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]

-1/8*(3*a-b)*(a+b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(3/2)/f-1/4*(a+b-b*cos(f*x+e)^2)^(
3/2)*cot(f*x+e)*csc(f*x+e)^3/a/f-1/8*(3*a-b)*cot(f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/a/f

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Rubi [A]  time = 0.13, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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 ^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*}

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Mathematica [A]  time = 0.51, 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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fricas [A]  time = 0.73, size = 520, normalized size = 3.64 \[ \left [-\frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \]

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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giac [B]  time = 0.55, size = 962, normalized size = 6.73 \[ \text {result too large to display} \]

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]

1/64*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)*(tan(1/2*f*
x + 1/2*e)^2 + (7*a + 2*b)/a) + 8*(3*a^2 + 2*a*b - b^2)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1
/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*a) - 4*(
3*a^(5/2) + 2*a^(3/2)*b - sqrt(a)*b^2)*log(abs(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^
4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a - a^(3/2) - 2*sqrt(a)*b))/a^2 + 4*(4*(sqrt
(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/
2*e)^2 + a))^3*a^2 + 8*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2
*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a*b + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*
e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^2 + 5*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2
- sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2) + 4
*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*
x + 1/2*e)^2 + a))^2*a^(3/2)*b - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1
/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^3 - 4*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*
f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*b + 2*(sqrt(a)*tan(1/2*f*x
+ 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b
^2 - 3*a^(7/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2
+ 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)^2*a))/f

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maple [B]  time = 2.20, size = 379, normalized size = 2.65 \[ -\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (3 a^{3} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+2 b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right ) a^{2}-\ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) b^{2} \left (\sin ^{4}\left (f x +e \right )\right ) a +6 \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (\sin ^{2}\left (f x +e \right )\right ) a^{\frac {5}{2}}+2 b \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (\sin ^{2}\left (f x +e \right )\right ) a^{\frac {3}{2}}+4 \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, a^{\frac {5}{2}}\right )}{16 \sin \left (f x +e \right )^{4} a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{5}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^5} \,d x \]

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)^5,x)

[Out]

int((a + b*sin(e + f*x)^2)^(1/2)/sin(e + f*x)^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{5}{\left (e + f x \right )}\, dx \]

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]

Integral(sqrt(a + b*sin(e + f*x)**2)*csc(e + f*x)**5, x)

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