3.385 \(\int \frac{(b \csc (e+f x))^m}{\sqrt{d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \sqrt [4]{\cos ^2(e+f x)} \sqrt{d \tan (e+f x)} (b \csc (e+f x))^m \, _2F_1\left (\frac{1}{4},\frac{1}{4} (1-2 m);\frac{1}{4} (5-2 m);\sin ^2(e+f x)\right )}{d f (1-2 m)} \]

[Out]

(2*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[1/4, (1 - 2*m)/4, (5 - 2*m)/4, Sin[e + f*x]^2]*
Sqrt[d*Tan[e + f*x]])/(d*f*(1 - 2*m))

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Rubi [A]  time = 0.150533, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2618, 2602, 2577} \[ \frac{2 \sqrt [4]{\cos ^2(e+f x)} \sqrt{d \tan (e+f x)} (b \csc (e+f x))^m \, _2F_1\left (\frac{1}{4},\frac{1}{4} (1-2 m);\frac{1}{4} (5-2 m);\sin ^2(e+f x)\right )}{d f (1-2 m)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^m/Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[1/4, (1 - 2*m)/4, (5 - 2*m)/4, Sin[e + f*x]^2]*
Sqrt[d*Tan[e + f*x]])/(d*f*(1 - 2*m))

Rule 2618

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^FracPart[m]*(Sin[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), 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]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \frac{(b \csc (e+f x))^m}{\sqrt{d \tan (e+f x)}} \, dx &=\left ((b \csc (e+f x))^m \left (\frac{\sin (e+f x)}{b}\right )^m\right ) \int \frac{\left (\frac{\sin (e+f x)}{b}\right )^{-m}}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{\left (\sqrt{\cos (e+f x)} (b \csc (e+f x))^{1+m} \left (\frac{\sin (e+f x)}{b}\right )^{\frac{1}{2}+m} \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \left (\frac{\sin (e+f x)}{b}\right )^{-\frac{1}{2}-m} \, dx}{b d}\\ &=\frac{2 \sqrt [4]{\cos ^2(e+f x)} (b \csc (e+f x))^{1+m} \, _2F_1\left (\frac{1}{4},\frac{1}{4} (1-2 m);\frac{1}{4} (5-2 m);\sin ^2(e+f x)\right ) \sin (e+f x) \sqrt{d \tan (e+f x)}}{b d f (1-2 m)}\\ \end{align*}

Mathematica [A]  time = 1.22483, size = 87, normalized size = 1.1 \[ -\frac{2 \sqrt{d \tan (e+f x)} \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \, _2F_1\left (\frac{1}{4} (1-2 m),1-\frac{m}{2};\frac{1}{4} (5-2 m);-\tan ^2(e+f x)\right )}{d f (2 m-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csc[e + f*x])^m/Sqrt[d*Tan[e + f*x]],x]

[Out]

(-2*(b*Csc[e + f*x])^m*Hypergeometric2F1[(1 - 2*m)/4, 1 - m/2, (5 - 2*m)/4, -Tan[e + f*x]^2]*Sqrt[d*Tan[e + f*
x]])/(d*f*(-1 + 2*m)*(Sec[e + f*x]^2)^(m/2))

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Maple [F]  time = 0.168, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\csc \left ( fx+e \right ) \right ) ^{m}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*csc(f*x+e))^m/(d*tan(f*x+e))^(1/2),x)

[Out]

int((b*csc(f*x+e))^m/(d*tan(f*x+e))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csc(f*x + e))^m/sqrt(d*tan(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m/(d*tan(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))**m/(d*tan(f*x+e))**(1/2),x)

[Out]

Integral((b*csc(e + f*x))**m/sqrt(d*tan(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csc(f*x + e))^m/sqrt(d*tan(f*x + e)), x)