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

Optimal. Leaf size=171 \[ -\frac{b (13 a-2 b) \sec (e+f x)}{6 a f (a+b)^3 \sqrt{a+b \sec ^2(e+f x)}}-\frac{5 b \sec (e+f x)}{6 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(a-4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f (a+b)^{7/2}}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[Out]

-((a - 4*b)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*(a + b)^(7/2)*f) - (Cot[e + f*x
]*Csc[e + f*x])/(2*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (5*b*Sec[e + f*x])/(6*(a + b)^2*f*(a + b*Sec[e +
f*x]^2)^(3/2)) - ((13*a - 2*b)*b*Sec[e + f*x])/(6*a*(a + b)^3*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.206449, antiderivative size = 171, 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 = {4134, 471, 527, 12, 377, 207} \[ -\frac{b (13 a-2 b) \sec (e+f x)}{6 a f (a+b)^3 \sqrt{a+b \sec ^2(e+f x)}}-\frac{5 b \sec (e+f x)}{6 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(a-4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f (a+b)^{7/2}}-\frac{\cot (e+f x) \csc (e+f x)}{2 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 4*b)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*(a + b)^(7/2)*f) - (Cot[e + f*x
]*Csc[e + f*x])/(2*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (5*b*Sec[e + f*x])/(6*(a + b)^2*f*(a + b*Sec[e +
f*x]^2)^(3/2)) - ((13*a - 2*b)*b*Sec[e + f*x])/(6*a*(a + b)^3*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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && 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 ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (-1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a-4 b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{2 (a+b) f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a-2 b)-10 a b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{6 a (a+b)^2 f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-2 b) b \sec (e+f x)}{6 a (a+b)^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 (a-4 b)}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{6 a^2 (a+b)^3 f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-2 b) b \sec (e+f x)}{6 a (a+b)^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{(a-4 b) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 (a+b)^3 f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-2 b) b \sec (e+f x)}{6 a (a+b)^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{(a-4 b) \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 )}{2 (a+b)^3 f}\\ &=-\frac{(a-4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{7/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-2 b) b \sec (e+f x)}{6 a (a+b)^3 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.54846, size = 151, normalized size = 0.88 \[ -\frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left ((a+b) \csc ^2(e+f x) \left (\left (3 a^2+2 b^2\right ) \cos (2 (e+f x))+3 a^2+6 a b-2 b^2\right )-3 a (a-4 b) (a \cos (2 (e+f x))+a+2 b) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},1-\frac{a \sin ^2(e+f x)}{a+b}\right )\right )}{24 a f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*((a + b)*(3*a^2 + 6*a*b - 2*b^2 + (3*a^2 + 2*b^2)*Cos[2*(e + f*x)])*Csc[e + f
*x]^2 - 3*a*(a - 4*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Hypergeometric2F1[-1/2, 1, 1/2, 1 - (a*Sin[e + f*x]^2)/(a
+ b)])*Sec[e + f*x]^5)/(24*a*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [B]  time = 1.021, size = 11110, normalized size = 65. \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)^3/(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)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.31295, size = 2086, normalized size = 12.2 \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)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((a^4 - 4*a^3*b)*cos(f*x + e)^6 - (a^4 - 6*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 + 4*a*b^3 - (
2*a^3*b - 9*a^2*b^2 + 4*a*b^3)*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*(a^4 - 3*a^3*b - 4*a^2*b
^2)*cos(f*x + e)^5 + 2*(9*a^3*b + 4*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^3 + (13*a^2*b^2 + 11*a*b^3 - 2*b^4)*
cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*
f*cos(f*x + e)^6 - (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^4 - (2*a^6*b
+ 7*a^5*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b^6)*f*cos(f*x + e)^2 - (a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4
+ 4*a^2*b^5 + a*b^6)*f), 1/6*(3*((a^4 - 4*a^3*b)*cos(f*x + e)^6 - (a^4 - 6*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 -
a^2*b^2 + 4*a*b^3 - (2*a^3*b - 9*a^2*b^2 + 4*a*b^3)*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*(a^4 - 3*a^3*b - 4*a^2*b^2)*cos(f*x + e)^5 + 2*
(9*a^3*b + 4*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^3 + (13*a^2*b^2 + 11*a*b^3 - 2*b^4)*cos(f*x + e))*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (a^7
+ 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^4 - (2*a^6*b + 7*a^5*b^2 + 8*a^4*b^
3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b^6)*f*cos(f*x + e)^2 - (a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*b^5 + a*b^6)*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)**3/(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 )^{3}}{{\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)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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