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

Optimal. Leaf size=129 \[ \frac {a^{3/2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{f (a+b)^3}-\frac {\left (3 a^2-6 a b-b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^3}-\frac {\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b)}-\frac {(3 a-b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2} \]

[Out]

-1/8*(3*a^2-6*a*b-b^2)*arctanh(cos(f*x+e))/(a+b)^3/f-1/8*(3*a-b)*cot(f*x+e)*csc(f*x+e)/(a+b)^2/f-1/4*cot(f*x+e
)*csc(f*x+e)^3/(a+b)/f+a^(3/2)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/(a+b)^3/f

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, 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, 471, 527, 522, 206, 205} \[ -\frac {\left (3 a^2-6 a b-b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f (a+b)^3}+\frac {a^{3/2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{f (a+b)^3}-\frac {\cot (e+f x) \csc ^3(e+f x)}{4 f (a+b)}-\frac {(3 a-b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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 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 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
]

Rubi steps

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

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Mathematica [C]  time = 4.70, size = 549, normalized size = 4.26 \[ -\frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-64 a^{3/2} \sqrt {b} \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )-64 a^{3/2} \sqrt {b} \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+a^2 \csc ^4\left (\frac {1}{2} (e+f x)\right )+6 a^2 \csc ^2\left (\frac {1}{2} (e+f x)\right )-a^2 \sec ^4\left (\frac {1}{2} (e+f x)\right )-6 a^2 \sec ^2\left (\frac {1}{2} (e+f x)\right )-24 a^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+24 a^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+2 a b \csc ^4\left (\frac {1}{2} (e+f x)\right )+4 a b \csc ^2\left (\frac {1}{2} (e+f x)\right )-2 a b \sec ^4\left (\frac {1}{2} (e+f x)\right )-4 a b \sec ^2\left (\frac {1}{2} (e+f x)\right )+48 a b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-48 a b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+b^2 \csc ^4\left (\frac {1}{2} (e+f x)\right )-2 b^2 \csc ^2\left (\frac {1}{2} (e+f x)\right )-b^2 \sec ^4\left (\frac {1}{2} (e+f x)\right )+2 b^2 \sec ^2\left (\frac {1}{2} (e+f x)\right )+8 b^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{128 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/128*((a + 2*b + a*Cos[2*(e + f*x)])*(-64*a^(3/2)*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]] - 64*a^(3/2)*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]] + 6*a^2*Csc[(e + f*x)/2]^
2 + 4*a*b*Csc[(e + f*x)/2]^2 - 2*b^2*Csc[(e + f*x)/2]^2 + a^2*Csc[(e + f*x)/2]^4 + 2*a*b*Csc[(e + f*x)/2]^4 +
b^2*Csc[(e + f*x)/2]^4 + 24*a^2*Log[Cos[(e + f*x)/2]] - 48*a*b*Log[Cos[(e + f*x)/2]] - 8*b^2*Log[Cos[(e + f*x)
/2]] - 24*a^2*Log[Sin[(e + f*x)/2]] + 48*a*b*Log[Sin[(e + f*x)/2]] + 8*b^2*Log[Sin[(e + f*x)/2]] - 6*a^2*Sec[(
e + f*x)/2]^2 - 4*a*b*Sec[(e + f*x)/2]^2 + 2*b^2*Sec[(e + f*x)/2]^2 - a^2*Sec[(e + f*x)/2]^4 - 2*a*b*Sec[(e +
f*x)/2]^4 - b^2*Sec[(e + f*x)/2]^4)*Sec[e + f*x]^2)/((a + b)^3*f*(a + b*Sec[e + f*x]^2))

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fricas [B]  time = 0.84, size = 693, normalized size = 5.37 \[ \left [\frac {2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (a \cos \left (f x + e\right )^{4} - 2 \, a \cos \left (f x + e\right )^{2} + a\right )} \sqrt {-a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a b} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (5 \, a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, \frac {2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 16 \, {\left (a \cos \left (f x + e\right )^{4} - 2 \, a \cos \left (f x + e\right )^{2} + a\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \cos \left (f x + e\right )}{b}\right ) - 2 \, {\left (5 \, a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^3 + 8*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^2 + a)*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)) - 2*(5*a^2 + 6*a*b + b^2)*cos(f*x +
e) - ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 - 6*a*b - b^2)*lo
g(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x + e)^2 + 3
*a^2 - 6*a*b - b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 +
3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f), 1/16*(2*(3*a^2 + 2*a*b - b^2)
*cos(f*x + e)^3 + 16*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^2 + a)*sqrt(a*b)*arctan(sqrt(a*b)*cos(f*x + e)/b) -
2*(5*a^2 + 6*a*b + b^2)*cos(f*x + e) - ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 6*a*b - b^2)*cos(f*x
+ e)^2 + 3*a^2 - 6*a*b - b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2
- 6*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 - 6*a*b - b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((32*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b+32*((1-
cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+256*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a)/(4096*b^2+8192*b*a+40
96*a^2)+(6*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2+36*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b*a-
18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2-8*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a-8*(1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1)))*a^2-b^2-2*b*a-a^2)/(128*b^3+384*b^2*a+384*b*a^2+128*a^3)/((1-cos(f*x+exp(1)))/(1+
cos(f*x+exp(1))))^2-2*b*a^2*1/4/(b^3+3*b^2*a+3*b*a^2+a^3)/sqrt(a*b)*atan((-a*cos(f*x+exp(1))+b)/(sqrt(a*b)*cos
(f*x+exp(1))+sqrt(a*b)))+(-b^2-6*b*a+3*a^2)/(32*b^3+96*b^2*a+96*b*a^2+32*a^3)*ln(abs(1-cos(f*x+exp(1)))/abs(1+
cos(f*x+exp(1)))))

