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

Optimal. Leaf size=147 \[ \frac {(a-b) \cos (e+f x)}{2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {\sqrt {b} (3 a-b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 \sqrt {a} f (a+b)^3}-\frac {(a-3 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^3}-\frac {\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )} \]

[Out]

-1/2*(a-3*b)*arctanh(cos(f*x+e))/(a+b)^3/f+1/2*(a-b)*cos(f*x+e)/(a+b)^2/f/(b+a*cos(f*x+e)^2)-1/2*cot(f*x+e)*cs
c(f*x+e)/(a+b)/f/(b+a*cos(f*x+e)^2)+1/2*(3*a-b)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/(a+b)^3/f/a^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 147, 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, 470, 527, 522, 206, 205} \[ \frac {(a-b) \cos (e+f x)}{2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {\sqrt {b} (3 a-b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 \sqrt {a} f (a+b)^3}-\frac {(a-3 b) \tanh ^{-1}(\cos (e+f x))}{2 f (a+b)^3}-\frac {\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a \cos ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 1.94, size = 468, normalized size = 3.18 \[ \frac {\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left ((a+b) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)-4 (a-3 b) \sec (e+f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-\left ((a+b) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)\right )+4 (a-3 b) \sec (e+f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-\frac {4 \sqrt {b} (b-3 a) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{\sqrt {a}}-\frac {4 \sqrt {b} (b-3 a) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{\sqrt {a}}-8 b (a+b)\right )}{32 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(-8*b*(a + b) - (4*Sqrt[b]*(-3*a + b)*ArcTan[((-Sqrt[a] - I*Sqr
t[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)])*Sec[e + f*x])/Sqrt[a] - (4*Sqrt[b]*(-3*a + b)*A
rcTan[((-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)])*Sec[e + f*x])/Sqrt[a] -
(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[(e + f*x)/2]^2*Sec[e + f*x] - 4*(a - 3*b)*(a + 2*b + a*Cos[2*(e +
f*x)])*Log[Cos[(e + f*x)/2]]*Sec[e + f*x] + 4*(a - 3*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Sin[(e + f*x)/2]]*S
ec[e + f*x] + (a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^2*Sec[e + f*x]))/(32*(a + b)^3*f*(a + b*
Sec[e + f*x]^2)^2)

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fricas [B]  time = 0.59, size = 698, normalized size = 4.75 \[ \left [\frac {2 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left ({\left (3 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a \cos \left (f x + e\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 4 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )}}, \frac {2 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left ({\left (3 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 4 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )}}\right ] \]

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]

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

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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)^3/(a+b*sec(f*x+e)^2)^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*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))/(16*b^2+32*b*a+16
*a^2)+(6*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b^2+4*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b*a-2*(
(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^2-15*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2+2*((1-cos(f*
x+exp(1)))/(1+cos(f*x+exp(1))))^2*b*a+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2-24*(1-cos(f*x+exp(1)))/(
1+cos(f*x+exp(1)))*b^2-20*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
)*a^2-3*b^2-6*b*a-3*a^2)/(48*b^3+144*b^2*a+144*b*a^2+48*a^3)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b+((
1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a+2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b-2*((1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1))))^2*a+(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
)*a)+(-3*b+a)/(8*b^3+24*b^2*a+24*b*a^2+8*a^3)*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))+(b^2-3*b*a)*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))))

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maple [A]  time = 1.31, size = 250, normalized size = 1.70 \[ -\frac {b \cos \left (f x +e \right ) a}{2 f \left (a +b \right )^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{2} \cos \left (f x +e \right )}{2 f \left (a +b \right )^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {3 b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right ) a}{2 f \left (a +b \right )^{3} \sqrt {a b}}-\frac {b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \left (a +b \right )^{3} \sqrt {a b}}+\frac {1}{4 f \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) a}{4 f \left (a +b \right )^{3}}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{4 f \left (a +b \right )^{3}}+\frac {1}{4 f \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (1+\cos \left (f x +e \right )\right ) a}{4 f \left (a +b \right )^{3}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{4 f \left (a +b \right )^{3}} \]

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*b/(a+b)^3*cos(f*x+e)/(b+a*cos(f*x+e)^2)*a-1/2/f*b^2/(a+b)^3*cos(f*x+e)/(b+a*cos(f*x+e)^2)+3/2/f*b/(a+b)
^3/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))*a-1/2/f*b^2/(a+b)^3/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2
))+1/4/f/(a+b)^2/(-1+cos(f*x+e))+1/4/f/(a+b)^3*ln(-1+cos(f*x+e))*a-3/4/f/(a+b)^3*ln(-1+cos(f*x+e))*b+1/4/f/(a+
b)^2/(1+cos(f*x+e))-1/4/f/(a+b)^3*ln(1+cos(f*x+e))*a+3/4/f/(a+b)^3*ln(1+cos(f*x+e))*b

