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

Optimal. Leaf size=110 \[ -\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^2 f (a-b)^{3/2}}-\frac {\tanh ^{-1}(\cos (e+f x))}{a^2 f}-\frac {b \sec (e+f x)}{2 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 110, 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 = {3664, 414, 522, 207, 205} \[ -\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^2 f (a-b)^{3/2}}-\frac {\tanh ^{-1}(\cos (e+f x))}{a^2 f}-\frac {b \sec (e+f x)}{2 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a - 2*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*a^2*(a - b)^(3/2)*f) - ArcTanh[Cos[e + f*
x]]/(a^2*f) - (b*Sec[e + f*x])/(2*a*(a - b)*f*(a - b + b*Sec[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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, 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 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.90, size = 184, normalized size = 1.67 \[ \frac {-\frac {2 a b \cos (e+f x)}{(a-b) ((a-b) \cos (2 (e+f x))+a+b)}+\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{3/2}}+\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{3/2}}+2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*a - 2*b)*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(3/2) + ((3*a - 2*b)*S
qrt[b]*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(3/2) - (2*a*b*Cos[e + f*x])/((a - b)
*(a + b + (a - b)*Cos[2*(e + f*x)])) - 2*Log[Cos[(e + f*x)/2]] + 2*Log[Sin[(e + f*x)/2]])/(2*a^2*f)

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fricas [B]  time = 0.62, size = 470, normalized size = 4.27 \[ \left [-\frac {2 \, a b \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b - 2 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 2 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - a^{2} b^{2}\right )} f\right )}}, -\frac {a b \cos \left (f x + e\right ) + {\left ({\left (3 \, a^{2} - 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b - 2 \, b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - a^{2} b^{2}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*b*cos(f*x + e) - ((3*a^2 - 5*a*b + 2*b^2)*cos(f*x + e)^2 + 3*a*b - 2*b^2)*sqrt(-b/(a - b))*log(((a
- b)*cos(f*x + e)^2 + 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 2*((a^2 - 2
*a*b + b^2)*cos(f*x + e)^2 + a*b - b^2)*log(1/2*cos(f*x + e) + 1/2) - 2*((a^2 - 2*a*b + b^2)*cos(f*x + e)^2 +
a*b - b^2)*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 - a^2*b^2)*f), -
1/2*(a*b*cos(f*x + e) + ((3*a^2 - 5*a*b + 2*b^2)*cos(f*x + e)^2 + 3*a*b - 2*b^2)*sqrt(b/(a - b))*arctan(-(a -
b)*sqrt(b/(a - b))*cos(f*x + e)/b) + ((a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a*b - b^2)*log(1/2*cos(f*x + e) + 1
/2) - ((a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a*b - b^2)*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 - a^2*b^2)*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*tan(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*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))
