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

Optimal. Leaf size=166 \[ -\frac {\tanh ^{-1}(\cos (e+f x))}{a^3 f}-\frac {b (7 a-4 b) \sec (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^3 f (a-b)^{5/2}}-\frac {b \sec (e+f x)}{4 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

[Out]

-arctanh(cos(f*x+e))/a^3/f-1/4*b*sec(f*x+e)/a/(a-b)/f/(a-b+b*sec(f*x+e)^2)^2-1/8*(7*a-4*b)*b*sec(f*x+e)/a^2/(a
-b)^2/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a^2-20*a*b+8*b^2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^3/(a-b
)^(5/2)/f

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Rubi [A]  time = 0.22, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3664, 414, 527, 522, 207, 205} \[ -\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^3 f (a-b)^{5/2}}-\frac {b (7 a-4 b) \sec (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\tanh ^{-1}(\cos (e+f x))}{a^3 f}-\frac {b \sec (e+f x)}{4 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*a^3*(a - b)^(5/2)*f) - ArcT
anh[Cos[e + f*x]]/(a^3*f) - (b*Sec[e + f*x])/(4*a*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) - ((7*a - 4*b)*b*Sec
[e + f*x])/(8*a^2*(a - b)^2*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 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 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 )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {b \sec (e+f x)}{4 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {4 a-b-3 b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 a (a-b) f}\\ &=-\frac {b \sec (e+f x)}{4 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \sec (e+f x)}{8 a^2 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {8 a^2-9 a b+4 b^2-(7 a-4 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 a^2 (a-b)^2 f}\\ &=-\frac {b \sec (e+f x)}{4 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \sec (e+f x)}{8 a^2 (a-b)^2 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^3 f}-\frac {\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 a^3 (a-b)^2 f}\\ &=-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^3 (a-b)^{5/2} f}-\frac {\tanh ^{-1}(\cos (e+f x))}{a^3 f}-\frac {b \sec (e+f x)}{4 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \sec (e+f x)}{8 a^2 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 3.44, size = 247, normalized size = 1.49 \[ \frac {\frac {8 a^2 b^2 \cos (e+f x)}{(a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}-\frac {2 a b (9 a-4 b) \cos (e+f x)}{(a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)}+8 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-8 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(5/2) +
(Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(5/2) + (
8*a^2*b^2*Cos[e + f*x])/((a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])^2) - (2*a*(9*a - 4*b)*b*Cos[e + f*x])/((
a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])) - 8*Log[Cos[(e + f*x)/2]] + 8*Log[Sin[(e + f*x)/2]])/(8*a^3*f)

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fricas [B]  time = 0.79, size = 1050, normalized size = 6.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/16*(2*(9*a^3*b - 13*a^2*b^2 + 4*a*b^3)*cos(f*x + e)^3 - ((15*a^4 - 50*a^3*b + 63*a^2*b^2 - 36*a*b^3 + 8*b^
4)*cos(f*x + e)^4 + 15*a^2*b^2 - 20*a*b^3 + 8*b^4 + 2*(15*a^3*b - 35*a^2*b^2 + 28*a*b^3 - 8*b^4)*cos(f*x + e)^
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*(7*a^2*b^2 - 4*a*b^3)*cos(f*x + e) + 8*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x
+ e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2)*log(1/2*cos(f*x + e)
+ 1/2) - 8*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b -
3*a^2*b^2 + 3*a*b^3 - b^4)*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)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*
a^4*b^3 + a^3*b^4)*f), -1/8*((9*a^3*b - 13*a^2*b^2 + 4*a*b^3)*cos(f*x + e)^3 + ((15*a^4 - 50*a^3*b + 63*a^2*b^
2 - 36*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 - 20*a*b^3 + 8*b^4 + 2*(15*a^3*b - 35*a^2*b^2 + 28*a*b^3 - 8
*b^4)*cos(f*x + e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + (7*a^2*b^2 - 4*a*b^3)*
cos(f*x + e) + 4*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^
3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) - 4*((a^4 - 4*a^3*b + 6*a^2*b^2 -
4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*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)^4 + 2*(a^6*
b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^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)/(a+b*tan(f*x+e)^2)^3,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^3*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))
+(-15*a^2*b+20*a*b^2-8*b^3)*1/4/(4*a^5-8*a^4*b+4*a^3*b^2)/sqrt(-b^2+a*b)*atan((-a*cos(f*x+exp(1))+b*cos(f*x+ex
p(1))+b)/(sqrt(-b^2+a*b)*cos(f*x+exp(1))+sqrt(-b^2+a*b)))+(9*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^3*b
-28*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^2*b^2+16*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a*b^3-2
7*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3*b+90*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2*b^2-120
*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b^3+48*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^4+27*(1-co
s(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^3*b-68*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2*b^2+32*(1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1)))*a*b^3-9*a^3*b+6*a^2*b^2)/(8*a^5-16*a^4*b+8*a^3*b^2)/(((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)^2)

