3.133 \(\int \frac{\csc ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=187 \[ -\frac{b (13 a-15 b) \sec (e+f x)}{8 a^3 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{7/2} f}-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

(-3*(a - 5*b)*(a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(8*a^(7/2)*f) - (5*(a -
b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*Sqrt[a - b + b*Sec[e + f*x]^2]) - (Cot[e + f*x]^3*Csc[e + f*x])/(4*a*f*
Sqrt[a - b + b*Sec[e + f*x]^2]) - ((13*a - 15*b)*b*Sec[e + f*x])/(8*a^3*f*Sqrt[a - b + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.236033, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3664, 470, 527, 12, 377, 207} \[ -\frac{b (13 a-15 b) \sec (e+f x)}{8 a^3 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{7/2} f}-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(-3*(a - 5*b)*(a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(8*a^(7/2)*f) - (5*(a -
b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*Sqrt[a - b + b*Sec[e + f*x]^2]) - (Cot[e + f*x]^3*Csc[e + f*x])/(4*a*f*
Sqrt[a - b + b*Sec[e + f*x]^2]) - ((13*a - 15*b)*b*Sec[e + f*x])/(8*a^3*f*Sqrt[a - b + b*Sec[e + f*x]^2])

Rule 3664

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

Rule 470

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

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{-a+b-4 (a-b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{-(3 a-5 b) (a-b)+10 (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int -\frac{3 (a-5 b) (a-b)^2}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 a^3 (a-b) f}\\ &=-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{(3 (a-5 b) (a-b)) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 a^3 f}\\ &=-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{(3 (a-5 b) (a-b)) \operatorname{Subst}\left (\int \frac{1}{-1+a x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{8 a^3 f}\\ &=-\frac{3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{8 a^{7/2} f}-\frac{5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.75673, size = 350, normalized size = 1.87 \[ \frac{\frac{\csc ^4(e+f x) \sec (e+f x) \left (\left (-8 a^2+52 a b-60 b^2\right ) \cos (2 (e+f x))+(a-b) (3 (a-5 b) \cos (4 (e+f x))-11 a-45 b)\right )}{4 \sqrt{2} a^3 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}-\frac{3 (a-5 b) (a-b) \cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{2 a^{7/2} \sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((((-8*a^2 + 52*a*b - 60*b^2)*Cos[2*(e + f*x)] + (a - b)*(-11*a - 45*b + 3*(a - 5*b)*Cos[4*(e + f*x)]))*Csc[e
+ f*x]^4*Sec[e + f*x])/(4*Sqrt[2]*a^3*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]) - (3*(a - 5*b)*
(a - b)*(ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*
x)/2]^2)^2])] + ArcTanh[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan
[(e + f*x)/2]^2)^2])])*Cos[e + f*x]*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]
)/(2*a^(7/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4]))/(8*f)

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Maple [B]  time = 0.228, size = 10582, normalized size = 56.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

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Maxima [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)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.79066, size = 1656, normalized size = 8.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*((a^3 - 7*a^2*b + 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15*a^2*b + 28*a*b^2 - 15*b^3)*cos(f*x +
e)^4 + a^2*b - 6*a*b^2 + 5*b^3 + (a^3 - 9*a^2*b + 23*a*b^2 - 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*
cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x +
e)^2 - 1)) + 2*(3*(a^3 - 6*a^2*b + 5*a*b^2)*cos(f*x + e)^5 - (5*a^3 - 31*a^2*b + 30*a*b^2)*cos(f*x + e)^3 - (1
3*a^2*b - 15*a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^5 - a^4*b)*f*cos(f*x
+ e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2), 1/8*(3*((a^3 - 7*a^
2*b + 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15*a^2*b + 28*a*b^2 - 15*b^3)*cos(f*x + e)^4 + a^2*b - 6*a*b
^2 + 5*b^3 + (a^3 - 9*a^2*b + 23*a*b^2 - 15*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + (3*(a^3 - 6*a^2*b + 5*a*b^2)*cos(f*x + e)^5 - (5*a^3 - 31*a^2*
b + 30*a*b^2)*cos(f*x + e)^3 - (13*a^2*b - 15*a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x +
e)^2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos
(f*x + e)^2)]

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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)**5/(a+b*tan(f*x+e)**2)**(3/2),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 )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^5/(b*tan(f*x + e)^2 + a)^(3/2), x)