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

Optimal. Leaf size=127 \[ -\frac{3 b \sec (e+f x)}{2 a^2 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{(a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 a^{5/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

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

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Rubi [A]  time = 0.157731, antiderivative size = 127, 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, 471, 527, 12, 377, 207} \[ -\frac{3 b \sec (e+f x)}{2 a^2 f \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{(a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 a^{5/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 4.09678, size = 308, normalized size = 2.43 \[ -\frac{\frac{\csc ^2(e+f x) \sec (e+f x) ((a-3 b) \cos (2 (e+f x))+a+3 b)}{\sqrt{2} a^2 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}+\frac{(a-3 b) \cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{2 a^{5/2} \sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + 3*b + (a - 3*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2*Sec[e + f*x])/(Sqrt[2]*a^2*Sqrt[(a + b + (a - b)*Cos[
2*(e + f*x)])*Sec[e + f*x]^2]) + ((a - 3*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^(5/2)*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 = 0.261, size = 5633, normalized size = 44.4 \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)^(3/2),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)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 3.54322, size = 1087, normalized size = 8.56 \begin{align*} \left [-\frac{{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, 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 ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \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}}}}{4 \,{\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f -{\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac{{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, 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 ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \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}}}}{2 \,{\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f -{\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(((a^2 - 4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 - a*b + 3*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*((a^2 - 3*a*b)*cos(f*x + e)^3 + 3*a*b*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 +
 b)/cos(f*x + e)^2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*x + e)^2), 1/2*(((a^2
 - 4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 - a*b + 3*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) + ((a^2 - 3*a*b)*cos(f*x + e)^3 + 3*a*b*co
s(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^
4 - 2*a^3*b)*f*cos(f*x + e)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral(csc(e + f*x)**3/(a + b*tan(e + f*x)**2)**(3/2), x)

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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