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maple [B]  time = 0.96, size = 296, normalized size = 2.29 \[ \frac {a^{2} b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \left (a +b \right )^{3} \sqrt {a b}}-\frac {1}{2 f \left (8 a +8 b \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}}+\frac {3 a}{16 f \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}-\frac {b}{16 f \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) a^{2}}{16 f \left (a +b \right )^{3}}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) a b}{8 f \left (a +b \right )^{3}}-\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{16 f \left (a +b \right )^{3}}+\frac {1}{2 f \left (8 a +8 b \right ) \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {3 a}{16 f \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {b}{16 f \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) a^{2}}{16 f \left (a +b \right )^{3}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) a b}{8 f \left (a +b \right )^{3}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{16 f \left (a +b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*sec(f*x+e)^2),x)

[Out]

1/f*a^2*b/(a+b)^3/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))-1/2/f/(8*a+8*b)/(-1+cos(f*x+e))^2+3/16/f/(a+b)^
2/(-1+cos(f*x+e))*a-1/16/f/(a+b)^2/(-1+cos(f*x+e))*b+3/16/f/(a+b)^3*ln(-1+cos(f*x+e))*a^2-3/8/f/(a+b)^3*ln(-1+
cos(f*x+e))*a*b-1/16/f/(a+b)^3*ln(-1+cos(f*x+e))*b^2+1/2/f/(8*a+8*b)/(1+cos(f*x+e))^2+3/16/f/(a+b)^2/(1+cos(f*
x+e))*a-1/16/f/(a+b)^2/(1+cos(f*x+e))*b-3/16/f/(a+b)^3*ln(1+cos(f*x+e))*a^2+3/8/f/(a+b)^3*ln(1+cos(f*x+e))*a*b
+1/16/f/(a+b)^3*ln(1+cos(f*x+e))*b^2

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maxima [B]  time = 0.46, size = 231, normalized size = 1.79 \[ \frac {\frac {16 \, a^{2} b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {{\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left ({\left (3 \, a - b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a + b\right )} \cos \left (f x + e\right )\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/16*(16*a^2*b*arctan(a*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 -
b^2)*log(cos(f*x + e) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (3*a^2 - 6*a*b - b^2)*log(cos(f*x + e) - 1)/(a^3
+ 3*a^2*b + 3*a*b^2 + b^3) + 2*((3*a - b)*cos(f*x + e)^3 - (5*a + b)*cos(f*x + e))/((a^2 + 2*a*b + b^2)*cos(f
*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2))/f

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mupad [B]  time = 7.90, size = 870, normalized size = 6.74 \[ \frac {3\,a^2\,{\cos \left (e+f\,x\right )}^3-b^2\,\cos \left (e+f\,x\right )-5\,a^2\,\cos \left (e+f\,x\right )-b^2\,{\cos \left (e+f\,x\right )}^3-3\,a^2\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )+b^2\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )-6\,a\,b\,\cos \left (e+f\,x\right )+6\,a^2\,{\cos \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )-3\,a^2\,{\cos \left (e+f\,x\right )}^4\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )-2\,b^2\,{\cos \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )+b^2\,{\cos \left (e+f\,x\right )}^4\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )+2\,a\,b\,{\cos \left (e+f\,x\right )}^3+6\,a\,b\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )-12\,a\,b\,{\cos \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )+6\,a\,b\,{\cos \left (e+f\,x\right )}^4\,\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )+\mathrm {atan}\left (\frac {a^5\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,9{}\mathrm {i}+a^2\,b^3\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,12{}\mathrm {i}+a^3\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,30{}\mathrm {i}+a\,b^4\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,1{}\mathrm {i}+a^4\,b\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,28{}\mathrm {i}}{9\,a^6\,b+28\,a^5\,b^2+30\,a^4\,b^3+12\,a^3\,b^4+a^2\,b^5}\right )\,\sqrt {-a^3\,b}\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {a^5\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,9{}\mathrm {i}+a^2\,b^3\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,12{}\mathrm {i}+a^3\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,30{}\mathrm {i}+a\,b^4\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,1{}\mathrm {i}+a^4\,b\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,28{}\mathrm {i}}{9\,a^6\,b+28\,a^5\,b^2+30\,a^4\,b^3+12\,a^3\,b^4+a^2\,b^5}\right )\,{\cos \left (e+f\,x\right )}^2\,\sqrt {-a^3\,b}\,16{}\mathrm {i}+\mathrm {atan}\left (\frac {a^5\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,9{}\mathrm {i}+a^2\,b^3\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,12{}\mathrm {i}+a^3\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,30{}\mathrm {i}+a\,b^4\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,1{}\mathrm {i}+a^4\,b\,\cos \left (e+f\,x\right )\,\sqrt {-a^3\,b}\,28{}\mathrm {i}}{9\,a^6\,b+28\,a^5\,b^2+30\,a^4\,b^3+12\,a^3\,b^4+a^2\,b^5}\right )\,{\cos \left (e+f\,x\right )}^4\,\sqrt {-a^3\,b}\,8{}\mathrm {i}}{8\,f\,a^3\,{\cos \left (e+f\,x\right )}^4-16\,f\,a^3\,{\cos \left (e+f\,x\right )}^2+8\,f\,a^3+24\,f\,a^2\,b\,{\cos \left (e+f\,x\right )}^4-48\,f\,a^2\,b\,{\cos \left (e+f\,x\right )}^2+24\,f\,a^2\,b+24\,f\,a\,b^2\,{\cos \left (e+f\,x\right )}^4-48\,f\,a\,b^2\,{\cos \left (e+f\,x\right )}^2+24\,f\,a\,b^2+8\,f\,b^3\,{\cos \left (e+f\,x\right )}^4-16\,f\,b^3\,{\cos \left (e+f\,x\right )}^2+8\,f\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)),x)

