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

Optimal. Leaf size=124 \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{7/2}}-\frac {15 \cot (e+f x)}{8 f (a+b)^3}+\frac {5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

-15/8*cot(f*x+e)/(a+b)^3/f-15/8*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/(a+b)^(7/2)/f+1/4*cot(f*x+e)/(a
+b)/f/(a+b+b*tan(f*x+e)^2)^2+5/8*cot(f*x+e)/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.11, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4132, 290, 325, 205} \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{7/2}}-\frac {15 \cot (e+f x)}{8 f (a+b)^3}+\frac {5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*(a + b)^(7/2)*f) - (15*Cot[e + f*x])/(8*(a + b)^3*
f) + Cot[e + f*x]/(4*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) + (5*Cot[e + f*x])/(8*(a + b)^2*f*(a + b + b*Tan[
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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

\begin {align*} \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a+b) f}\\ &=\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^2 f}\\ &=-\frac {15 \cot (e+f x)}{8 (a+b)^3 f}+\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^3 f}\\ &=-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{7/2} f}-\frac {15 \cot (e+f x)}{8 (a+b)^3 f}+\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 6.82, size = 987, normalized size = 7.96 \[ \frac {(\cos (2 e+2 f x) a+a+2 b)^3 \left (\frac {15 b \tan ^{-1}\left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {15 i b \tan ^{-1}\left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) \sec ^6(e+f x)}{(a+b)^3 \left (b \sec ^2(e+f x)+a\right )^3}+\frac {(\cos (2 e+2 f x) a+a+2 b) \csc (e) \csc (e+f x) \sec (2 e) \left (-32 \sin (f x) a^4+32 \sin (3 f x) a^4-48 \sin (2 e-f x) a^4+48 \sin (2 e+f x) a^4-32 \sin (4 e+f x) a^4-8 \sin (2 e+3 f x) a^4+32 \sin (4 e+3 f x) a^4-8 \sin (6 e+3 f x) a^4+8 \sin (2 e+5 f x) a^4+8 \sin (6 e+5 f x) a^4-64 b \sin (f x) a^3+46 b \sin (3 f x) a^3-128 b \sin (2 e-f x) a^3+146 b \sin (2 e+f x) a^3-82 b \sin (4 e+f x) a^3+18 b \sin (2 e+3 f x) a^3+73 b \sin (4 e+3 f x) a^3-9 b \sin (6 e+3 f x) a^3-9 b \sin (2 e+5 f x) a^3+9 b \sin (4 e+5 f x) a^3+22 b^2 \sin (f x) a^2-54 b^2 \sin (3 f x) a^2-106 b^2 \sin (2 e-f x) a^2+182 b^2 \sin (2 e+f x) a^2-54 b^2 \sin (4 e+f x) a^2+54 b^2 \sin (2 e+3 f x) a^2+24 b^2 \sin (4 e+3 f x) a^2-24 b^2 \sin (6 e+3 f x) a^2-2 b^2 \sin (2 e+5 f x) a^2+2 b^2 \sin (4 e+5 f x) a^2+80 b^3 \sin (f x) a-8 b^3 \sin (3 f x) a+80 b^3 \sin (2 e-f x) a+80 b^3 \sin (2 e+f x) a-80 b^3 \sin (4 e+f x) a+8 b^3 \sin (2 e+3 f x) a+8 b^3 \sin (4 e+3 f x) a-8 b^3 \sin (6 e+3 f x) a+16 b^4 \sin (f x)+16 b^4 \sin (2 e-f x)+16 b^4 \sin (2 e+f x)-16 b^4 \sin (4 e+f x)\right ) \sec ^6(e+f x)}{512 a^2 (a+b)^3 f \left (b \sec ^2(e+f x)+a\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((15*b*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*
e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin
[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(64*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - (((15*I)/64)*b*ArcTa
n[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Co
s[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin[2*e])/(Sqrt[a + b]*f*Sqrt[b*Co
s[4*e] - I*b*Sin[4*e]])))/((a + b)^3*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a*Cos[2*e + 2*f*x])*Csc[e]*Csc[e
+ f*x]*Sec[2*e]*Sec[e + f*x]^6*(-32*a^4*Sin[f*x] - 64*a^3*b*Sin[f*x] + 22*a^2*b^2*Sin[f*x] + 80*a*b^3*Sin[f*x]
+ 16*b^4*Sin[f*x] + 32*a^4*Sin[3*f*x] + 46*a^3*b*Sin[3*f*x] - 54*a^2*b^2*Sin[3*f*x] - 8*a*b^3*Sin[3*f*x] - 48
*a^4*Sin[2*e - f*x] - 128*a^3*b*Sin[2*e - f*x] - 106*a^2*b^2*Sin[2*e - f*x] + 80*a*b^3*Sin[2*e - f*x] + 16*b^4
*Sin[2*e - f*x] + 48*a^4*Sin[2*e + f*x] + 146*a^3*b*Sin[2*e + f*x] + 182*a^2*b^2*Sin[2*e + f*x] + 80*a*b^3*Sin
[2*e + f*x] + 16*b^4*Sin[2*e + f*x] - 32*a^4*Sin[4*e + f*x] - 82*a^3*b*Sin[4*e + f*x] - 54*a^2*b^2*Sin[4*e + f
*x] - 80*a*b^3*Sin[4*e + f*x] - 16*b^4*Sin[4*e + f*x] - 8*a^4*Sin[2*e + 3*f*x] + 18*a^3*b*Sin[2*e + 3*f*x] + 5
4*a^2*b^2*Sin[2*e + 3*f*x] + 8*a*b^3*Sin[2*e + 3*f*x] + 32*a^4*Sin[4*e + 3*f*x] + 73*a^3*b*Sin[4*e + 3*f*x] +
24*a^2*b^2*Sin[4*e + 3*f*x] + 8*a*b^3*Sin[4*e + 3*f*x] - 8*a^4*Sin[6*e + 3*f*x] - 9*a^3*b*Sin[6*e + 3*f*x] - 2
4*a^2*b^2*Sin[6*e + 3*f*x] - 8*a*b^3*Sin[6*e + 3*f*x] + 8*a^4*Sin[2*e + 5*f*x] - 9*a^3*b*Sin[2*e + 5*f*x] - 2*
a^2*b^2*Sin[2*e + 5*f*x] + 9*a^3*b*Sin[4*e + 5*f*x] + 2*a^2*b^2*Sin[4*e + 5*f*x] + 8*a^4*Sin[6*e + 5*f*x]))/(5
12*a^2*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^3)