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maxima [A]  time = 0.45, size = 231, normalized size = 1.57 \[ -\frac {\frac {{\left (a - 3 \, b\right )} \log \left (\cos \left (f x + e\right ) + 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {{\left (a - 3 \, b\right )} \log \left (\cos \left (f x + e\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (3 \, a b - b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} - \frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{4} - a^{2} b - 2 \, a b^{2} - b^{3} - {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{2}}}{4 \, f} \]

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]

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

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mupad [B]  time = 5.65, size = 1845, normalized size = 12.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^2),x)

[Out]

- ((cos(e + f*x)^3*(a - b))/(2*(2*a*b + a^2 + b^2)) + (b*cos(e + f*x))/(2*a*b + a^2 + b^2))/(f*(b - a*cos(e +
f*x)^4 + cos(e + f*x)^2*(a - b))) - (log(cos(e + f*x) - 1)*(b/(a + b)^3 - 1/(4*(a + b)^2)))/f - (log(cos(e + f
*x) + 1)*(a - 3*b))/(4*f*(a + b)^3) - (atan((((-a*b)^(1/2)*((cos(e + f*x)*(a*b^4 - 6*a^4*b + a^5 - 6*a^2*b^3 +
18*a^3*b^2))/(2*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) + ((-a*b)^(1/2)*((4*a^8*b + 4*a^2*b^7 + 24*a^3*b
^6 + 60*a^4*b^5 + 80*a^5*b^4 + 60*a^6*b^3 + 24*a^7*b^2)/(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b
^3 + 15*a^4*b^2) - (cos(e + f*x)*(-a*b)^(1/2)*(3*a - b)*(80*a^8*b + 16*a^9 - 16*a^2*b^7 - 80*a^3*b^6 - 144*a^4
*b^5 - 80*a^5*b^4 + 80*a^6*b^3 + 144*a^7*b^2))/(8*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)*(4*a*b^3 + 4*a^3*b + a^4
+ b^4 + 6*a^2*b^2)))*(3*a - b))/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)))*(3*a - b)*1i)/(4*(a*b^3 + 3*a^3*b +
a^4 + 3*a^2*b^2)) + ((-a*b)^(1/2)*((cos(e + f*x)*(a*b^4 - 6*a^4*b + a^5 - 6*a^2*b^3 + 18*a^3*b^2))/(2*(4*a*b^3
+ 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) - ((-a*b)^(1/2)*((4*a^8*b + 4*a^2*b^7 + 24*a^3*b^6 + 60*a^4*b^5 + 80*a^5*
b^4 + 60*a^6*b^3 + 24*a^7*b^2)/(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2) + (cos(e
+ f*x)*(-a*b)^(1/2)*(3*a - b)*(80*a^8*b + 16*a^9 - 16*a^2*b^7 - 80*a^3*b^6 - 144*a^4*b^5 - 80*a^5*b^4 + 80*a^
6*b^3 + 144*a^7*b^2))/(8*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)))*(3*
a - b))/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)))*(3*a - b)*1i)/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)))/(((3*a
*b^4)/4 - (3*a^4*b)/4 - (13*a^2*b^3)/4 + (13*a^3*b^2)/4)/(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*
b^3 + 15*a^4*b^2) + ((-a*b)^(1/2)*((cos(e + f*x)*(a*b^4 - 6*a^4*b + a^5 - 6*a^2*b^3 + 18*a^3*b^2))/(2*(4*a*b^3
+ 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) + ((-a*b)^(1/2)*((4*a^8*b + 4*a^2*b^7 + 24*a^3*b^6 + 60*a^4*b^5 + 80*a^5*
b^4 + 60*a^6*b^3 + 24*a^7*b^2)/(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2) - (cos(e
+ f*x)*(-a*b)^(1/2)*(3*a - b)*(80*a^8*b + 16*a^9 - 16*a^2*b^7 - 80*a^3*b^6 - 144*a^4*b^5 - 80*a^5*b^4 + 80*a^
6*b^3 + 144*a^7*b^2))/(8*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)))*(3*
a - b))/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)))*(3*a - b))/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)) - ((-a*b)^
(1/2)*((cos(e + f*x)*(a*b^4 - 6*a^4*b + a^5 - 6*a^2*b^3 + 18*a^3*b^2))/(2*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a
^2*b^2)) - ((-a*b)^(1/2)*((4*a^8*b + 4*a^2*b^7 + 24*a^3*b^6 + 60*a^4*b^5 + 80*a^5*b^4 + 60*a^6*b^3 + 24*a^7*b^
2)/(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2) + (cos(e + f*x)*(-a*b)^(1/2)*(3*a -
b)*(80*a^8*b + 16*a^9 - 16*a^2*b^7 - 80*a^3*b^6 - 144*a^4*b^5 - 80*a^5*b^4 + 80*a^6*b^3 + 144*a^7*b^2))/(8*(a*
b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)))*(3*a - b))/(4*(a*b^3 + 3*a^3*b
+ a^4 + 3*a^2*b^2)))*(3*a - b))/(4*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2))))*(-a*b)^(1/2)*(3*a - b)*1i)/(2*f*(a*b
^3 + 3*a^3*b + a^4 + 3*a^2*b^2))

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

[Out]

Timed out

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