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

Optimal. Leaf size=72 \[ -\frac{2 \cos (a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0971656, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2601, 2570, 2572, 2639} \[ -\frac{2 \cos (a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x])/(b*Sqrt[d*Tan[a + b*x]]) - (2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*Sqrt[Sin[2*a + 2
*b*x]]*Sqrt[d*Tan[a + b*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 2570

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((b*Cos[e + f
*x])^(n + 1)*(a*Sin[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2572

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

Rule 2639

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

Rubi steps

\begin{align*} \int \frac{\csc (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx &=\frac{\sqrt{\sin (a+b x)} \int \frac{\sqrt{\cos (a+b x)}}{\sin ^{\frac{3}{2}}(a+b x)} \, dx}{\sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \cos (a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{\left (2 \sqrt{\sin (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{\sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \cos (a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{(2 \sin (a+b x)) \int \sqrt{\sin (2 a+2 b x)} \, dx}{\sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \cos (a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{2 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.306319, size = 69, normalized size = 0.96 \[ -\frac{2 \cos (a+b x) \left (2 \tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+3\right )}{3 b \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]*(3 + 2*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]^2)
)/(3*b*Sqrt[d*Tan[a + b*x]])

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Maple [B]  time = 0.164, size = 490, normalized size = 6.8 \begin{align*}{\frac{\sqrt{2}}{b\cos \left ( bx+a \right ) } \left ( 2\,\cos \left ( bx+a \right ){\it EllipticE} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-\cos \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}+2\,{\it EllipticE} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-{\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-\cos \left ( bx+a \right ) \sqrt{2} \right ){\frac{1}{\sqrt{{\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }}}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\sqrt{d \tan \left (b x + a\right )}}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \csc \left (b x + a\right )}{d \tan \left (b x + a\right )}, x\right ) \end{align*}

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)/(d*tan(b*x + a)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (a + b x \right )}}{\sqrt{d \tan{\left (a + b x \right )}}}\, dx \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\sqrt{d \tan \left (b x + a\right )}}\,{d x} \end{align*}

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