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

Optimal. Leaf size=177 \[ -\frac{b (13 a-15 b) \sec (e+f x)}{6 a^3 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(a-5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 a^{7/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

-((a - 5*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*a^(7/2)*f) - (Cot[e + f*x]*Csc[
e + f*x])/(2*a*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (5*b*Sec[e + f*x])/(6*a^2*f*(a - b + b*Sec[e + f*x]^2)^(3
/2)) - ((13*a - 15*b)*b*Sec[e + f*x])/(6*a^3*(a - b)*f*Sqrt[a - b + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.20946, antiderivative size = 177, 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, 471, 527, 12, 377, 207} \[ -\frac{b (13 a-15 b) \sec (e+f x)}{6 a^3 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(a-5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 a^{7/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 5*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*a^(7/2)*f) - (Cot[e + f*x]*Csc[
e + f*x])/(2*a*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (5*b*Sec[e + f*x])/(6*a^2*f*(a - b + b*Sec[e + f*x]^2)^(3
/2)) - ((13*a - 15*b)*b*Sec[e + f*x])/(6*a^3*(a - b)*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 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 \tan ^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+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 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a-b-4 b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\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 )}{6 a^2 (a-b) f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) 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 )}{6 a^3 (a-b)^2 f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{(a-5 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 )}{2 a^3 f}\\ &=-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{(a-5 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 )}{2 a^3 f}\\ &=-\frac{(a-5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 4.25862, size = 385, normalized size = 2.18 \[ \frac{\frac{\sqrt{\frac{(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (8 a b^2 \cos (e+f x)-24 b (a-b) \cos (e+f x) ((a-b) \cos (2 (e+f x))+a+b)-3 (a-b) \cot (e+f x) \csc (e+f x) ((a-b) \cos (2 (e+f x))+a+b)^2\right )}{3 a^3 (a-b) ((a-b) \cos (2 (e+f x))+a+b)^2}-\frac{(a-5 b) \cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{2 a^{7/2} \sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(8*a*b^2*Cos[e + f*x] - 24*(a - b)*b*Cos[e +
f*x]*(a + b + (a - b)*Cos[2*(e + f*x)]) - 3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2*Cot[e + f*x]*Csc[e +
f*x]))/(3*a^3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2) - ((a - 5*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]))/(2*f)

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Maple [B]  time = 2.254, size = 38486, normalized size = 217.4 \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*tan(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*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.421, size = 2014, normalized size = 11.38 \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*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 16*a*b^3 + 5*b^4)*cos(f*x + e)^6 - (a^4 - 10*a^3*b + 32*a^2*b^2 - 38*
a*b^3 + 15*b^4)*cos(f*x + e)^4 - a^2*b^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b^2 + 28*a*b^3 - 15*b^4)*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^4 - 7*a^3*b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2*
(9*a^3*b - 23*a^2*b^2 + 15*a*b^3)*cos(f*x + e)^3 + (13*a^2*b^2 - 15*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2))/((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5
*b^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^2 - a^4*b^3)*
f), 1/6*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 16*a*b^3 + 5*b^4)*cos(f*x + e)^6 - (a^4 - 10*a^3*b + 32*a^2*b^2 - 38
*a*b^3 + 15*b^4)*cos(f*x + e)^4 - a^2*b^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b^2 + 28*a*b^3 - 15*b^4)*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^
4 - 7*a^3*b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2*(9*a^3*b - 23*a^2*b^2 + 15*a*b^3)*cos(f*x + e)^3 + (13*
a^2*b^2 - 15*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 3*a^6*b + 3*a^5*b
^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b
^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^2 - a^4*b^3)*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*tan(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 \tan \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*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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