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

Optimal. Leaf size=136 \[ -\frac{b (5 a-3 b) \sec (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{a^{5/2} f}-\frac{b \sec (e+f x)}{3 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.145504, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3664, 414, 527, 12, 377, 207} \[ -\frac{b (5 a-3 b) \sec (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{a^{5/2} f}-\frac{b \sec (e+f x)}{3 a f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]/(a^(5/2)*f)) - (b*Sec[e + f*x])/(3*a*(a - b)*
f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((5*a - 3*b)*b*Sec[e + f*x])/(3*a^2*(a - b)^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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, 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 (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 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 )}{3 a (a-b) f}\\ &=-\frac{b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}+\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 )}{3 a^2 (a-b)^2 f}\\ &=-\frac{b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{a^2 f}\\ &=-\frac{b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\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 )}{a^2 f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{a^{5/2} f}-\frac{b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 4.70739, size = 305, normalized size = 2.24 \[ \frac{\cos (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-\frac{2 \sqrt{2} \sqrt{a} b \left (3 \left (2 a^2-3 a b+b^2\right ) \cos (2 (e+f x))+6 a^2+a b-3 b^2\right )}{(a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}-\frac{3 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{\sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}\right )}{6 a^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*((-2*Sqrt[2]*Sqrt[a]*b*(6*a^2 + a*b - 3*b^2 + 3*(2*a^2 - 3*a*b + b^2)*Cos[2*(e + f*x)]))/((a - b
)^2*(a + b + (a - b)*Cos[2*(e + f*x)])^2) - (3*(ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*T
an[(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]*Sq
rt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])])*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])/(6*a^(5/2)*f)

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Maple [B]  time = 1.415, size = 27448, normalized size = 201.8 \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)/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

result too large to display

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

[Out]

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

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Fricas [B]  time = 3.21182, size = 1561, normalized size = 11.48 \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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a
^2*b^2 + 3*a*b^3 - 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*(2*a^3*b - 3*a^2*b^2 + a*b^3
)*cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3)*cos(f*x + e))*sqrt(((a - b)*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)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*
f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*f), 1/3*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos
(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - 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*(2*a^3*b - 3*a^2*b^2 + a*b^3)*
cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3)*cos(f*x + e))*sqrt(((a - b)*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)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*
cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*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)/(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 )}{{\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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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