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maple [B]  time = 0.79, size = 408, normalized size = 2.46 \[ -\frac {9 b \left (\cos ^{3}\left (f x +e \right )\right )}{8 f a \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a -b \right )}+\frac {b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{2 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a -b \right )}-\frac {7 b^{2} \cos \left (f x +e \right )}{8 f a \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {b^{3} \cos \left (f x +e \right )}{2 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {15 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{8 f a \left (a^{2}-2 a b +b^{2}\right ) \sqrt {\left (a -b \right ) b}}-\frac {5 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f \,a^{2} \left (a^{2}-2 a b +b^{2}\right ) \sqrt {\left (a -b \right ) b}}+\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{3} \left (a^{2}-2 a b +b^{2}\right ) \sqrt {\left (a -b \right ) b}}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*tan(f*x+e)^2)^3,x)

[Out]

-9/8/f*b/a/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2/(a-b)*cos(f*x+e)^3+1/2/f*b^2/a^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b
+b)^2/(a-b)*cos(f*x+e)^3-7/8/f*b^2/a/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2/(a^2-2*a*b+b^2)*cos(f*x+e)+1/2/f*b^3/
a^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2/(a^2-2*a*b+b^2)*cos(f*x+e)+15/8/f*b/a/(a^2-2*a*b+b^2)/((a-b)*b)^(1/2)*
arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-5/2/f*b^2/a^2/(a^2-2*a*b+b^2)/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)
/((a-b)*b)^(1/2))+1/f*b^3/a^3/(a^2-2*a*b+b^2)/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))+1/2/f/a
^3*ln(-1+cos(f*x+e))-1/2/f/a^3*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)^3,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 = 16.27, size = 1844, normalized size = 11.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(e/2 + (f*x)/2))/(a^3*f) - ((3*(3*a*b - 2*b^2))/(4*a*(a^2 - 2*a*b + b^2)) + (3*tan(e/2 + (f*x)/2)^4*(40
*a*b^3 + 9*a^3*b - 16*b^4 - 30*a^2*b^2))/(4*a^3*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f*x)/2)^6*(9*a^2*b - 28*a*b
^2 + 16*b^3))/(4*a^2*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f*x)/2)^2*(27*a^2*b - 68*a*b^2 + 32*b^3))/(4*a^2*(a^2
- 2*a*b + b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(8*a*b - 4*a^2) + tan(e/2 + (f*x)/2)^6*(8*a*b - 4*a^2) + a^2*tan(e/2
+ (f*x)/2)^8 + tan(e/2 + (f*x)/2)^4*(6*a^2 - 16*a*b + 16*b^2) + a^2)) + (b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*
((((b^(3/2)*(15*a^2 - 20*a*b + 8*b^2)^3*(4096*a^15*b - 128*a^16 + 6144*a^7*b^9 - 46080*a^8*b^8 + 150784*a^9*b^
7 - 281216*a^10*b^6 + 327168*a^11*b^5 - 243584*a^12*b^4 + 113920*a^13*b^3 - 31104*a^14*b^2))/(32768*a^9*(a - b
