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

Optimal. Leaf size=143 \[ -\frac{3 (a-b)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{5/2} f}-\frac{(5 a-3 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{8 a^2 f}-\frac{\cot ^3(e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{4 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.160042, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 (a-b)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{8 a^{5/2} f}-\frac{(5 a-3 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{8 a^2 f}-\frac{\cot ^3(e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 4.33127, size = 278, normalized size = 1.94 \[ \frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-\sqrt{2} \sqrt{a} \cot (e+f x) \csc (e+f x) \left (2 a \csc ^2(e+f x)+3 a-3 b\right )-\frac{3 (a-b)^2 \cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{\sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}\right )}{16 a^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-(Sqrt[2]*Sqrt[a]*Cot[e + f*x]*Csc[e + f*x]*(3*a - 3*b + 2*a*Csc[e + f*x]^2)) - (3*(a - b)^2*(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]^4])*Sqrt[(a + b + (a
- b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(16*a^(5/2)*f)

________________________________________________________________________________________

Maple [B]  time = 0.238, size = 6334, normalized size = 44.3 \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)^(1/2),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)^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 2.69862, size = 1071, normalized size = 7.49 \begin{align*} \left [\frac{3 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt{a} \log \left (-\frac{2 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \,{\left (3 \,{\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{16 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac{3 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) +{\left (3 \,{\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \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)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + b^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^2 - a*b)*cos(f*x + e)^3 - (5*a^2 - 3*a*b)*cos(f*x + e))*sqrt(((a - b)*
cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/8*(3*((a^2 - 2
*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + b^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^2 - a*b)*cos(f*x + e)^3 - (5*a^2 -
3*a*b)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f
*x + e)^2 + a^3*f)]

________________________________________________________________________________________

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)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{5}}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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