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

Optimal. Leaf size=234 \[ -\frac{\left (3 a^2-24 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{9/2}}-\frac{5 b (11 a-10 b) \sec (e+f x)}{24 f (a+b)^4 \sqrt{a+b \sec ^2(e+f x)}}-\frac{b (23 a-12 b) \sec (e+f x)}{24 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[Out]

-((3*a^2 - 24*a*b + 8*b^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*(a + b)^(9/2)*f)
- ((5*a - 2*b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (Cot[e + f*x]^3*Csc[
e + f*x])/(4*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - ((23*a - 12*b)*b*Sec[e + f*x])/(24*(a + b)^3*f*(a + b*S
ec[e + f*x]^2)^(3/2)) - (5*(11*a - 10*b)*b*Sec[e + f*x])/(24*(a + b)^4*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.323945, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4134, 470, 527, 12, 377, 207} \[ -\frac{\left (3 a^2-24 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{9/2}}-\frac{5 b (11 a-10 b) \sec (e+f x)}{24 f (a+b)^4 \sqrt{a+b \sec ^2(e+f x)}}-\frac{b (23 a-12 b) \sec (e+f x)}{24 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2 - 24*a*b + 8*b^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*(a + b)^(9/2)*f)
- ((5*a - 2*b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (Cot[e + f*x]^3*Csc[
e + f*x])/(4*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - ((23*a - 12*b)*b*Sec[e + f*x])/(24*(a + b)^3*f*(a + b*S
ec[e + f*x]^2)^(3/2)) - (5*(11*a - 10*b)*b*Sec[e + f*x])/(24*(a + b)^4*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

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 \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-a-2 (2 a-b) x^2}{\left (-1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-a (3 a-4 b)+4 (5 a-2 b) b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{8 (a+b)^2 f}\\ &=-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-a^2 (9 a-26 b)+2 a (23 a-12 b) b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{24 a (a+b)^3 f}\\ &=-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int -\frac{3 a^2 \left (3 a^2-24 a b+8 b^2\right )}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{24 a^2 (a+b)^4 f}\\ &=-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\left (3 a^2-24 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 (a+b)^4 f}\\ &=-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\left (3 a^2-24 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-(-a-b) x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^4 f}\\ &=-\frac{\left (3 a^2-24 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{9/2} f}-\frac{(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.92286, size = 129, normalized size = 0.55 \[ -\frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (3 (a+b) \csc ^4(e+f x) ((a+8 b) \cos (2 (e+f x))+3 a-4 b)-2 \left (3 a^2-24 a b+8 b^2\right ) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},1-\frac{a \sin ^2(e+f x)}{a+b}\right )\right )}{96 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(3*(a + b)*(3*a - 4*b + (a + 8*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^4 - 2*(3*a^2
- 24*a*b + 8*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, 1 - (a*Sin[e + f*x]^2)/(a + b)])*Sec[e + f*x]^5)/(96*(a +
b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [B]  time = 1.572, size = 15551, normalized size = 66.5 \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*sec(f*x+e)^2)^(5/2),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*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.80722, size = 2969, normalized size = 12.69 \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*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*((3*a^4 - 24*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(3*a^4 - 27*a^3*b + 32*a^2*b^2 - 8*a*b^3)*cos(f*x
+ e)^6 + (3*a^4 - 36*a^3*b + 107*a^2*b^2 - 56*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 24*a*b^3 + 8*b^4 + 2
*(3*a^3*b - 27*a^2*b^2 + 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + 2*(3*(3*a^4 - 21*
a^3*b - 16*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 117*a^3*b + 4*a^2*b^2 + 104*a*b^3 - 32*b^4)*cos(f*x +
e)^5 - (78*a^3*b - 71*a^2*b^2 - 61*a*b^3 + 88*b^4)*cos(f*x + e)^3 - 5*(11*a^2*b^2 + a*b^3 - 10*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), 1/24*(3*((3*a^4 - 24*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(3*a^4 - 27*a^3*b + 32*a^2*b^2
- 8*a*b^3)*cos(f*x + e)^6 + (3*a^4 - 36*a^3*b + 107*a^2*b^2 - 56*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 -
24*a*b^3 + 8*b^4 + 2*(3*a^3*b - 27*a^2*b^2 + 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(sqrt(-a - b
)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + (3*(3*a^4 - 21*a^3*b - 16*a^2*b^2 + 8*a*
b^3)*cos(f*x + e)^7 - (15*a^4 - 117*a^3*b + 4*a^2*b^2 + 104*a*b^3 - 32*b^4)*cos(f*x + e)^5 - (78*a^3*b - 71*a^
2*b^2 - 61*a*b^3 + 88*b^4)*cos(f*x + e)^3 - 5*(11*a^2*b^2 + a*b^3 - 10*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)]

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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*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 )^{5}}{{\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)^5/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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