3.131 \(\int \frac {\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=226 \[ -\frac {8 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 f (a+b)^5 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {4 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\left (5 a^2-10 a b+b^2\right ) \cot (e+f x)}{5 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

-8/15*b*(5*a^2-10*a*b+b^2)*tan(f*x+e)/(a+b)^5/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/5*(5*a^2-10*a*b+b^2)*cot(f*x+e)/(
a+b)^3/f/(a+b+b*tan(f*x+e)^2)^(3/2)-2/15*(5*a+b)*cot(f*x+e)^3/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/5*cot(f*x
+e)^5/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-4/15*b*(5*a^2-10*a*b+b^2)*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^(
3/2)

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Rubi [A]  time = 0.24, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4132, 462, 453, 271, 192, 191} \[ -\frac {8 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 f (a+b)^5 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {4 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\left (5 a^2-10 a b+b^2\right ) \cot (e+f x)}{5 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((5*a^2 - 10*a*b + b^2)*Cot[e + f*x])/(5*(a + b)^3*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (2*(5*a + b)*Cot[e +
f*x]^3)/(15*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - Cot[e + f*x]^5/(5*(a + b)*f*(a + b + b*Tan[e + f*
x]^2)^(3/2)) - (4*b*(5*a^2 - 10*a*b + b^2)*Tan[e + f*x])/(15*(a + b)^4*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (
8*b*(5*a^2 - 10*a*b + b^2)*Tan[e + f*x])/(15*(a + b)^5*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 271

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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 ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {2 (5 a+b)+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (5 a^2-10 a b+b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b)^2 f}\\ &=-\frac {\left (5 a^2-10 a b+b^2\right ) \cot (e+f x)}{5 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (4 b \left (5 a^2-10 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b)^3 f}\\ &=-\frac {\left (5 a^2-10 a b+b^2\right ) \cot (e+f x)}{5 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (8 b \left (5 a^2-10 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^4 f}\\ &=-\frac {\left (5 a^2-10 a b+b^2\right ) \cot (e+f x)}{5 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a+b) \cot ^3(e+f x)}{15 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-10 a b+b^2\right ) \tan (e+f x)}{15 (a+b)^5 f \sqrt {a+b+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 7.41, size = 173, normalized size = 0.77 \[ \frac {\tan (e+f x) \sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (\left (-8 a^2+50 a b-15 b^2\right ) \csc ^2(e+f x)+\frac {20 a b^2 (a+b)}{(a \cos (2 (e+f x))+a+2 b)^2}+\frac {10 a b (5 b-6 a)}{a \cos (2 (e+f x))+a+2 b}-3 (a+b)^2 \csc ^6(e+f x)+2 (a+b) (5 b-2 a) \csc ^4(e+f x)\right )}{120 f (a+b)^5 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*((20*a*b^2*(a + b))/(a + 2*b + a*Cos[2*(e + f*x)])^2 + (10*a*b*(-6*a + 5*b))
/(a + 2*b + a*Cos[2*(e + f*x)]) + (-8*a^2 + 50*a*b - 15*b^2)*Csc[e + f*x]^2 + 2*(a + b)*(-2*a + 5*b)*Csc[e + f
*x]^4 - 3*(a + b)^2*Csc[e + f*x]^6)*Sec[e + f*x]^4*Tan[e + f*x])/(120*(a + b)^5*f*(a + b*Sec[e + f*x]^2)^(5/2)
)

