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

Optimal. Leaf size=219 \[ -\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (5 a^2-20 a b+16 b^2\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

-8/15*b*(5*a^2-20*a*b+16*b^2)*tan(f*x+e)/a^5/f/(a+b*tan(f*x+e)^2)^(1/2)-1/5*(5*a^2-20*a*b+16*b^2)*cot(f*x+e)/a
^3/f/(a+b*tan(f*x+e)^2)^(3/2)-2/15*(5*a-4*b)*cot(f*x+e)^3/a^2/f/(a+b*tan(f*x+e)^2)^(3/2)-1/5*cot(f*x+e)^5/a/f/
(a+b*tan(f*x+e)^2)^(3/2)-4/15*b*(5*a^2-20*a*b+16*b^2)*tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)^(3/2)

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Rubi [A]  time = 0.23, antiderivative size = 219, 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 = {3663, 462, 453, 271, 192, 191} \[ -\frac {8 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (5 a^2-20 a b+16 b^2\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*a^2 - 20*a*b + 16*b^2)*Cot[e + f*x])/(5*a^3*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (2*(5*a - 4*b)*Cot[e + f*x]
^3)/(15*a^2*f*(a + b*Tan[e + f*x]^2)^(3/2)) - Cot[e + f*x]^5/(5*a*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (4*b*(5*a^
2 - 20*a*b + 16*b^2)*Tan[e + f*x])/(15*a^4*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (8*b*(5*a^2 - 20*a*b + 16*b^2)*Ta
n[e + f*x])/(15*a^5*f*Sqrt[a + 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 3663

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

Rubi steps

\begin {align*} \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^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 x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {2 (5 a-4 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (-15 a^2+12 (5 a-4 b) b\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (4 b \left (-15 a^2+12 (5 a-4 b) b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (8 b \left (-15 a^2+12 (5 a-4 b) b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{45 a^4 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.34, size = 174, normalized size = 0.79 \[ \frac {\sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac {5 b (b-a) \sin (2 (e+f x)) \left (\left (6 a^2-17 a b+11 b^2\right ) \cos (2 (e+f x))+6 a^2-7 a b-11 b^2\right )}{((a-b) \cos (2 (e+f x))+a+b)^2}-\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+2 a (2 a-7 b) \csc ^2(e+f x)-66 a b+73 b^2\right )\right )}{15 \sqrt {2} a^5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-(Cot[e + f*x]*(8*a^2 - 66*a*b + 73*b^2 + 2*a*(2*a -
7*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4)) + (5*b*(-a + b)*(6*a^2 - 7*a*b - 11*b^2 + (6*a^2 - 17*a*b + 11*b
^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)])^2))/(15*Sqrt[2]*a^5*f)

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fricas [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*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.10, size = 371, normalized size = 1.69 \[ -\frac {\left (8 \left (\cos ^{8}\left (f x +e \right )\right ) a^{4}-112 \left (\cos ^{8}\left (f x +e \right )\right ) a^{3} b +328 \left (\cos ^{8}\left (f x +e \right )\right ) a^{2} b^{2}-352 \left (\cos ^{8}\left (f x +e \right )\right ) a \,b^{3}+128 \left (\cos ^{8}\left (f x +e \right )\right ) b^{4}-20 \left (\cos ^{6}\left (f x +e \right )\right ) a^{4}+292 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3} b -976 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b^{2}+1216 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{3}-512 \left (\cos ^{6}\left (f x +e \right )\right ) b^{4}+15 \left (\cos ^{4}\left (f x +e \right )\right ) a^{4}-240 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3} b +1008 a^{2} b^{2} \left (\cos ^{4}\left (f x +e \right )\right )-1536 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{3}+768 \left (\cos ^{4}\left (f x +e \right )\right ) b^{4}+60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b -400 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}+832 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{3}-512 \left (\cos ^{2}\left (f x +e \right )\right ) b^{4}+40 a^{2} b^{2}-160 a \,b^{3}+128 b^{4}\right ) \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \left (\cos ^{5}\left (f x +e \right )\right )}{15 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{4} \sin \left (f x +e \right )^{5} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^4*(8*cos(f*x+e)^8*a^4-112*cos(f*x+e)^8*a^3*b+328*cos(f*x+e)^8*a^2*b^
2-352*cos(f*x+e)^8*a*b^3+128*cos(f*x+e)^8*b^4-20*cos(f*x+e)^6*a^4+292*cos(f*x+e)^6*a^3*b-976*cos(f*x+e)^6*a^2*
b^2+1216*cos(f*x+e)^6*a*b^3-512*cos(f*x+e)^6*b^4+15*cos(f*x+e)^4*a^4-240*cos(f*x+e)^4*a^3*b+1008*a^2*b^2*cos(f
*x+e)^4-1536*cos(f*x+e)^4*a*b^3+768*cos(f*x+e)^4*b^4+60*cos(f*x+e)^2*a^3*b-400*cos(f*x+e)^2*a^2*b^2+832*cos(f*
x+e)^2*a*b^3-512*cos(f*x+e)^2*b^4+40*a^2*b^2-160*a*b^3+128*b^4)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^
2)^(5/2)*cos(f*x+e)^5/sin(f*x+e)^5/a^5

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maxima [A]  time = 0.73, size = 337, normalized size = 1.54 \[ -\frac {\frac {40 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {20 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {160 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {128 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{5}} + \frac {64 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{4}} + \frac {15}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )} - \frac {60 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )} + \frac {48 \, b^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3} \tan \left (f x + e\right )} + \frac {10}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{3}} - \frac {8 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(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)*a^3) + 20*b*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^
2) - 160*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^4) - 80*b^2*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)
*a^3) + 128*b^3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^5) + 64*b^3*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3
/2)*a^4) + 15/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)) - 60*b/((b*tan(f*x + e)^2 + a)^(3/2)*a^2*tan(f*x +
e)) + 48*b^2/((b*tan(f*x + e)^2 + a)^(3/2)*a^3*tan(f*x + e)) + 10/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e
)^3) - 8*b/((b*tan(f*x + e)^2 + a)^(3/2)*a^2*tan(f*x + e)^3) + 3/((b*tan(f*x + e)^2 + a)^(3/2)*a*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*tan(e + f*x)^2)^(5/2)),x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(e + f*x)**6/(a + b*tan(e + f*x)**2)**(5/2), x)

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