+(-3*a*b+2*b^2)*1/4/(a^3-a^2*b)/sqrt(-b^2+a*b)*atan((-a*cos(f*x+exp(1))+b*cos(f*x+exp(1))+b)/(sqrt(-b^2+a*b)*c
os(f*x+exp(1))+sqrt(-b^2+a*b)))+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+
exp(1)))*b^2-a*b)/(2*a^3-2*a^2*b)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(
f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a))

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maple [A]  time = 0.64, size = 179, normalized size = 1.63 \[ -\frac {b \cos \left (f x +e \right )}{2 f a \left (a -b \right ) \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {3 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f a \left (a -b \right ) \sqrt {\left (a -b \right ) b}}-\frac {b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{2} \left (a -b \right ) \sqrt {\left (a -b \right ) b}}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \,a^{2}}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 13.72, size = 1140, normalized size = 10.36 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f}-\frac {\frac {b}{a\,\left (a-b\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a\,b-2\,b^2\right )}{a^2\,\left (a-b\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (4\,b-2\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\left (\frac {b^{3/2}\,{\left (3\,a-2\,b\right )}^3\,\left (2\,a^{10}-58\,a^9\,b+306\,a^8\,b^2-686\,a^7\,b^3+772\,a^6\,b^4-432\,a^5\,b^5+96\,a^4\,b^6\right )}{8\,a^6\,{\left (a-b\right )}^{9/2}\,\left (-a^5+3\,a^4\,b-3\,a^3\,b^2+a^2\,b^3\right )}+\frac {2\,\sqrt {b}\,\left (3\,a-2\,b\right )\,\left (9\,a^5\,b-63\,a^4\,b^2+158\,a^3\,b^3-188\,a^2\,b^4+108\,a\,b^5-24\,b^6\right )}{a^2\,{\left (a-b\right )}^{3/2}\,\left (-a^5+3\,a^4\,b-3\,a^3\,b^2+a^2\,b^3\right )}\right )\,\left (27\,a^5-259\,a^4\,b+820\,a^3\,b^2-1164\,a^2\,b^3+768\,a\,b^4-192\,b^5\right )}{2\,a^5\,{\left (a-b\right )}^{9/2}\,\left (16\,a^3-39\,a^2\,b+36\,a\,b^2-12\,b^3\right )}-\frac {\left (\frac {8\,\left (9\,a^2\,b^2-12\,a\,b^3+4\,b^4\right )}{-a^5+3\,a^4\,b-3\,a^3\,b^2+a^2\,b^3}-\frac {b\,{\left (3\,a-2\,b\right )}^2\,\left (2\,a^8-35\,a^7\,b+234\,a^6\,b^2-611\,a^5\,b^3+746\,a^4\,b^4-432\,a^3\,b^5+96\,a^2\,b^6\right )}{2\,a^4\,{\left (a-b\right )}^3\,\left (-a^5+3\,a^4\,b-3\,a^3\,b^2+a^2\,b^3\right )}\right )\,\left (2\,a^4-47\,a^3\,b+186\,a^2\,b^2-240\,a\,b^3+96\,b^4\right )}{a^5\,\sqrt {b}\,{\left (a-b\right )}^3\,\left (16\,a^3-39\,a^2\,b+36\,a\,b^2-12\,b^3\right )}\right )+\frac {\left (\frac {\sqrt {b}\,\left (3\,a-2\,b\right )\,\left (12\,a^5\,b-53\,a^4\,b^2+60\,a^3\,b^3-20\,a^2\,b^4\right )}{a^2\,{\left (a-b\right )}^{3/2}\,\left (a^5-2\,a^4\,b+a^3\,b^2\right )}+\frac {b^{3/2}\,{\left (3\,a-2\,b\right )}^3\,\left (4\,a^{10}-24\,a^9\,b+52\,a^8\,b^2-48\,a^7\,b^3+16\,a^6\,b^4\right )}{16\,a^6\,{\left (a-b\right )}^{9/2}\,\left (a^5-2\,a^4\,b+a^3\,b^2\right )}\right )\,\left (27\,a^5-259\,a^4\,b+820\,a^3\,b^2-1164\,a^2\,b^3+768\,a\,b^4-192\,b^5\right )}{2\,a^5\,{\left (a-b\right )}^{9/2}\,\left (16\,a^3-39\,a^2\,b+36\,a\,b^2-12\,b^3\right )}-\frac {\left (\frac {4\,\left (9\,a^2\,b^2-12\,a\,b^3+4\,b^4\right )}{a^5-2\,a^4\,b+a^3\,b^2}-\frac {b\,{\left (3\,a-2\,b\right )}^2\,\left (4\,a^8-36\,a^7\,b+96\,a^6\,b^2-96\,a^5\,b^3+32\,a^4\,b^4\right )}{4\,a^4\,{\left (a-b\right )}^3\,\left (a^5-2\,a^4\,b+a^3\,b^2\right )}\right )\,\left (2\,a^4-47\,a^3\,b+186\,a^2\,b^2-240\,a\,b^3+96\,b^4\right )}{a^5\,\sqrt {b}\,{\left (a-b\right )}^3\,\left (16\,a^3-39\,a^2\,b+36\,a\,b^2-12\,b^3\right )}\right )\,\left (4\,a^7\,{\left (a-b\right )}^{9/2}-12\,a^6\,b\,{\left (a-b\right )}^{9/2}-4\,a^4\,b^3\,{\left (a-b\right )}^{9/2}+12\,a^5\,b^2\,{\left (a-b\right )}^{9/2}\right )}{9\,a^2\,b-12\,a\,b^2+4\,b^3}\right )\,\left (3\,a-2\,b\right )}{2\,a^2\,f\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(e/2 + (f*x)/2))/(a^2*f) - (b/(a*(a - b)) - (tan(e/2 + (f*x)/2)^2*(a*b - 2*b^2))/(a^2*(a - b)))/(f*(a -
tan(e/2 + (f*x)/2)^2*(2*a - 4*b) + a*tan(e/2 + (f*x)/2)^4)) + (b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*((((b^(3/2
)*(3*a - 2*b)^3*(2*a^10 - 58*a^9*b + 96*a^4*b^6 - 432*a^5*b^5 + 772*a^6*b^4 - 686*a^7*b^3 + 306*a^8*b^2))/(8*a
^6*(a - b)^(9/2)*(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2)) + (2*b^(1/2)*(3*a - 2*b)*(108*a*b^5 + 9*a^5*b - 24*b^6
- 188*a^2*b^4 + 158*a^3*b^3 - 63*a^4*b^2))/(a^2*(a - b)^(3/2)*(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2)))*(768*a*
b^4 - 259*a^4*b + 27*a^5 - 192*b^5 - 1164*a^2*b^3 + 820*a^3*b^2))/(2*a^5*(a - b)^(9/2)*(36*a*b^2 - 39*a^2*b +
16*a^3 - 12*b^3)) - (((8*(4*b^4 - 12*a*b^3 + 9*a^2*b^2))/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (b*(3*a - 2*b
)^2*(2*a^8 - 35*a^7*b + 96*a^2*b^6 - 432*a^3*b^5 + 746*a^4*b^4 - 611*a^5*b^3 + 234*a^6*b^2))/(2*a^4*(a - b)^3*
(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2)))*(2*a^4 - 47*a^3*b - 240*a*b^3 + 96*b^4 + 186*a^2*b^2))/(a^5*b^(1/2)*(a
- b)^3*(36*a*b^2 - 39*a^2*b + 16*a^3 - 12*b^3))) + (((b^(1/2)*(3*a - 2*b)*(12*a^5*b - 20*a^2*b^4 + 60*a^3*b^3
- 53*a^4*b^2))/(a^2*(a - b)^(3/2)*(a^5 - 2*a^4*b + a^3*b^2)) + (b^(3/2)*(3*a - 2*b)^3*(4*a^10 - 24*a^9*b + 16
*a^6*b^4 - 48*a^7*b^3 + 52*a^8*b^2))/(16*a^6*(a - b)^(9/2)*(a^5 - 2*a^4*b + a^3*b^2)))*(768*a*b^4 - 259*a^4*b
+ 27*a^5 - 192*b^5 - 1164*a^2*b^3 + 820*a^3*b^2))/(2*a^5*(a - b)^(9/2)*(36*a*b^2 - 39*a^2*b + 16*a^3 - 12*b^3)
) - (((4*(4*b^4 - 12*a*b^3 + 9*a^2*b^2))/(a^5 - 2*a^4*b + a^3*b^2) - (b*(3*a - 2*b)^2*(4*a^8 - 36*a^7*b + 32*a
^4*b^4 - 96*a^5*b^3 + 96*a^6*b^2))/(4*a^4*(a - b)^3*(a^5 - 2*a^4*b + a^3*b^2)))*(2*a^4 - 47*a^3*b - 240*a*b^3
+ 96*b^4 + 186*a^2*b^2))/(a^5*b^(1/2)*(a - b)^3*(36*a*b^2 - 39*a^2*b + 16*a^3 - 12*b^3)))*(4*a^7*(a - b)^(9/2)
- 12*a^6*b*(a - b)^(9/2) - 4*a^4*b^3*(a - b)^(9/2) + 12*a^5*b^2*(a - b)^(9/2)))/(9*a^2*b - 12*a*b^2 + 4*b^3))
*(3*a - 2*b))/(2*a^2*f*(a - b)^(3/2))

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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*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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