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

Optimal. Leaf size=83 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0965129, antiderivative size = 83, 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 = {3186, 402, 217, 203, 377, 206} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/f) - (Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[e
+ f*x])/Sqrt[a + b - b*Cos[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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 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 (e+f x) \sqrt{a+b \sin ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b-b x^2}}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{f}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{a \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 )}{f}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{\cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{f}\\ \end{align*}

Mathematica [A]  time = 0.116469, size = 99, normalized size = 1.19 \[ \frac{\sqrt{-b} \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.883, size = 174, normalized size = 2.1 \begin{align*} -{\frac{1}{2\,f\cos \left ( fx+e \right ) }\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \sqrt{a}\ln \left ({\frac{1}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}-1} \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 ) -\sqrt{b}\arctan \left ({\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ) \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)*(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.5495, size = 2862, normalized size = 34.48 \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*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*c
os(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e
)^2 - 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 -
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)) + 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)))/f, 1/8*(4*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))) + sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)
*cos(f*x + e)^6 + 160*(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 - 3
2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e
)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*
x + e)^2 + a + b)*sqrt(-b)))/f, 1/4*(sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 +
a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x +
e)^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e))) + 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)))/f, 1/4*(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))) + sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b
*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b
^3)*cos(f*x + e))))/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{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*csc(e + f*x), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(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), x)