3.125 \(\int \csc ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {a} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f} \]

[Out]

-1/2*(a+b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f/a^(1/2)-1/2*cot(f*x+e)*csc(f*x+e)*(a+b-b*c
os(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3186, 378, 377, 206} \[ -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {a} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 378

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

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Mathematica [A]  time = 0.29, size = 100, normalized size = 1.19 \[ \frac {-2 (a+b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a-b \cos (2 (e+f x))+b}}\right )-\sqrt {2} \sqrt {a} \cot (e+f x) \csc (e+f x) \sqrt {2 a-b \cos (2 (e+f x))+b}}{4 \sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] - Sqrt[2]*Sqrt[a]*Sqrt[
2*a + b - b*Cos[2*(e + f*x)]]*Cot[e + f*x]*Csc[e + f*x])/(4*Sqrt[a]*f)

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fricas [A]  time = 0.54, size = 338, normalized size = 4.02 \[ \left [\frac {4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a \cos \left (f x + e\right ) + {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{8 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}, \frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a \cos \left (f x + e\right )}{4 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) + ((a + b)*cos(f*x + e)^2 - a - b)*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)*c
os(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)))/(a*f*cos(f*x + e)^2 - a*f), 1/4*(((a + b)*cos(f*x + e)^2 - a - b)*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*
sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e))/(a*f*cos(f*x + e)^2 - a*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)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n
ostep/2)Unable to divide, perhaps due to rounding error%%%{16,[4,4]%%%}+%%%{%%%{32,[1]%%%},[4,3]%%%}+%%%{%%%{1
6,[2]%%%},[4,2]%%%}+%%%{%%%{-32,[1]%%%},[2,4]%%%}+%%%{%%%{-64,[2]%%%},[2,3]%%%}+%%%{%%%{-32,[3]%%%},[2,2]%%%}+
%%%{%%%{16,[2]%%%},[0,4]%%%}+%%%{%%%{32,[3]%%%},[0,3]%%%}+%%%{%%%{16,[4]%%%},[0,2]%%%} / %%%{%%%{1,[1]%%%},[4,
0]%%%}+%%%{%%%{-2,[2]%%%},[2,0]%%%}+%%%{%%%{1,[3]%%%},[0,0]%%%} Error: Bad Argument Value

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maple [B]  time = 1.79, size = 227, normalized size = 2.70 \[ -\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (a \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )+b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\right )}{4 \sqrt {a}\, \sin \left (f x +e \right )^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(a*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*
x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^2+b*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*
x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^2+2*a^(1/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2))/a^(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)^(1/2)/sin(e + f*x)^3,x)

[Out]

int((a + b*sin(e + f*x)^2)^(1/2)/sin(e + f*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*csc(e + f*x)**3, x)

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