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

Optimal. Leaf size=99 \[ \frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{3/2} f (a+b)^2}-\frac {b \cos (e+f x)}{2 a f (a+b) \left (a \cos ^2(e+f x)+b\right )}-\frac {\tanh ^{-1}(\cos (e+f x))}{f (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4133, 470, 522, 206, 205} \[ \frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{3/2} f (a+b)^2}-\frac {b \cos (e+f x)}{2 a f (a+b) \left (a \cos ^2(e+f x)+b\right )}-\frac {\tanh ^{-1}(\cos (e+f x))}{f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*a^(3/2)*(a + b)^2*f) - ArcTanh[Cos[e + f*x]]/((a
+ b)^2*f) - (b*Cos[e + f*x])/(2*a*(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 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 (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 ) \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {b \cos (e+f x)}{2 a (a+b) f \left (b+a \cos ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {b+(-2 a-b) x^2}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{2 a (a+b) f}\\ &=-\frac {b \cos (e+f x)}{2 a (a+b) f \left (b+a \cos ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{(a+b)^2 f}+\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{2 a (a+b)^2 f}\\ &=\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{3/2} (a+b)^2 f}-\frac {\tanh ^{-1}(\cos (e+f x))}{(a+b)^2 f}-\frac {b \cos (e+f x)}{2 a (a+b) f \left (b+a \cos ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 1.27, size = 384, normalized size = 3.88 \[ \frac {\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac {\sqrt {b} (3 a+b) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{a^{3/2}}+\frac {\sqrt {b} (3 a+b) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{a^{3/2}}-2 \sec (e+f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)+2 \sec (e+f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a \cos (2 (e+f x))+a+2 b)-\frac {2 b (a+b)}{a}\right )}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*((-2*b*(a + b))/a + (Sqrt[b]*(3*a + b)*ArcTan[((-Sqrt[a] - I*Sq
rt[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*Si
n[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x])/a^(3/2) + (Sqrt[b]*(3*a + b)*Arc
Tan[((-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])/a^(3/2) - 2
*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2]]*Sec[e + f*x] + 2*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Sin[
(e + f*x)/2]]*Sec[e + f*x]))/(8*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 1.04, size = 390, normalized size = 3.94 \[ \left [\frac {{\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) - 2 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac {{\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) - {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) - {\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((3*a^2 + a*b)*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)) - 2*(a*b + b^2)*cos(f*x + e) - 2*(a^2*cos(f*x + e)^2 + a*b)*log(1/2*cos(f*x
+ e) + 1/2) + 2*(a^2*cos(f*x + e)^2 + a*b)*log(-1/2*cos(f*x + e) + 1/2))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x
+ e)^2 + (a^3*b + 2*a^2*b^2 + a*b^3)*f), 1/2*(((3*a^2 + a*b)*cos(f*x + e)^2 + 3*a*b + b^2)*sqrt(b/a)*arctan(a*
sqrt(b/a)*cos(f*x + e)/b) - (a*b + b^2)*cos(f*x + e) - (a^2*cos(f*x + e)^2 + a*b)*log(1/2*cos(f*x + e) + 1/2)
+ (a^2*cos(f*x + e)^2 + a*b)*log(-1/2*cos(f*x + e) + 1/2))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^3*
b + 2*a^2*b^2 + a*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)/(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/(4*a^2+8*a*b+4*b^2)*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos
(f*x+exp(1))))+(-3*a*b-b^2)*1/4/(a^3+2*a^2*b+a*b^2)/sqrt(a*b)*atan((-a*cos(f*x+exp(1))+b)/(sqrt(a*b)*cos(f*x+e
xp(1))+sqrt(a*b)))+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b-(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2-a*
b-b^2)/(2*a^3+4*a^2*b+2*a*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))))^2*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a+b))

