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

Optimal. Leaf size=134 \[ -\frac{b (a+3 b) \cos (e+f x)}{2 a^2 f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{(a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 a^{5/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \sqrt{a-b \cos ^2(e+f x)+b}} \]

[Out]

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

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Rubi [A]  time = 0.164457, antiderivative size = 134, 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 = {3186, 414, 527, 12, 377, 206} \[ -\frac{b (a+3 b) \cos (e+f x)}{2 a^2 f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{(a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 a^{5/2} f}-\frac{\cot (e+f x) \csc (e+f x)}{2 a f \sqrt{a-b \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.659131, size = 134, normalized size = 1. \[ \frac{\frac{\cot (e+f x) \csc (e+f x) \left (-2 a^2+b (a+3 b) \cos (2 (e+f x))-3 a b-3 b^2\right )}{\sqrt{2} a^2 (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}}-\frac{(a-3 b) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )}{a^{5/2}}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 2.581, size = 274, normalized size = 2. \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) }\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{3\,b}{4}\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,a+ \left ( -a+b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{{a}^{2} \left ( a+b \right ) }{\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{2\,{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{1}{4}\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,a+ \left ( -a+b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{3}{2}}}} \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(3/4/a^(5/2)*b*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/a^2*b^2*cos(f*x+e)^2/(a+b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/
2)-1/2/a^2/sin(f*x+e)^2*(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)-1/4/a^(3/2)*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))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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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*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.83483, size = 1474, normalized size = 11. \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)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5*a*b^2 - 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*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)/(co
s(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*((a^2*b + 3*a*b^2)*cos(f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*
x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*f*c
os(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)*f), 1/4*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5
*a*b^2 - 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*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*((a^2*b + 3*a*b
^2)*cos(f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + a^3*b^
2)*f*cos(f*x + e)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)*f)]

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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 \sin ^{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*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(csc(e + f*x)**3/(a + b*sin(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 \sin \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*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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