3.130 \(\int \frac{\csc ^4(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{8 b (a-b) \tan (e+f x)}{3 f (a+b)^4 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{4 b (a-b) \tan (e+f x)}{3 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{(a-b) \cot (e+f x)}{f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.160717, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4132, 453, 271, 192, 191} \[ -\frac{8 b (a-b) \tan (e+f x)}{3 f (a+b)^4 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{4 b (a-b) \tan (e+f x)}{3 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{(a-b) \cot (e+f x)}{f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

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

Mathematica [A]  time = 4.64024, size = 138, normalized size = 0.87 \[ \frac{\tan (e+f x) \sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (\frac{4 b^2 (a+b)}{(a \cos (2 (e+f x))+a+2 b)^2}+\frac{4 b (b-3 a)}{a \cos (2 (e+f x))+a+2 b}-(a+b) \csc ^4(e+f x)-2 (a-3 b) \csc ^2(e+f x)\right )}{24 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*((4*b^2*(a + b))/(a + 2*b + a*Cos[2*(e + f*x)])^2 + (4*b*(-3*a + b))/(a + 2*
b + a*Cos[2*(e + f*x)]) - 2*(a - 3*b)*Csc[e + f*x]^2 - (a + b)*Csc[e + f*x]^4)*Sec[e + f*x]^4*Tan[e + f*x])/(2
4*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [A]  time = 0.411, size = 225, normalized size = 1.4 \begin{align*}{\frac{ \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b+2\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{2}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}+21\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b-21\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}b+24\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{2}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{3}-8\,a{b}^{2}+8\,{b}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{3\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a+b \right ) ^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{4} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \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)^4/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

1/3/f/(a^2+2*a*b+b^2)/(a+b)^2/(b+a*cos(f*x+e)^2)^4*(2*cos(f*x+e)^6*a^3-12*cos(f*x+e)^6*a^2*b+2*cos(f*x+e)^6*a*
b^2-3*cos(f*x+e)^4*a^3+21*cos(f*x+e)^4*a^2*b-21*cos(f*x+e)^4*a*b^2+3*cos(f*x+e)^4*b^3-12*cos(f*x+e)^2*a^2*b+24
*cos(f*x+e)^2*a*b^2-12*cos(f*x+e)^2*b^3-8*a*b^2+8*b^3)*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)/si
n(f*x+e)^3

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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)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.20669, size = 701, normalized size = 4.44 \begin{align*} -\frac{{\left (2 \,{\left (a^{3} - 6 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{7} - 3 \,{\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \,{\left (a b^{2} - b^{3}\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}}}}{3 \,{\left ({\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{6} -{\left (a^{6} + 2 \, a^{5} b - 2 \, a^{4} b^{2} - 8 \, a^{3} b^{3} - 7 \, a^{2} b^{4} - 2 \, a b^{5}\right )} f \cos \left (f x + e\right )^{4} -{\left (2 \, a^{5} b + 7 \, a^{4} b^{2} + 8 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 2 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\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)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*(a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^7 - 3*(a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e)^5 - 12*(a^2*b
- 2*a*b^2 + b^3)*cos(f*x + e)^3 - 8*(a*b^2 - b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((
(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^6 - (a^6 + 2*a^5*b - 2*a^4*b^2 - 8*a^3*b^3 -
7*a^2*b^4 - 2*a*b^5)*f*cos(f*x + e)^4 - (2*a^5*b + 7*a^4*b^2 + 8*a^3*b^3 + 2*a^2*b^4 - 2*a*b^5 - b^6)*f*cos(f*
x + e)^2 - (a^4*b^2 + 4*a^3*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*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)**4/(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 )^{4}}{{\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)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(5/2), x)