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

Optimal. Leaf size=213 \[ \frac{\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{\sqrt{b} \left (15 a^2-10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{3/2} f (a+b)^4}-\frac{b (2 a-b) \cos (e+f x)}{4 a f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^2}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )^2}-\frac{(a-5 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^4} \]

[Out]

(Sqrt[b]*(15*a^2 - 10*a*b - b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(3/2)*(a + b)^4*f) - ((a - 5*b)*
ArcTanh[Cos[e + f*x]])/(2*(a + b)^4*f) - ((2*a - b)*b*Cos[e + f*x])/(4*a*(a + b)^2*f*(b + a*Cos[e + f*x]^2)^2)
+ ((4*a^2 - 9*a*b - b^2)*Cos[e + f*x])/(8*a*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) - (Cos[e + f*x]*Cot[e + f*x]^
2)/(2*(a + b)*f*(b + a*Cos[e + f*x]^2)^2)

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Rubi [A]  time = 0.31166, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4133, 470, 578, 527, 522, 206, 205} \[ \frac{\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{\sqrt{b} \left (15 a^2-10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{3/2} f (a+b)^4}-\frac{b (2 a-b) \cos (e+f x)}{4 a f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^2}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )^2}-\frac{(a-5 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(15*a^2 - 10*a*b - b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(3/2)*(a + b)^4*f) - ((a - 5*b)*
ArcTanh[Cos[e + f*x]])/(2*(a + b)^4*f) - ((2*a - b)*b*Cos[e + f*x])/(4*a*(a + b)^2*f*(b + a*Cos[e + f*x]^2)^2)
+ ((4*a^2 - 9*a*b - b^2)*Cos[e + f*x])/(8*a*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) - (Cos[e + f*x]*Cot[e + f*x]^
2)/(2*(a + b)*f*(b + a*Cos[e + f*x]^2)^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 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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 )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2 \left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 b+(-a+2 b) x^2\right )}{\left (1-x^2\right ) \left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{2 (a+b) f}\\ &=-\frac{(2 a-b) b \cos (e+f x)}{4 a (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{2 (2 a-b) b-2 \left (2 a^2-8 a b-b^2\right ) x^2}{\left (1-x^2\right ) \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{8 a (a+b)^2 f}\\ &=-\frac{(2 a-b) b \cos (e+f x)}{4 a (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-2 (11 a-b) b^2+2 b \left (4 a^2-9 a b-b^2\right ) x^2}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{16 a b (a+b)^3 f}\\ &=-\frac{(2 a-b) b \cos (e+f x)}{4 a (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{(a-5 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{2 (a+b)^4 f}+\frac{\left (b \left (15 a^2-10 a b-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a (a+b)^4 f}\\ &=\frac{\sqrt{b} \left (15 a^2-10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{3/2} (a+b)^4 f}-\frac{(a-5 b) \tanh ^{-1}(\cos (e+f x))}{2 (a+b)^4 f}-\frac{(2 a-b) b \cos (e+f x)}{4 a (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}\\ \end{align*}

Mathematica [C]  time = 3.65137, size = 532, normalized size = 2.5 \[ \frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\frac{\sqrt{b} \left (-15 a^2+10 a b+b^2\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{a^{3/2}}-\frac{\sqrt{b} \left (-15 a^2+10 a b+b^2\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{a^{3/2}}+\frac{8 b^2 (a+b)^2}{a}-\frac{2 b (9 a+b) (a+b) (a \cos (2 (e+f x))+a+2 b)}{a}+(a+b) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)^2-4 (a-5 b) \sec (e+f x) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)^2-(a+b) \csc ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)^2+4 (a-5 b) \sec (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)^2\right )}{64 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^5*((8*b^2*(a + b)^2)/a - (2*b*(a + b)*(9*a + b)*(a + 2*b + a*Cos[
2*(e + f*x)]))/a - (Sqrt[b]*(-15*a^2 + 10*a*b + b^2)*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*Sec[e + f*x])/a^(3/2) - (Sqrt[b]*(-15*a^2 + 10*a*b + b^2)*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*Sec[e + f*x])/a^(3/2) - (a + b)*(a + 2*
b + a*Cos[2*(e + f*x)])^2*Csc[(e + f*x)/2]^2*Sec[e + f*x] - 4*(a - 5*b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Log[C
os[(e + f*x)/2]]*Sec[e + f*x] + 4*(a - 5*b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Log[Sin[(e + f*x)/2]]*Sec[e + f*x
] + (a + b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[(e + f*x)/2]^2*Sec[e + f*x]))/(64*(a + b)^4*f*(a + b*Sec[e +
f*x]^2)^3)

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Maple [B]  time = 0.114, size = 430, normalized size = 2. \begin{align*}{\frac{1}{4\,f \left ( a+b \right ) ^{3} \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 ) ^{4}}}+{\frac{5\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{4}}}-{\frac{9\,b \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{8\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{5\,{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}a}{4\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{7\,a{b}^{2}\cos \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,{b}^{3}\cos \left ( fx+e \right ) }{4\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{4}\cos \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}+{\frac{15\,ab}{8\,f \left ( a+b \right ) ^{4}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{5\,{b}^{2}}{4\,f \left ( a+b \right ) ^{4}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{3}}{8\,f \left ( a+b \right ) ^{4}a}\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 ) ^{3} \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 ) ^{4}}}-{\frac{5\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) b}{4\,f \left ( a+b \right ) ^{4}}} \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)^3,x)

