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]

-((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])

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Rubi [A]  time = 0.23976, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 7.30988, 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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Maple [A]  time = 0.409, size = 324, normalized size = 1.4 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{4}-80\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{3}b+40\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{2}{b}^{2}-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{4}+212\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}b-220\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}{b}^{2}+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{3}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{4}-180\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}b+378\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}{b}^{2}-180\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{3}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{4}+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}b-220\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}{b}^{2}+212\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{3}-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{4}+40\,{a}^{2}{b}^{2}-80\,a{b}^{3}+8\,{b}^{4} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{15\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a+b \right ) ^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{4} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{5}{2}}}} \end{align*}

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/(a^2+2*a*b+b^2)/(a+b)^3/(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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 29.0847, size = 1033, normalized size = 4.57 \begin{align*} -\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 )} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)