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fricas [B]  time = 0.61, size = 615, normalized size = 4.96 \[ \left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (5 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \, b^{2} \cos \left (f x + e\right )}{32 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (5 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \, b^{2} \cos \left (f x + e\right )}{16 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*(4*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 + 20*(5*a*b - b^2)*cos(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 +
2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos
(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e
) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*b^2*cos(f*x + e))/(((a^5 + 3*a^4
*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 + (a^3
*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f)*sin(f*x + e)), -1/16*(2*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 + 10*(5*a*
b - b^2)*cos(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(b/(a + b))*arctan(1/2*((a
+ 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*b^2*cos(f*x + e))/
(((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x
+ e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f)*sin(f*x + e))]

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giac [A]  time = 0.50, size = 184, normalized size = 1.48 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {7 \, b^{2} \tan \left (f x + e\right )^{3} + 9 \, a b \tan \left (f x + e\right ) + 9 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b/((a^3 + 3*a^2*b + 3*
a*b^2 + b^3)*sqrt(a*b + b^2)) + (7*b^2*tan(f*x + e)^3 + 9*a*b*tan(f*x + e) + 9*b^2*tan(f*x + e))/((a^3 + 3*a^2
*b + 3*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)^2) + 8/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)))/f

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maple [A]  time = 1.07, size = 157, normalized size = 1.27 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a +b \right )^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 a b \tan \left (f x +e \right )}{8 \left (a +b \right )^{3} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b^{2} \tan \left (f x +e \right )}{8 f \left (a +b \right )^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{f \left (a +b \right )^{3} \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/f/(a+b)^3*b^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8*a*b*tan(f*x+e)/(a+b)^3/f/(a+b+b*tan(f*x+e)^2)^2-9/8
/f/(a+b)^3*b^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-15/8/f/(a+b)^3*b/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b
)^(1/2))-1/f/(a+b)^3/tan(f*x+e)

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maxima [B]  time = 0.44, size = 219, normalized size = 1.77 \[ -\frac {\frac {15 \, b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, b^{2} \tan \left (f x + e\right )^{4} + 25 \, {\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(15*b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt((a + b)*b)) + (15*b^2*
tan(f*x + e)^4 + 25*(a*b + b^2)*tan(f*x + e)^2 + 8*a^2 + 16*a*b + 8*b^2)/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5
)*tan(f*x + e)^5 + 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*tan(f*x + e)^3 + (a^5 + 5*a^4*b + 10*a^3*
b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*tan(f*x + e)))/f

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mupad [B]  time = 5.11, size = 146, normalized size = 1.18 \[ -\frac {\frac {1}{a+b}+\frac {25\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,{\left (a+b\right )}^2}+\frac {15\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{8\,{\left (a+b\right )}^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3\,\left (2\,b^2+2\,a\,b\right )+\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+2\,a\,b+b^2\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{8\,f\,{\left (a+b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1/(a + b) + (25*b*tan(e + f*x)^2)/(8*(a + b)^2) + (15*b^2*tan(e + f*x)^4)/(8*(a + b)^3))/(f*(tan(e + f*x)^3
*(2*a*b + 2*b^2) + tan(e + f*x)*(2*a*b + a^2 + b^2) + b^2*tan(e + f*x)^5)) - (15*b^(1/2)*atan((b^(1/2)*tan(e +
f*x)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(a + b)^(7/2)))/(8*f*(a + b)^(7/2))

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

[Out]

Timed out

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