3.277 \(\int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx\)

Optimal. Leaf size=92 \[ \frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {5 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{6 b} \]

[Out]

5/6*sec(b*x+a)/b/csc(b*x+a)^(1/2)+1/3*sec(b*x+a)^3/b/csc(b*x+a)^(1/2)-5/6*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/
sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticF(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b

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Rubi [A]  time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2626, 3771, 2641} \[ \frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {5 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^4,x]

[Out]

(5*Sec[a + b*x])/(6*b*Sqrt[Csc[a + b*x]]) + Sec[a + b*x]^3/(3*b*Sqrt[Csc[a + b*x]]) + (5*Sqrt[Csc[a + b*x]]*El
lipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(6*b)

Rule 2626

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*b*(a*Csc[
e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(n - 1)), x] + Dist[(b^2*(m + n - 2))/(n - 1), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx &=\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5}{6} \int \sqrt {\csc (a+b x)} \sec ^2(a+b x) \, dx\\ &=\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5}{12} \int \sqrt {\csc (a+b x)} \, dx\\ &=\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {1}{12} \left (5 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx\\ &=\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5 \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 64, normalized size = 0.70 \[ \frac {\sqrt {\csc (a+b x)} \left (\tan (a+b x) \left (2 \sec ^2(a+b x)+5\right )-5 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^4,x]

[Out]

(Sqrt[Csc[a + b*x]]*(-5*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]] + (5 + 2*Sec[a + b*x]^2)*Tan[a
+ b*x]))/(6*b)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^(1/2)*sec(b*x+a)^4,x, algorithm="fricas")

[Out]

integral(sqrt(csc(b*x + a))*sec(b*x + a)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^(1/2)*sec(b*x+a)^4,x, algorithm="giac")

[Out]

integrate(sqrt(csc(b*x + a))*sec(b*x + a)^4, x)

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maple [A]  time = 0.20, size = 168, normalized size = 1.83 \[ -\frac {\sqrt {\left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}\, \left (5 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticF \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (b x +a \right )\right )+10 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )+4 \sin \left (b x +a \right )\right )}{12 \left (\sin \left (b x +a \right )+1\right ) \sqrt {-\sin \left (b x +a \right ) \left (\sin \left (b x +a \right )-1\right ) \left (\sin \left (b x +a \right )+1\right )}\, \left (\sin \left (b x +a \right )-1\right ) \cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^(1/2)*sec(b*x+a)^4,x)

[Out]

-1/12*(cos(b*x+a)^2*sin(b*x+a))^(1/2)*(5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Elli
pticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))*cos(b*x+a)^2+10*cos(b*x+a)^2*sin(b*x+a)+4*sin(b*x+a))/(sin(b*x+a)+1)/(
-sin(b*x+a)*(sin(b*x+a)-1)*(sin(b*x+a)+1))^(1/2)/(sin(b*x+a)-1)/cos(b*x+a)/sin(b*x+a)^(1/2)/b

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^(1/2)*sec(b*x+a)^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}}{{\cos \left (a+b\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(a + b*x))^(1/2)/cos(a + b*x)^4,x)

[Out]

int((1/sin(a + b*x))^(1/2)/cos(a + b*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\csc {\left (a + b x \right )}} \sec ^{4}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**(1/2)*sec(b*x+a)**4,x)

[Out]

Integral(sqrt(csc(a + b*x))*sec(a + b*x)**4, x)

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