3.126 \(\int \csc ^5(e+f x) \sqrt{a+b \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=143 \[ -\frac{(3 a-b) (a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{8 a^{3/2} f}-\frac{\cot (e+f x) \csc ^3(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a f}-\frac{(3 a-b) \cot (e+f x) \csc (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{8 a f} \]

[Out]

-((3*a - b)*(a + b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(8*a^(3/2)*f) - ((3*a - b)
*Sqrt[a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc[e + f*x])/(8*a*f) - ((a + b - b*Cos[e + f*x]^2)^(3/2)*Cot[e +
f*x]*Csc[e + f*x]^3)/(4*a*f)

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Rubi [A]  time = 0.132382, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3186, 382, 378, 377, 206} \[ -\frac{(3 a-b) (a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{8 a^{3/2} f}-\frac{\cot (e+f x) \csc ^3(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a f}-\frac{(3 a-b) \cot (e+f x) \csc (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{8 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a - b)*(a + b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(8*a^(3/2)*f) - ((3*a - b)
*Sqrt[a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc[e + f*x])/(8*a*f) - ((a + b - b*Cos[e + f*x]^2)^(3/2)*Cot[e +
f*x]*Csc[e + f*x]^3)/(4*a*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 382

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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 ^5(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 )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 a f}-\frac{(3 a-b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b-b x^2}}{\left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 a f}\\ &=-\frac{(3 a-b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 a f}-\frac{((3 a-b) (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 )}{8 a f}\\ &=-\frac{(3 a-b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 a f}-\frac{((3 a-b) (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 )}{8 a f}\\ &=-\frac{(3 a-b) (a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{8 a^{3/2} f}-\frac{(3 a-b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 a f}\\ \end{align*}

Mathematica [A]  time = 0.517932, size = 127, normalized size = 0.89 \[ \frac{\left (-6 a^2-4 a b+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )-\sqrt{2} \sqrt{a} \cot (e+f x) \csc (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b} \left (2 a \csc ^2(e+f x)+3 a+b\right )}{16 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a^2 - 4*a*b + 2*b^2)*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]*(3*a + b + 2*a*Csc[e + f*x]^2))/(16*a^(3
/2)*f)

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Maple [B]  time = 1.583, size = 379, normalized size = 2.7 \begin{align*} -{\frac{1}{16\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}\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 ( 3\,{a}^{3}\ln \left ({\frac{ \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}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{4}+2\,b\ln \left ({\frac{ \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}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{4}{a}^{2}-\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 ){b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}a+6\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \sin \left ( fx+e \right ) \right ) ^{2}{a}^{5/2}+2\,b\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \sin \left ( fx+e \right ) \right ) ^{2}{a}^{3/2}+4\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }{a}^{5/2} \right ){a}^{-{\frac{5}{2}}}{\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)^5*(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(3*a^3*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*c
os(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4+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)^4*a^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)/sin(f*x+e)^2)*b^2*sin(f*x+e)^4*a+6*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x
+e)^2*a^(5/2)+2*b*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x+e)^2*a^(3/2)+4*(cos(f*x+e)^2*(a+b*sin(f*x+e)
^2))^(1/2)*a^(5/2))/sin(f*x+e)^4/a^(5/2)/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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(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)^5, x)

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Fricas [A]  time = 3.55323, size = 1239, normalized size = 8.66 \begin{align*} \left [-\frac{{\left ({\left (3 \, a^{2} + 2 \, 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} + 3 \, a^{2} + 2 \, a b - b^{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 (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{32 \,{\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac{{\left ({\left (3 \, a^{2} + 2 \, 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} + 3 \, a^{2} + 2 \, a b - b^{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 (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{16 \,{\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*(((3*a^2 + 2*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 + 2*a*b - 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)) - 4*((3*a^2 + a*b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^
2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f), 1/16*(((3*a^2 + 2*a*b - b^2)*cos(f*x + e)
^4 - 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 3*a^2 + 2*a*b - b^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*((3*a
^2 + a*b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^4
- 2*a^2*f*cos(f*x + e)^2 + a^2*f)]

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Sympy [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)**5*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(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)^5, x)