) + 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)))/f