[Out]

1/4/f/(a+b)^3/(1+cos(f*x+e))-1/4/f/(a+b)^4*ln(1+cos(f*x+e))*a+5/4/f/(a+b)^4*ln(1+cos(f*x+e))*b-9/8/f/(a+b)^4*b
/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3*a^2-5/4/f/(a+b)^4*b^2/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3*a-1/8/f/(a+b)^4*b^3
/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3-7/8/f/(a+b)^4*b^2/(b+a*cos(f*x+e)^2)^2*a*cos(f*x+e)-3/4/f/(a+b)^4*b^3/(b+a*
cos(f*x+e)^2)^2*cos(f*x+e)+1/8/f/(a+b)^4*b^4/(b+a*cos(f*x+e)^2)^2/a*cos(f*x+e)+15/8/f/(a+b)^4*b*a/(a*b)^(1/2)*
arctan(a*cos(f*x+e)/(a*b)^(1/2))-5/4/f/(a+b)^4*b^2/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))-1/8/f/(a+b)^4*
b^3/a/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+1/4/f/(a+b)^3/(-1+cos(f*x+e))+1/4/f/(a+b)^4*ln(-1+cos(f*x+e
))*a-5/4/f/(a+b)^4*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.16509, size = 2946, normalized size = 13.83 \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)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(4*a^4 - 5*a^3*b - 10*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + 2*(17*a^3*b + 11*a^2*b^2 - 5*a*b^3 + b^4)*cos
(f*x + e)^3 - ((15*a^4 - 10*a^3*b - a^2*b^2)*cos(f*x + e)^6 - (15*a^4 - 40*a^3*b + 19*a^2*b^2 + 2*a*b^3)*cos(f
*x + e)^4 - 15*a^2*b^2 + 10*a*b^3 + b^4 - (30*a^3*b - 35*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^2)*sqrt(-b/a)*l
og((a*cos(f*x + e)^2 - 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 2*(11*a^2*b^2 + 10*a*b^3 - b
^4)*cos(f*x + e) - 4*((a^4 - 5*a^3*b)*cos(f*x + e)^6 - (a^4 - 7*a^3*b + 10*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 +
5*a*b^3 - (2*a^3*b - 11*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 4*((a^4 - 5*a^3*b)*c
os(f*x + e)^6 - (a^4 - 7*a^3*b + 10*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 + 5*a*b^3 - (2*a^3*b - 11*a^2*b^2 + 5*a*
b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*
x + e)^6 - (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^4 - (2*a^6*b + 7*a^5
*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b^6)*f*cos(f*x + e)^2 - (a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*
b^5 + a*b^6)*f), 1/8*((4*a^4 - 5*a^3*b - 10*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + (17*a^3*b + 11*a^2*b^2 - 5*a*b^3
+ b^4)*cos(f*x + e)^3 + ((15*a^4 - 10*a^3*b - a^2*b^2)*cos(f*x + e)^6 - (15*a^4 - 40*a^3*b + 19*a^2*b^2 + 2*a
*b^3)*cos(f*x + e)^4 - 15*a^2*b^2 + 10*a*b^3 + b^4 - (30*a^3*b - 35*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^2)*s
qrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) + (11*a^2*b^2 + 10*a*b^3 - b^4)*cos(f*x + e) - 2*((a^4 - 5*a^3*b)*
cos(f*x + e)^6 - (a^4 - 7*a^3*b + 10*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 + 5*a*b^3 - (2*a^3*b - 11*a^2*b^2 + 5*a
*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 2*((a^4 - 5*a^3*b)*cos(f*x + e)^6 - (a^4 - 7*a^3*b + 10*a^
2*b^2)*cos(f*x + e)^4 - a^2*b^2 + 5*a*b^3 - (2*a^3*b - 11*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x
+ e) + 1/2))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (a^7 + 2*a^6*b - 2*a^5*b^2
- 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^4 - (2*a^6*b + 7*a^5*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b
^5 - a*b^6)*f*cos(f*x + e)^2 - (a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*b^5 + a*b^6)*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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.49216, size = 1068, normalized size = 5.01 \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)^3,x, algorithm="giac")

[Out]

1/8*(2*(a - 5*b)*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) - (15
*a^2*b - 10*a*b^2 - b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/((a^5 + 4*a^4*b +
6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*sqrt(a*b)) + (a + b - 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 10*b*(cos(f*x
+ e) - 1)/(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)*(cos(f*x + e) -
1)) - (cos(f*x + e) - 1)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(f*x + e) + 1)) - 2*(9*a^3*b + 17*a^2*b^2 + 7*a
*b^3 - b^4 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) -
23*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f
*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 21*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 29*a*b^3*(cos(f*x
+ e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) -
1)^3/(cos(f*x + e) + 1)^3 - 5*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 13*a*b^3*(cos(f*x + e) - 1)^
3/(cos(f*x + e) + 1)^3 + b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^
3 + a*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)^2))/f