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fricas [B]  time = 15.77, size = 460, normalized size = 2.04 \[ -\frac {{\left (8 \, {\left (a^{4} - 10 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{9} - 4 \, {\left (5 \, a^{4} - 53 \, a^{3} b + 55 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} - 60 \, a^{3} b + 126 \, a^{2} b^{2} - 60 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (15 \, a^{3} b - 55 \, a^{2} b^{2} + 53 \, a b^{3} - 5 \, b^{4}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} - 10 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{8} - 2 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 4 \, a^{2} b^{5} - a b^{6}\right )} f \cos \left (f x + e\right )^{6} + {\left (a^{7} + a^{6} b - 9 \, a^{5} b^{2} - 25 \, a^{4} b^{3} - 25 \, a^{3} b^{4} - 9 \, a^{2} b^{5} + a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 5 \, a^{2} b^{5} - 4 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7}\right )} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(8*(a^4 - 10*a^3*b + 5*a^2*b^2)*cos(f*x + e)^9 - 4*(5*a^4 - 53*a^3*b + 55*a^2*b^2 - 15*a*b^3)*cos(f*x +
e)^7 + 3*(5*a^4 - 60*a^3*b + 126*a^2*b^2 - 60*a*b^3 + 5*b^4)*cos(f*x + e)^5 + 4*(15*a^3*b - 55*a^2*b^2 + 53*a*
b^3 - 5*b^4)*cos(f*x + e)^3 + 8*(5*a^2*b^2 - 10*a*b^3 + b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2)/(((a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^8 - 2*(a^7 + 4*a^6*b
+ 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6*b - 9*a^5*b^2 - 25*a^4*b^3 - 25*a^
3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5*a^4*b^3 - 5*a^2*b^5 - 4*a*b^6 - b
^7)*f*cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*f)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)

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maple [A]  time = 1.74, size = 324, normalized size = 1.43 \[ -\frac {\left (8 \left (\cos ^{8}\left (f x +e \right )\right ) a^{4}-80 \left (\cos ^{8}\left (f x +e \right )\right ) a^{3} b +40 \left (\cos ^{8}\left (f x +e \right )\right ) a^{2} b^{2}-20 \left (\cos ^{6}\left (f x +e \right )\right ) a^{4}+212 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3} b -220 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b^{2}+60 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{3}+15 \left (\cos ^{4}\left (f x +e \right )\right ) a^{4}-180 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3} b +378 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b^{2}-180 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{3}+15 \left (\cos ^{4}\left (f x +e \right )\right ) b^{4}+60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b -220 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}+212 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{3}-20 \left (\cos ^{2}\left (f x +e \right )\right ) b^{4}+40 a^{2} b^{2}-80 a \,b^{3}+8 b^{4}\right ) \left (\cos ^{5}\left (f x +e \right )\right ) \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}}}{15 f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{4} \sin \left (f x +e \right )^{5} \left (a^{2}+2 a b +b^{2}\right ) \left (a +b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

-1/15/f/(b+a*cos(f*x+e)^2)^4*(8*cos(f*x+e)^8*a^4-80*cos(f*x+e)^8*a^3*b+40*cos(f*x+e)^8*a^2*b^2-20*cos(f*x+e)^6
*a^4+212*cos(f*x+e)^6*a^3*b-220*cos(f*x+e)^6*a^2*b^2+60*cos(f*x+e)^6*a*b^3+15*cos(f*x+e)^4*a^4-180*cos(f*x+e)^
4*a^3*b+378*cos(f*x+e)^4*a^2*b^2-180*cos(f*x+e)^4*a*b^3+15*cos(f*x+e)^4*b^4+60*cos(f*x+e)^2*a^3*b-220*cos(f*x+
e)^2*a^2*b^2+212*cos(f*x+e)^2*a*b^3-20*cos(f*x+e)^2*b^4+40*a^2*b^2-80*a*b^3+8*b^4)*cos(f*x+e)^5*((b+a*cos(f*x+
e)^2)/cos(f*x+e)^2)^(5/2)/sin(f*x+e)^5/(a^2+2*a*b+b^2)/(a+b)^3

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maxima [A]  time = 0.38, size = 373, normalized size = 1.65 \[ -\frac {\frac {40 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {20 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2}} - \frac {160 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{3}} + \frac {128 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{5}} + \frac {64 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{4}} + \frac {15}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {60 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {48 \, b^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {8 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/15*(40*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 20*b*tan(f*x + e)/((b*tan(f*x + e)^2 + a
+ b)^(3/2)*(a + b)^2) - 160*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^4) - 80*b^2*tan(f*x + e)
/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^3) + 128*b^3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^5
) + 64*b^3*tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^4) + 15/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a
+ b)*tan(f*x + e)) - 60*b/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*tan(f*x + e)) + 48*b^2/((b*tan(f*x + e)
^2 + a + b)^(3/2)*(a + b)^3*tan(f*x + e)) + 10/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)^3) - 8*b
/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*tan(f*x + e)^3) + 3/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan
(f*x + e)^5))/f

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)),x)

[Out]

\text{Hanged}

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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)**6/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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