3.61 \(\int \csc (a+b x) \sqrt {d \tan (a+b x)} \, dx\)

Optimal. Leaf size=47 \[ \frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{b} \]

[Out]

-csc(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^
(1/2)*(d*tan(b*x+a))^(1/2)/b

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Rubi [A]  time = 0.07, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2601, 2573, 2641} \[ \frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Sqrt[d*Tan[a + b*x]],x]

[Out]

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

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]

Rubi steps

\begin {align*} \int \csc (a+b x) \sqrt {d \tan (a+b x)} \, dx &=\frac {\left (\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{\sqrt {\sin (a+b x)}}\\ &=\left (\csc (a+b x) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {\csc (a+b x) F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{b}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 73, normalized size = 1.55 \[ -\frac {2 \sqrt [4]{-1} \cos (a+b x) \sqrt {\sec ^2(a+b x)} \sqrt {d \tan (a+b x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\right |-1\right )}{b \sqrt {\tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*(-1)^(1/4)*Cos[a + b*x]*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Sec[a + b*x]^2]*Sqrt[
d*Tan[a + b*x]])/(b*Sqrt[Tan[a + b*x]])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} \csc \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(b*x + a))*csc(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a), x)

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maple [B]  time = 0.50, size = 157, normalized size = 3.34 \[ -\frac {\sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{b \sin \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)*(d*tan(b*x+a))^(1/2),x)

[Out]

-1/b*(d*sin(b*x+a)/cos(b*x+a))^(1/2)*(-1+cos(b*x+a))*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x
+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(
b*x+a))^(1/2),1/2*2^(1/2))*(cos(b*x+a)+1)^2/sin(b*x+a)^3*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(1/2)/sin(a + b*x),x)

[Out]

int((d*tan(a + b*x))^(1/2)/sin(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)*(d*tan(b*x+a))**(1/2),x)

[Out]

Integral(sqrt(d*tan(a + b*x))*csc(a + b*x), x)

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