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]

-((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)

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Rubi [A]  time = 0.100486, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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 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 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 \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*}

Mathematica [A]  time = 0.277023, 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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Maple [B]  time = 1.444, size = 227, normalized size = 2.7 \begin{align*} -{\frac{1}{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( a\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( a-b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+a+b \right ) } \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}+b\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( a-b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+a+b \right ) } \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \right ){\frac{1}{\sqrt{a}}}{\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)^(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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sin \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{3}\,{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)^(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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Fricas [A]  time = 2.25762, size = 849, normalized size = 10.11 \begin{align*} \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 ] \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)^(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{3}{\left (e + f x \right )}\, 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)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sin \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{3}\,{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)^(1/2),x, algorithm="giac")

[Out]

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