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

Optimal. Leaf size=237 \[ -\frac{\left (3 a^2-30 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{9/2} f}-\frac{5 b (11 a-21 b) \sec (e+f x)}{24 a^4 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{b (23 a-35 b) \sec (e+f x)}{24 a^3 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

-((3*a^2 - 30*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(8*a^(9/2)*f) - ((
5*a - 7*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((23*a - 35*b)*b*Sec[e + f*x])/(24*a^3*f*(a - b + b*Sec[e + f*x]
^2)^(3/2)) - (5*(11*a - 21*b)*b*Sec[e + f*x])/(24*a^4*f*Sqrt[a - b + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.324343, antiderivative size = 237, 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 = {3664, 470, 527, 12, 377, 207} \[ -\frac{\left (3 a^2-30 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{9/2} f}-\frac{5 b (11 a-21 b) \sec (e+f x)}{24 a^4 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{b (23 a-35 b) \sec (e+f x)}{24 a^3 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2 - 30*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(8*a^(9/2)*f) - ((
5*a - 7*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((23*a - 35*b)*b*Sec[e + f*x])/(24*a^3*f*(a - b + b*Sec[e + f*x]
^2)^(3/2)) - (5*(11*a - 21*b)*b*Sec[e + f*x])/(24*a^4*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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3 \left (a-b+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 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-a+b-2 (2 a-3 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-(3 a-7 b) (a-b)+4 (5 a-7 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-(9 a-35 b) (a-b)^2+2 (23 a-35 b) (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 )}{24 a^3 (a-b) f}\\ &=-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int -\frac{3 (a-b)^2 \left (3 a^2-30 a b+35 b^2\right )}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{24 a^4 (a-b)^2 f}\\ &=-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\left (3 a^2-30 a b+35 b^2\right ) \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^4 f}\\ &=-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\left (3 a^2-30 a b+35 b^2\right ) \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^4 f}\\ &=-\frac{\left (3 a^2-30 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{8 a^{9/2} f}-\frac{(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 6.81356, size = 1142, normalized size = 4.82 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((4*b^2*Cos[e + f*x])/(3*a^3*(
a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])^2) - (2*(2*a*b*Cos[e + f*x] - 3*b^2*Cos[e + f*x]))/(a^4*(a +
b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) + ((-3*a*Cos[e + f*x] + 11*b*Cos[e + f*x])*Csc[e + f*x]^2)/(8*a^
4) - (Cot[e + f*x]*Csc[e + f*x]^3)/(4*a^3)))/f + ((3*a^2 - 30*a*b + 35*b^2)*(-((2*Sqrt[a]*ArcTanh[(Sqrt[b]*(1
+ Tan[(e + f*x)/2]^2))/Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] - Sqrt[b]*(ArcTanh[(a - a
*Tan[(e + f*x)/2]^2 + 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])]))*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x])^2]*Sqrt[(a + b + (a - b)*C
os[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(4*b*Tan[(e +
f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a + b + (a -
b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^
2)^2]) + ((2*Sqrt[a]*ArcTanh[(Sqrt[b]*(1 + Tan[(e + f*x)/2]^2))/Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e +
f*x)/2]^2)^2]] + Sqrt[b]*(ArcTanh[(a - a*Tan[(e + f*x)/2]^2 + 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])]))*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 +
Cos[e + f*x])^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(-1 + Tan[(e + f*x)/2]^2)*(1
+ Tan[(e + f*x)/2]^2)*Sqrt[(4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^
2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(
e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])))/(8*a^4*f)

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Maple [B]  time = 4.138, size = 49917, normalized size = 210.6 \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)^(5/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 4.1483, size = 2449, normalized size = 10.33 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*((3*a^4 - 36*a^3*b + 98*a^2*b^2 - 100*a*b^3 + 35*b^4)*cos(f*x + e)^8 - 2*(3*a^4 - 39*a^3*b + 131*a^2*
b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e)^6 + (3*a^4 - 48*a^3*b + 233*a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)
^4 + 3*a^2*b^2 - 30*a*b^3 + 35*b^4 + 2*(3*a^3*b - 36*a^2*b^2 + 95*a*b^3 - 70*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*(3*a^4 - 33*a^3*b + 65*a^2*b^2 - 35*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 177*a^3*b
+ 445*a^2*b^2 - 315*a*b^3)*cos(f*x + e)^5 - (78*a^3*b - 305*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^3 - 5*(11*a^2*b^
2 - 21*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 2*a^6*b + a^5*b^2)*f*co
s(f*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(
f*x + e)^4 + 2*(a^6*b - 2*a^5*b^2)*f*cos(f*x + e)^2), 1/24*(3*((3*a^4 - 36*a^3*b + 98*a^2*b^2 - 100*a*b^3 + 35
*b^4)*cos(f*x + e)^8 - 2*(3*a^4 - 39*a^3*b + 131*a^2*b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e)^6 + (3*a^4 - 48*a^
3*b + 233*a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 30*a*b^3 + 35*b^4 + 2*(3*a^3*b - 36*a^2*
b^2 + 95*a*b^3 - 70*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*(3*a^4 - 33*a^3*b + 65*a^2*b^2 - 35*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 177*a^3*b + 4
45*a^2*b^2 - 315*a*b^3)*cos(f*x + e)^5 - (78*a^3*b - 305*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^3 - 5*(11*a^2*b^2 -
21*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 2*a^6*b + a^5*b^2)*f*cos(f
*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(f*x
+ e)^4 + 2*(a^6*b - 2*a^5*b^2)*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)**(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 \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)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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