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maple [A]  time = 1.19, size = 172, normalized size = 1.74 \[ -\frac {b \cos \left (f x +e \right )}{2 f \left (a +b \right )^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{2} \cos \left (f x +e \right )}{2 f \left (a +b \right )^{2} a \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {3 b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \left (a +b \right )^{2} \sqrt {a b}}+\frac {b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \left (a +b \right )^{2} a \sqrt {a b}}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \left (a +b \right )^{2}}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 138, normalized size = 1.39 \[ -\frac {\frac {b \cos \left (f x + e\right )}{a^{2} b + a b^{2} + {\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{2}} - \frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} + \frac {\log \left (\cos \left (f x + e\right ) + 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {\log \left (\cos \left (f x + e\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.95, size = 2188, normalized size = 22.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((3*a + b)*(-a^3*b)^(1/2)*((cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 + 9*a^2*b^2))/(2*(a*b^2 + 2*a^2*b + a^3
)) + ((3*a + b)*(-a^3*b)^(1/2)*((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a^5*b^2)/(a*b^3 + 3*a^3*b +
a^4 + 3*a^2*b^2) - (cos(e + f*x)*(3*a + b)*(-a^3*b)^(1/2)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a^4*b^4 - 32*a^
5*b^3 + 32*a^6*b^2))/(8*(a*b^2 + 2*a^2*b + a^3)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*a^4*b + a^5 + a^3*b^2)))*1i
)/(4*(2*a^4*b + a^5 + a^3*b^2)) + ((3*a + b)*(-a^3*b)^(1/2)*((cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 + 9*a^2*b^2)
)/(2*(a*b^2 + 2*a^2*b + a^3)) - ((3*a + b)*(-a^3*b)^(1/2)*((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a
^5*b^2)/(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2) + (cos(e + f*x)*(3*a + b)*(-a^3*b)^(1/2)*(48*a^7*b + 16*a^8 - 16*a
^3*b^5 - 48*a^4*b^4 - 32*a^5*b^3 + 32*a^6*b^2))/(8*(a*b^2 + 2*a^2*b + a^3)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*
a^4*b + a^5 + a^3*b^2)))*1i)/(4*(2*a^4*b + a^5 + a^3*b^2)))/(((5*a*b^2)/2 + 3*a^2*b + b^3/2)/(a*b^3 + 3*a^3*b
+ a^4 + 3*a^2*b^2) - ((3*a + b)*(-a^3*b)^(1/2)*((cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 + 9*a^2*b^2))/(2*(a*b^2 +
2*a^2*b + a^3)) + ((3*a + b)*(-a^3*b)^(1/2)*((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a^5*b^2)/(a*b^
3 + 3*a^3*b + a^4 + 3*a^2*b^2) - (cos(e + f*x)*(3*a + b)*(-a^3*b)^(1/2)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a
^4*b^4 - 32*a^5*b^3 + 32*a^6*b^2))/(8*(a*b^2 + 2*a^2*b + a^3)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*a^4*b + a^5 +
a^3*b^2))))/(4*(2*a^4*b + a^5 + a^3*b^2)) + ((3*a + b)*(-a^3*b)^(1/2)*((cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 +
9*a^2*b^2))/(2*(a*b^2 + 2*a^2*b + a^3)) - ((3*a + b)*(-a^3*b)^(1/2)*((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^
4*b^3 + 8*a^5*b^2)/(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2) + (cos(e + f*x)*(3*a + b)*(-a^3*b)^(1/2)*(48*a^7*b + 16
*a^8 - 16*a^3*b^5 - 48*a^4*b^4 - 32*a^5*b^3 + 32*a^6*b^2))/(8*(a*b^2 + 2*a^2*b + a^3)*(2*a^4*b + a^5 + a^3*b^2
))))/(4*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*a^4*b + a^5 + a^3*b^2))))*(3*a + b)*(-a^3*b)^(1/2)*1i)/(2*f*(2*a^4*
b + a^5 + a^3*b^2)) - (atan((((((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a^5*b^2)/(2*(a*b^3 + 3*a^3*b
+ a^4 + 3*a^2*b^2)) - (cos(e + f*x)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a^4*b^4 - 32*a^5*b^3 + 32*a^6*b^2))/
(8*(a + b)^2*(a*b^2 + 2*a^2*b + a^3)))*1i)/(2*(a + b)^2) + (cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 + 9*a^2*b^2)*1
i)/(4*(a*b^2 + 2*a^2*b + a^3)))/(a + b)^2 - ((((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a^5*b^2)/(2*(
a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)) + (cos(e + f*x)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a^4*b^4 - 32*a^5*b^3
+ 32*a^6*b^2))/(8*(a + b)^2*(a*b^2 + 2*a^2*b + a^3)))*1i)/(2*(a + b)^2) - (cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4
+ 9*a^2*b^2)*1i)/(4*(a*b^2 + 2*a^2*b + a^3)))/(a + b)^2)/((((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8
*a^5*b^2)/(2*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)) - (cos(e + f*x)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a^4*b^4
- 32*a^5*b^3 + 32*a^6*b^2))/(8*(a + b)^2*(a*b^2 + 2*a^2*b + a^3)))/(2*(a + b)^2) + (cos(e + f*x)*(6*a*b^3 + 4
*a^4 + b^4 + 9*a^2*b^2))/(4*(a*b^2 + 2*a^2*b + a^3)))/(a + b)^2 - ((5*a*b^2)/2 + 3*a^2*b + b^3/2)/(a*b^3 + 3*a
^3*b + a^4 + 3*a^2*b^2) + (((2*a^6*b + 2*a^2*b^5 + 8*a^3*b^4 + 12*a^4*b^3 + 8*a^5*b^2)/(2*(a*b^3 + 3*a^3*b + a
^4 + 3*a^2*b^2)) + (cos(e + f*x)*(48*a^7*b + 16*a^8 - 16*a^3*b^5 - 48*a^4*b^4 - 32*a^5*b^3 + 32*a^6*b^2))/(8*(
a + b)^2*(a*b^2 + 2*a^2*b + a^3)))/(2*(a + b)^2) - (cos(e + f*x)*(6*a*b^3 + 4*a^4 + b^4 + 9*a^2*b^2))/(4*(a*b^
2 + 2*a^2*b + a^3)))/(a + b)^2))*1i)/(f*(a + b)^2) - (b*cos(e + f*x))/(2*a*f*(a + b)*(b + a*cos(e + f*x)^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)/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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