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

Optimal. Leaf size=129 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{5/2} f}+\frac {b (5 a+3 b) \cos (e+f x)}{3 a^2 f (a+b)^2 \sqrt {a-b \cos ^2(e+f x)+b}}+\frac {b \cos (e+f x)}{3 a f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

[Out]

-arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(5/2)/f+1/3*b*cos(f*x+e)/a/(a+b)/f/(a+b-b*cos(f*x+e)
^2)^(3/2)+1/3*b*(5*a+3*b)*cos(f*x+e)/a^2/(a+b)^2/f/(a+b-b*cos(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, 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 = {3186, 414, 527, 12, 377, 206} \[ \frac {b (5 a+3 b) \cos (e+f x)}{3 a^2 f (a+b)^2 \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{5/2} f}+\frac {b \cos (e+f x)}{3 a f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 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 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 127, normalized size = 0.98 \[ \frac {\frac {\sqrt {2} b \cos (e+f x) \left (12 a^2-b (5 a+3 b) \cos (2 (e+f x))+13 a b+3 b^2\right )}{3 a^2 (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a-b \cos (2 (e+f x))+b}}\right )}{a^{5/2}}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]]/a^(5/2)) + (Sqrt[2]*b*Cos[e + f*
x]*(12*a^2 + 13*a*b + 3*b^2 - b*(5*a + 3*b)*Cos[2*(e + f*x)]))/(3*a^2*(a + b)^2*(2*a + b - b*Cos[2*(e + f*x)])
^(3/2)))/f

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fricas [B]  time = 0.90, size = 752, normalized size = 5.83 \[ \left [\frac {3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left ({\left (5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left ({\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) - 2 \, {\left ({\left (5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left ({\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*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^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b
- b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt
(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*((5*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 -
3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^5*b^2 + 2*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^7 + 4*a^6*b + 6*a^5*b^2
+ 4*a^4*b^3 + a^3*b^4)*f), 1/6*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*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(-1/2*((a - b)*cos(f*x + e)^
2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e))) - 2*((5*a
^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 - 3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a +
b))/((a^5*b^2 + 2*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^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[-81,22]Warning, need to choose a branch for the root of a polynomial with parameters. Thi
s might be wrong.The choice was done assuming [a,b]=[53,42]Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_no
step/2)Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by
` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable s
hould perhaps be purged.Evaluation time: 0.66Error: Bad Argument Type

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maple [B]  time = 3.71, size = 249, normalized size = 1.93 \[ \frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a^{2} \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{2 a^{\frac {5}{2}}}+\frac {b \left (2 b \left (\sin ^{2}\left (f x +e \right )\right )+3 a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 a \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (a^{2}+2 a b +b^{2}\right )}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(1/a^2*b*cos(f*x+e)^2/(a+b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2
)-1/2/a^(5/2)*ln((2*a+(-a+b)*sin(f*x+e)^2+2*a^(1/2)*(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2))/sin(f*x+e)^2)+1
/3/a*b*(2*b*sin(f*x+e)^2+3*a+b)*cos(f*x+e)^2/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)/(a+b*sin(f*x+e)^2)/(a^2
+2*a*b+b^2))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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maxima [B]  time = 0.53, size = 306, normalized size = 2.37 \[ \frac {\frac {4 \, b^{3} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{3} b^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b^{3} + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b^{4}} + \frac {2 \, b^{2} \cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} a^{2} b + {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} a b^{2}} + \frac {6 \, b^{2} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{3} b + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b^{2}} - \frac {3 \, \log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right )}{a^{\frac {5}{2}}} + \frac {3 \, \log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{a^{\frac {5}{2}}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)*(a + b*sin(e + f*x)^2)^(5/2)),x)

[Out]

int(1/(sin(e + f*x)*(a + b*sin(e + f*x)^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(e + f*x)/(a + b*sin(e + f*x)**2)**(5/2), x)

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