)^(15/2)*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6*b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)) - (b^(1/2)*(15*a^2 -
20*a*b + 8*b^2)*(1536*a*b^9 + 720*a^9*b - 11520*a^2*b^8 + 37760*a^3*b^7 - 70400*a^4*b^6 + 81384*a^5*b^5 - 5956
4*a^6*b^4 + 26864*a^7*b^3 - 6780*a^8*b^2))/(128*a^3*(a - b)^(5/2)*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6*b^5 + 15*
a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)))*(3072*a*b^4 - 1090*a^4*b + 111*a^5 - 768*b^5 - 4752*a^2*b^3 + 3424*a^3*b^
2))/(2*a^5*(a - b)^(13/2)*(960*a*b^4 - 1055*a^4*b + 256*a^5 - 192*b^5 - 1920*a^2*b^3 + 1960*a^3*b^2)) + (((576
*a*b^6 - 64*b^7 - 1920*a^2*b^5 + 3160*a^3*b^4 - 2625*a^4*b^3 + 900*a^5*b^2)/(8*(a^11 - 6*a^10*b + a^5*b^6 - 6*
a^6*b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)) + (b*(15*a^2 - 20*a*b + 8*b^2)^2*(2768*a^12*b - 128*a^13 + 61
44*a^4*b^9 - 46080*a^5*b^8 + 150656*a^6*b^7 - 279104*a^7*b^6 + 318672*a^8*b^5 - 228160*a^9*b^4 + 99424*a^10*b^
3 - 24192*a^11*b^2))/(2048*a^6*(a - b)^5*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6*b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15
*a^9*b^2)))*(1344*a*b^4 - 205*a^4*b + 8*a^5 - 384*b^5 - 1752*a^2*b^3 + 980*a^3*b^2))/(a^5*b^(1/2)*(a - b)^6*(9
60*a*b^4 - 1055*a^4*b + 256*a^5 - 192*b^5 - 1920*a^2*b^3 + 1960*a^3*b^2))) + (((b^(1/2)*(15*a^2 - 20*a*b + 8*b
^2)*(240*a^8*b - 320*a^3*b^6 + 1600*a^4*b^5 - 3232*a^5*b^4 + 3208*a^6*b^3 - 1505*a^7*b^2))/(64*a^3*(a - b)^(5/
2)*(a^10 - 4*a^9*b + a^6*b^4 - 4*a^7*b^3 + 6*a^8*b^2)) + (b^(3/2)*(15*a^2 - 20*a*b + 8*b^2)^3*(64*a^15 - 512*a
^14*b + 256*a^9*b^6 - 1280*a^10*b^5 + 2624*a^11*b^4 - 2816*a^12*b^3 + 1664*a^13*b^2))/(16384*a^9*(a - b)^(15/2
)*(a^10 - 4*a^9*b + a^6*b^4 - 4*a^7*b^3 + 6*a^8*b^2)))*(3072*a*b^4 - 1090*a^4*b + 111*a^5 - 768*b^5 - 4752*a^2
*b^3 + 3424*a^3*b^2))/(2*a^5*(a - b)^(13/2)*(960*a*b^4 - 1055*a^4*b + 256*a^5 - 192*b^5 - 1920*a^2*b^3 + 1960*
a^3*b^2)) - (((64*b^6 - 320*a*b^5 + 640*a^2*b^4 - 600*a^3*b^3 + 225*a^4*b^2)/(4*(a^10 - 4*a^9*b + a^6*b^4 - 4*
a^7*b^3 + 6*a^8*b^2)) - (b*(15*a^2 - 20*a*b + 8*b^2)^2*(64*a^12 - 752*a^11*b + 512*a^6*b^6 - 2560*a^7*b^5 + 52
16*a^8*b^4 - 5424*a^9*b^3 + 2944*a^10*b^2))/(1024*a^6*(a - b)^5*(a^10 - 4*a^9*b + a^6*b^4 - 4*a^7*b^3 + 6*a^8*
b^2)))*(1344*a*b^4 - 205*a^4*b + 8*a^5 - 384*b^5 - 1752*a^2*b^3 + 980*a^3*b^2))/(a^5*b^(1/2)*(a - b)^6*(960*a*
b^4 - 1055*a^4*b + 256*a^5 - 192*b^5 - 1920*a^2*b^3 + 1960*a^3*b^2)))*(256*a^13*(a - b)^(15/2) - 1536*a^12*b*(
a - b)^(15/2) + 256*a^7*b^6*(a - b)^(15/2) - 1536*a^8*b^5*(a - b)^(15/2) + 3840*a^9*b^4*(a - b)^(15/2) - 5120*
a^10*b^3*(a - b)^(15/2) + 3840*a^11*b^2*(a - b)^(15/2)))/(225*a^4*b - 320*a*b^4 + 64*b^5 + 640*a^2*b^3 - 600*a
^3*b^2))*(15*a^2 - 20*a*b + 8*b^2))/(8*a^3*f*(a - b)^(5/2))

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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)**3,x)

[Out]

Timed out

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