[Out]

(atan((a^5*cos(e + f*x)*(-a^3*b)^(1/2)*9i + a^2*b^3*cos(e + f*x)*(-a^3*b)^(1/2)*12i + a^3*b^2*cos(e + f*x)*(-a
^3*b)^(1/2)*30i + a*b^4*cos(e + f*x)*(-a^3*b)^(1/2)*1i + a^4*b*cos(e + f*x)*(-a^3*b)^(1/2)*28i)/(9*a^6*b + a^2
*b^5 + 12*a^3*b^4 + 30*a^4*b^3 + 28*a^5*b^2))*(-a^3*b)^(1/2)*8i - 5*a^2*cos(e + f*x) - b^2*cos(e + f*x) + 3*a^
2*cos(e + f*x)^3 - b^2*cos(e + f*x)^3 - 3*a^2*atanh(cos(e + f*x)) + b^2*atanh(cos(e + f*x)) - atan((a^5*cos(e
+ f*x)*(-a^3*b)^(1/2)*9i + a^2*b^3*cos(e + f*x)*(-a^3*b)^(1/2)*12i + a^3*b^2*cos(e + f*x)*(-a^3*b)^(1/2)*30i +
a*b^4*cos(e + f*x)*(-a^3*b)^(1/2)*1i + a^4*b*cos(e + f*x)*(-a^3*b)^(1/2)*28i)/(9*a^6*b + a^2*b^5 + 12*a^3*b^4
+ 30*a^4*b^3 + 28*a^5*b^2))*cos(e + f*x)^2*(-a^3*b)^(1/2)*16i + atan((a^5*cos(e + f*x)*(-a^3*b)^(1/2)*9i + a^
2*b^3*cos(e + f*x)*(-a^3*b)^(1/2)*12i + a^3*b^2*cos(e + f*x)*(-a^3*b)^(1/2)*30i + a*b^4*cos(e + f*x)*(-a^3*b)^
(1/2)*1i + a^4*b*cos(e + f*x)*(-a^3*b)^(1/2)*28i)/(9*a^6*b + a^2*b^5 + 12*a^3*b^4 + 30*a^4*b^3 + 28*a^5*b^2))*
cos(e + f*x)^4*(-a^3*b)^(1/2)*8i - 6*a*b*cos(e + f*x) + 6*a^2*cos(e + f*x)^2*atanh(cos(e + f*x)) - 3*a^2*cos(e
+ f*x)^4*atanh(cos(e + f*x)) - 2*b^2*cos(e + f*x)^2*atanh(cos(e + f*x)) + b^2*cos(e + f*x)^4*atanh(cos(e + f*
x)) + 2*a*b*cos(e + f*x)^3 + 6*a*b*atanh(cos(e + f*x)) - 12*a*b*cos(e + f*x)^2*atanh(cos(e + f*x)) + 6*a*b*cos
(e + f*x)^4*atanh(cos(e + f*x)))/(8*a^3*f + 8*b^3*f - 16*a^3*f*cos(e + f*x)^2 + 8*a^3*f*cos(e + f*x)^4 - 16*b^
3*f*cos(e + f*x)^2 + 8*b^3*f*cos(e + f*x)^4 + 24*a*b^2*f + 24*a^2*b*f - 48*a*b^2*f*cos(e + f*x)^2 - 48*a^2*b*f
*cos(e + f*x)^2 + 24*a*b^2*f*cos(e + f*x)^4 + 24*a^2*b*f*cos(e + f*x)^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**5/(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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